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I have an assignment to solve using dynamic programming the following problem:
There is a rectangular sheet and a set of rectangular elements of given dimensions and value. The task is to divide the sheet into elements of given dimensions, so that the sum of values of the elements is maximum. Find this sum and a tree of consequent cuts.
There are following conditions:
It is NOT possible to rotate the given elements.
It is possible to cut out unlimited number of certain types of
elements.
It is possible that some parts of the sheet will remain unused.
The only possible way to cut the sheet is by a straight
line, so that you again obtain two smaller rectangles.
The problem is solved. Solution can be found below.
==========================================================================
I understand the problem for one dimension, which comes to the rod cutting problem. You divide the rod into the smallest possible pieces, take the first one and check if you can build it with the given segments. Remember the weight you'll get with building the part this way and move on to a bigger part containing the previous one. You go back by the length of the segment you're trying at the moment and check if using this segment plus the weight of the previously build part will make up to better sum of the weight for the current part.
Supposedly, the cutting wood problem is no different, but you add the 2-dimension, additional loop somewhere in the middle. Unfortunately, I can't imagine how to store the values and how to go back for the 2-dimensions.
I've tried doing like:
1. Loop on one dimension
2. Loop on second dimension
3. Loop on all the segments you can use
4. Check if you can fit the current segment depending on 1. and 2.
5. If yes, go back the length of the segment to see if weight of the segment + what's stored there gives you a greater result; do the same for the width
6. Store the result in the cell you're currently on
7. Go through the array and find the greatest result
Here is the code I produced after many debugging tries:
public int Cut((int length, int width) sheet, (int length, int width, int price)[] elements, out Cut cuts)
{
int[,] tmpSheetArr = new int[sheet.length + 1, sheet.width + 1];
for (int i = 1; i < tmpSheetArr.GetLength(0); i++)
{
for (int j = 1; j < tmpSheetArr.GetLength(1); j++)
{
tmpSheetArr[i, j] = Int32.MinValue;
}
}
for (int i = 1; i < tmpSheetArr.GetLength(0); i++) //columns
{
for (int j = 1; j < tmpSheetArr.GetLength(1); j++) //rows
{
for (int e = 0; e < elements.Length; e++)
{
(int length, int width, int price) elem = elements[e];
if (i >= elem.length && j >= elem.width)
{
int tmpJ, tmpI, tmpVal;
tmpJ = j - elem.width;
tmpI = i;
while (0 < tmpI)
{
if(tmpI > i - elem.length && tmpI <= i && tmpJ > j - elem.width && tmpJ <= j)
{
tmpJ -= 1;
if (-1 == tmpJ)
{
tmpJ = tmpSheetArr.GetLength(1) - 1;
tmpI -= 1;
}
continue;
}
tmpVal = tmpSheetArr[tmpI, tmpJ] == Int32.MinValue ? 0 : tmpSheetArr[tmpI, tmpJ];
if (tmpSheetArr[i, j] < elem.price + tmpVal)
{
tmpSheetArr[i, j] = elem.price + tmpVal;
}
tmpJ -= 1;
if(-1 == tmpJ)
{
tmpJ = tmpSheetArr.GetLength(1) - 1;
tmpI -= 1;
}
}
}
}
}
}
int tmpMax = 0;
for (int i = 1; i < tmpSheetArr.GetLength(0); i++)
{
for (int j = 1; j < tmpSheetArr.GetLength(1); j++)
{
if (tmpMax < tmpSheetArr[i, j])
tmpMax = tmpSheetArr[i, j];
}
}
cuts = null;
return tmpMax;
}
It doesn't work, gives too big results in some cases and gets stuck on bigger problems. I think the main problem is about going back - with only the weight stored I don't know what size of the block was used and if it will overlap with the current one.
I decided to write it from the beginning, but really can't find another approach. I have a code for the 1D problem:
int cutRod(int[] price, int n)
{
int[] val = new int[n + 1];
val[0] = 0;
int i, j;
// Build the table val[] and return the last entry
// from the table
for (i = 1; i <= n; i++)
{
int max_val = Int32.MinValue;
for (j = 0; j < i; j++)
max_val = Math.Max(max_val, price[j] + val[i - j - 1]);
val[i] = max_val;
}
return val[n];
}
How do I change it so it works for 2D problem?
I tried to explain my limited understanding and way of thinking the best I could. I would appreciate any help on this matter.
Make your dynamic state at x be a dictionary mapping a particular "skyline" of what blocks placed before x look like after x. You start with a flat skyline (no blocks so far, clean edge), and you're looking for a flat skyline at the other end (didn't go off the edge of the sheet).
As you advance you "lower" your skyline by 1, start looking at ways to cut out new blocks, and get new possible skylines.
The number of possible skylines will grow exponentially with the width of the rectangle.
The solution:
Build an array of maximum values that can be obtained from given piece of dimensions 1x1 up to the size of the board. Maximum value for given piece is stored under index of [(length of the piece) - 1, (width of the piece) - 1]. To find the maximum value, check how the current piece can be formed with previous pieces and cuts.
To construct the tree of cuts, build a second array of the best cuts for the current piece. Root of the cuts tree for the current piece is stored under index of [(length of the piece) - 1, (width of the piece) - 1].
Cuts class:
public class Cut
{
public int length; // vertical dimension (before cut)
public int width; // horizontal dimension (before cut)
public int price; // sum of the values of the two elements resulting from the cut
public bool vertical; // true for vertical cut, false otherwise
public int n; // distance from left side (for vertical cut) or top (for horizontal cut) of the current piece
// price 0 means there was no cut, topleft and bottomright are null,
public Cut topleft; // top/left resulting piece after cut
public Cut bottomright; // bottom/right resulting piece after cut
public Cut(int length, int width, int price, bool vertical=true, int n=0, Cut topleft=null, Cut bottomright=null)
{
this.length = length;
this.width = width;
this.price = price;
this.vertical = vertical;
this.n = n;
this.topleft = topleft;
this.bottomright = bottomright;
}
}
Function finding the maximum value and a tree of cuts:
public int Cut((int length, int width) sheet, (int length, int width, int price)[] elements, out Cut cuts)
{
int[,] sheetArr = new int[sheet.length, sheet.width]; //contains best values of current pieces that can be formed
Cut[,] cutsArr = new Cut[sheet.length, sheet.width]; //contains references for cuts used to form pieces of the best value,
for (int l = 0; l < sheet.length; l++) //loop on length
{
for (int w = 0; w < sheet.width; w++) //loop on width
{
foreach ((int length, int width, int price) elem in elements) //loop on elements
{
if (elem.length == l + 1 && elem.width == w + 1) //check if current piece can be build with one of the given elements
{
sheetArr[l, w] = elem.price;
cutsArr[l, w] = new Cut(elem.length, elem.width, elem.price); //piece is exactly one of the elements (no cut)
break; //no 2 elements of the same size in the given elements
}
cutsArr[l, w] = new Cut(l + 1, w + 1, 0); //piece can not be formed from given elements, price = 0 (no cut)
}
for (int i = 1; i < Math.Floor((decimal)(l + 1) / 2) + 1; i++) //go back on length
{
if (sheetArr[i - 1, w] + sheetArr[l - i, w] > sheetArr[l, w])
{
sheetArr[l, w] = sheetArr[i - 1, w] + sheetArr[l - i, w];
cutsArr[l, w] = new Cut(l + 1, w + 1, sheetArr[l, w], false, i, cutsArr[i - 1, w], cutsArr[l - i, w]);
}
}
for (int i = 1; i < Math.Floor((decimal)(w + 1) / 2) + 1; i++) //go back on width
{
if (sheetArr[l, i - 1] + sheetArr[l, w - i] > sheetArr[l, w])
{
sheetArr[l, w] = sheetArr[l, i - 1] + sheetArr[l, w - i];
cutsArr[l, w] = new Cut(l + 1, w + 1, sheetArr[l, w], true, i, cutsArr[l, i - 1], cutsArr[l, w - i]);
}
}
}
}
cuts = cutsArr[sheet.length - 1, sheet.width - 1];
return sheetArr[sheet.length - 1, sheet.width - 1];
}
Given an unsorted array, find the max j - i difference between indices such that j > i and a[j] > a[i] in O(n). I am able to find j and i using trivial methods in O(n^2) complexity but would like to know how to do this in O(n)?
Input: {9, 2, 3, 4, 5, 6, 7, 8, 18, 0}
Output: 8 ( j = 8, i = 0)
Input: {1, 2, 3, 4, 5, 6}
Output: 5 (j = 5, i = 0)
For brevity's sake I am going to assume all the elements are unique. The algorithm can be extended to handle non-unique element case.
First, observe that if x and y are your desired max and min locations respectively, then there can not be any a[i] > a[x] and i > x, and similarly, no a[j] < a[y] and j < y.
So we scan along the array a and build an array S such that S[i] holds the index of the minimum element in a[0:i]. Similarly an array T which holds the index of the maximum element in a[n-1:i] (i.e., backwards).
Now we can see that a[S[i]] and a[T[i]] are necessarily decreasing sequences, since they were the minimum till i and maximum from n till i respectively.
So now we try to do a merge-sort like procedure. At each step, if a[S[head]] < a[T[head]], we pop off an element from T, otherwise we pop off an element from S. At each such step, we record the difference in the head of S and T if a[S[head]] < a[T[head]]. The maximum such difference gives you your answer.
EDIT: Here is a simple code in Python implementing the algorithm.
def getMaxDist(arr):
# get minima going forward
minimum = float("inf")
minima = collections.deque()
for i in range(len(arr)):
if arr[i] < minimum:
minimum = arr[i]
minima.append((arr[i], i))
# get maxima going back
maximum = float("-inf")
maxima = collections.deque()
for i in range(len(arr)-1,0,-1):
if arr[i] > maximum:
maximum = arr[i]
maxima.appendleft((arr[i], i))
# do merge between maxima and minima
maxdist = 0
while len(maxima) and len(minima):
if maxima[0][0] > minima[0][0]:
if maxima[0][1] - minima[0][1] > maxdist:
maxdist = maxima[0][1] - minima[0][1]
maxima.popleft()
else:
minima.popleft()
return maxdist
Let's make this simple observation: If we have 2 elements a[i], a[j] with i < j and a[i] < a[j] then we can be sure that j won't be part of the solution as the first element (he can be the second but that's a second story) because i would be a better alternative.
What this tells us is that if we build greedily a decreasing sequence from the elements of a the left part of the answer will surely come from there.
For example for : 12 3 61 23 51 2 the greedily decreasing sequence is built like this:
12 -> 12 3 -> we ignore 61 because it's worse than 3 -> we ignore 23 because it's worse than 3 -> we ignore 51 because it's worse than 3 -> 12 3 2.
So the answer would contain on the left side 12 3 or 2.
Now on a random case this has O(log N) length so you can binary search on it for each element as the right part of the answer and you would get O(N log log N) which is good, and if you apply the same logic on the right part of the string on a random case you could get O(log^2 N + N(from the reading)) which is O(N). But we can do O(N) on a non-random case too.
Suppose we have this decreasing sequence. We start from the right of the string and do the following while we can pair the last of the decreasing sequence with the current number
1) If we found a better solution by taking the last of the decreasing sequence and the current number than we update the answer
2) Even if we updated the answer or not we pop the last element of the decreasing sequence because we are it's perfect match (any other match would be to the left and would give an answer with smaller j - i)
3) Repeat while we can pair these 2
Example Code:
#include <iostream>
#include <vector>
using namespace std;
int main() {
int N; cin >> N;
vector<int> A(N + 1);
for (int i = 1; i <= N; ++i)
cin >> A[i];
// let's solve the problem
vector<int> decreasing;
pair<int, int> answer;
// build the decreasing sequence
decreasing.push_back(1);
for (int i = 1; i <= N; ++i)
if (A[i] < A[decreasing.back()])
decreasing.push_back(i); // we work with indexes because we might have equal values
for (int i = N; i > 0; --i) {
while (decreasing.size() and A[decreasing.back()] < A[i]) { // while we can pair these 2
pair<int, int> current_pair(decreasing.back(), i);
if (current_pair.second - current_pair.first > answer.second - answer.first)
answer = current_pair;
decreasing.pop_back();
}
}
cout << "Best pair found: (" << answer.first << ", " << answer.second << ") with values (" << A[answer.first] << ", " << A[answer.second] << ")\n";
}
Later Edit:
I see you gave an example: I indexed from 1 to make it clearer and I print (i, j) instead of (j, i). You can alter it as you see fit.
We can avoid checking the whole array by starting from the maximum difference of j-i and comparing arr[j]>arr[i] for all the possible combinations j and i for that particular maximum difference
Whenever we get a combination of (j,i) with arr[j]>arr[i] we can exit the loop
Example : In an array of {2,3,4,5,8,1}
first code will check for maximum difference 5(5-0) i.e (arr[0],arr[5]), if arr[5]>arr[0] function will exit else will take combinations of max diff 4 (5,1) and (4,0) i.e arr[5],arr[1] and arr[4],arr[0]
int maxIndexDiff(int arr[], int n)
{
int maxDiff = n-1;
int i, j;
while (maxDiff>0)
{
j=n-1;
while(j>=maxDiff)
{
i=j - maxDiff;
if(arr[j]>arr[i])
{
return maxDiff;
}
j=j-1;
}
maxDiff=maxDiff-1;
}
return -1;
}`
https://ide.geeksforgeeks.org/cjCW3wXjcj
Here is a very simple O(n) Python implementation of the merged down-sequence idea. The implementation works even in the case of duplicate values:
downs = [0]
for i in range(N):
if ar[i] < ar[downs[-1]]:
downs.append(i)
best = 0
i, j = len(downs)-1, N-1
while i >= 0:
if ar[downs[i]] <= ar[j]:
best = max(best, j-downs[i])
i -= 1
else:
j -= 1
print best
To solve this problem, we need to get two optimum indexes of arr[]: left index i and right index j. For an element arr[i], we do not need to consider arr[i] for left index if there is an element smaller than arr[i] on left side of arr[i]. Similarly, if there is a greater element on right side of arr[j] then we do not need to consider this j for right index. So we construct two auxiliary arrays LMin[] and RMax[] such that LMin[i] holds the smallest element on left side of arr[i] including arr[i], and RMax[j] holds the greatest element on right side of arr[j] including arr[j]. After constructing these two auxiliary arrays, we traverse both of these arrays from left to right. While traversing LMin[] and RMa[] if we see that LMin[i] is greater than RMax[j], then we must move ahead in LMin[] (or do i++) because all elements on left of LMin[i] are greater than or equal to LMin[i]. Otherwise we must move ahead in RMax[j] to look for a greater j – i value. Here is the c code running in O(n) time:
#include <stdio.h>
#include <stdlib.h>
/* Utility Functions to get max and minimum of two integers */
int max(int x, int y)
{
return x > y? x : y;
}
int min(int x, int y)
{
return x < y? x : y;
}
/* For a given array arr[], returns the maximum j – i such that
arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
int maxDiff;
int i, j;
int *LMin = (int *)malloc(sizeof(int)*n);
int *RMax = (int *)malloc(sizeof(int)*n);
/* Construct LMin[] such that LMin[i] stores the minimum value
from (arr[0], arr[1], ... arr[i]) */
LMin[0] = arr[0];
for (i = 1; i < n; ++i)
LMin[i] = min(arr[i], LMin[i-1]);
/* Construct RMax[] such that RMax[j] stores the maximum value
from (arr[j], arr[j+1], ..arr[n-1]) */
RMax[n-1] = arr[n-1];
for (j = n-2; j >= 0; --j)
RMax[j] = max(arr[j], RMax[j+1]);
/* Traverse both arrays from left to right to find optimum j - i
This process is similar to merge() of MergeSort */
i = 0, j = 0, maxDiff = -1;
while (j < n && i < n)
{
if (LMin[i] < RMax[j])
{
maxDiff = max(maxDiff, j-i);
j = j + 1;
}
else
i = i+1;
}
return maxDiff;
}
/* Driver program to test above functions */
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6};
int n = sizeof(arr)/sizeof(arr[0]);
int maxDiff = maxIndexDiff(arr, n);
printf("\n %d", maxDiff);
getchar();
return 0;
}
Simplified version of Subhasis Das answer using auxiliary arrays.
def maxdistance(nums):
n = len(nums)
minima ,maxima = [None]*n, [None]*n
minima[0],maxima[n-1] = nums[0],nums[n-1]
for i in range(1,n):
minima[i] = min(nums[i],minima[i-1])
for i in range(n-2,-1,-1):
maxima[i]= max(nums[i],maxima[i+1])
i,j,maxdist = 0,0,-1
while(i<n and j<n):
if minima[i] <maxima[j]:
maxdist = max(j-i,maxdist)
j = j+1
else:
i += 1
print maxdist
I can think of improvement over O(n^2), but need to verify if this is O(n) in worse case or not.
Create a variable BestSoln=0; and traverse the array for first element
and store the best solution for first element i.e bestSoln=k;.
Now for 2nd element consider only elements which are k distances away
from the second element.
If BestSoln in this case is better than first iteration then replace
it otherwise let it be like that. Keep iterating for other elements.
It can be improved further if we store max element for each subarray starting from i to end.
This can be done in O(n) by traversing the array from end.
If a particular element is more than it's local max then there is no need to do evaluation for this element.
Input:
{9, 2, 3, 4, 5, 6, 7, 8, 18, 0}
create local max array for this array:
[18,18,18,18,18,18,18,0,0] O(n).
Now, traverse the array for 9 ,here best solution will be i=0,j=8.
Now for second element or after it, we don't need to evaluate. and best solution is i=0,j=8.
But suppose array is Input:
{19, 2, 3, 4, 5, 6, 7, 8, 18, 0,4}
Local max array [18,18,18,18,18,18,18,0,0] then in first iteration we don't need to evaluate as local max is less than current elem.
Now for second iteration best solution is, i=1,j=10. Now for other elements we don't need to consider evaluation as they can't give best solution.
Let me know your view your use case to which my solution is not applicable.
This is a very simple solution for O(2n) of speed and additional ~O(2n) of space (in addition to the input array). The following implementation is in C:
int findMaxDiff(int array[], int size) {
int index = 0;
int maxima[size];
int indexes[size];
while (index < size) {
int max = array[index];
int i;
for (i = index; i < size; i++) {
if (array[i] > max) {
max = array[i];
indexes[index] = i;
}
}
maxima[index] = max;
index++;
}
int j;
int result;
for (j = 0; j < size; j++) {
int max2 = 0;
if (maxima[j] - array[j] > max2) {
max2 = maxima[j] - array[j];
result = indexes[j];
}
}
return result;
}
The first loop scan the array once, finding for each element the maximum of the remaining elements to its right. We store also the relative index in a separate array.
The second loops finds the maximum between each element and the correspondent right-hand-side maximum, and returns the right index.
My Solution with in O(log n) (Please correct me here if I am wrong in calculating this complexity)time ...
Idea is to insert into a BST and then search for node and if the node has a right child then traverse through the right sub tree to calculate the node with maximum index..
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t1 = Integer.parseInt(br.readLine());
for(int j=0;j<t1;j++){
int size = Integer.parseInt(br.readLine());
String input = br.readLine();
String[] t = input.split(" ");
Node root = new Node(Integer.parseInt(t[0]),0);
for(int i=1;i<size;i++){
Node addNode = new Node(Integer.parseInt(t[i]),i);
insertIntoBST(root,addNode);
}
for(String s: t){
Node nd = findNode(root,Integer.parseInt(s));
if(nd.right != null){
int i = nd.index;
int j1 = calculate(nd.right);
mVal = max(mVal,j1-i);
}
}
System.out.println(mVal);
mVal=0;
}
}
static int mVal =0;
public static int calculate (Node root){
if(root==null){
return -1;
}
int i = max(calculate(root.left),calculate(root.right));
return max(root.index,i);
}
public static Node findNode(Node root,int n){
if(root==null){
return null;
}
if(root.value == n){
return root;
}
Node result = findNode(root.left,n);
if(result ==null){
result = findNode(root.right,n);
}
return result;
}
public static int max(int a , int b){
return a<b?b:a;
}
public static class Node{
Node left;
Node right;
int value;
int index;
public Node(int value,int index){
this.value = value;
this.index = index;
}
}
public static void insertIntoBST(Node root, Node addNode){
if(root.value< addNode.value){
if(root.right!=null){
insertIntoBST(root.right,addNode);
}else{
root.right = addNode;
}
}
if(root.value>=addNode.value){
if(root.left!=null){
insertIntoBST(root.left,addNode);
}else{
root.left =addNode;
}
}
}
}
A simplified algorithm from Subhasis Das's answer:
# assume list is not empty
max_dist = 0
acceptable_min = (0, arr[0])
acceptable_max = (0, arr[0])
min = (0, arr[0])
for i in range(len(arr)):
if arr[i] < min[1]:
min = (i, arr[i])
elif arr[i] - min[1] > max_dist:
max_dist = arr[i] - min[1]
acceptable_min = min
acceptable_max = (i, arr[i])
# acceptable_min[0] is the i
# acceptable_max[0] is the j
# max_dist is the max difference
Below is a C++ solution for the condition a[i] <= a[j]. It needs a slight modification to handle the case a[i] < a[j].
template<typename T>
std::size_t max_dist_sorted_pair(const std::vector<T>& seq)
{
const auto n = seq.size();
const auto less = [&seq](std::size_t i, std::size_t j)
{ return seq[i] < seq[j]; };
// max_right[i] is the position of the rightmost
// largest element in the suffix seq[i..]
std::vector<std::size_t> max_right(n);
max_right.back() = n - 1;
for (auto i = n - 1; i > 0; --i)
max_right[i - 1] = std::max(max_right[i], i - 1, less);
std::size_t max_dist = 0;
for (std::size_t i = 0, j = 0; i < n; ++i)
while (!less(max_right[j], i))
{
j = max_right[j];
max_dist = std::max(max_dist, j - i);
if (++j == n)
return max_dist;
}
return max_dist;
}
Please review this solution and cases where it might fail:
def maxIndexDiff(arr, n):
j = n-1
for i in range(0,n):
if j > i:
if arr[j] >= arr[i]:
return j-i
elif arr[j-1] >= arr[i]:
return (j-1) - i
elif arr[j] >= arr[i+1]:
return j - (i+1)
j -= 1
return -1
int maxIndexDiff(int arr[], int n)
{
// Your code here
vector<int> rightMax(n);
rightMax[n-1] = arr[n-1];
for(int i =n-2;i>=0;i--){
rightMax[i] = max(rightMax[i+1],arr[i]);
}
int i = 0,j=0,maxDis = 0;
while(i<n &&j<n){
if(rightMax[j]>=arr[i]){
maxDis = max(maxDis,j-i);
j++;
} else
i++;
}
return maxDis;
}
There is concept of keeping leftMin and rightMax but leftMin is not really required and leftMin will do the work anyways.
We are choosing rightMax and traversing from start till we get a smaller value than that!
Create Arraylist of pairs where is key is array element and value is the index. Sort this arraylist of pairs. Traverse this arraylist of pairs to get the maximum gap between(maxj-i). Also keep a track of maxj and update when new maxj is found. Please find my java solution which takes O(nlogn) time complexity and O(n) space complexity.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
class MaxDistanceSolution {
private class Pair implements Comparable<Pair> {
int key;
int value;
public int getKey() {
return key;
}
public int getValue() {
return value;
}
Pair(int key, int value) {
this.key = key;
this.value = value;
}
#Override
public int compareTo(Pair o) {
return this.getKey() - o.getKey();
}
}
public int maximumGap(final ArrayList<Integer> A) {
int n = A.size();
ArrayList<Pair> B = new ArrayList<>();
for (int i = 0 ; i < n; i++)
B.add(new Pair(A.get(i), i));
Collections.sort(B);
int maxJ = B.get(n-1).getValue();
int gaps = 0;
for (int i = n - 2; i >= 0; i--) {
gaps = Math.max(gaps, maxJ - B.get(i).getValue());
maxJ = Math.max(maxJ, B.get(i).getValue());
}
return gaps;
}
}
public class MaxDistance {
public static void main(String[] args) {
MaxDistanceSolution sol = new MaxDistanceSolution();
ArrayList<Integer> A = new ArrayList<>(Arrays.asList(3, 5, 4, 2));
int gaps = sol.maximumGap(A);
System.out.println(gaps);
}
}
I have solved this question here.
https://github.com/nagendra547/coding-practice/blob/master/src/arrays/FindMaxIndexDifference.java
Putting code here too. Thanks.
private static int findMaxIndexDifferenceOptimal(int[] a) {
int n = a.length;
// array containing minimums
int A[] = new int[n];
A[0] = a[0];
for (int i = 1; i < n; i++) {
A[i] = Math.min(a[i], A[i - 1]);
}
// array containing maximums
int B[] = new int[n];
B[n - 1] = a[n - 1];
for (int j = n - 2; j >= 0; j--) {
B[j] = Math.max(a[j], B[j + 1]);
}
int i = 0, maxDiff = -1;
int j = 0;
while (i < n && j < n) {
if (B[j] > A[i]) {
maxDiff = Math.max(j - i, maxDiff);
j++;
} else {
i++;
}
}
return maxDiff;
}
Given an array A of N integers we draw N discs in a 2D plane, such that i-th disc has center in (0,i) and a radius A[i]. We say that k-th disc and j-th disc intersect, if k-th and j-th discs have at least one common point.
Write a function
int number_of_disc_intersections(int[] A);
which given an array A describing N discs as explained above, returns the number of pairs of intersecting discs. For example, given N=6 and
A[0] = 1
A[1] = 5
A[2] = 2
A[3] = 1
A[4] = 4
A[5] = 0
there are 11 pairs of intersecting discs:
0th and 1st
0th and 2nd
0th and 4th
1st and 2nd
1st and 3rd
1st and 4th
1st and 5th
2nd and 3rd
2nd and 4th
3rd and 4th
4th and 5th
so the function should return 11.
The function should return -1 if the number of intersecting pairs exceeds 10,000,000. The function may assume that N does not exceed 10,000,000.
O(N) complexity and O(N) memory solution.
private static int Intersections(int[] a)
{
int result = 0;
int[] dps = new int[a.length];
int[] dpe = new int[a.length];
for (int i = 0, t = a.length - 1; i < a.length; i++)
{
int s = i > a[i]? i - a[i]: 0;
int e = t - i > a[i]? i + a[i]: t;
dps[s]++;
dpe[e]++;
}
int t = 0;
for (int i = 0; i < a.length; i++)
{
if (dps[i] > 0)
{
result += t * dps[i];
result += dps[i] * (dps[i] - 1) / 2;
if (10000000 < result) return -1;
t += dps[i];
}
t -= dpe[i];
}
return result;
}
So you want to find the number of intersections of the intervals [i-A[i], i+A[i]].
Maintain a sorted array (call it X) containing the i-A[i] (also have some extra space which has the value i+A[i] in there).
Now walk the array X, starting at the leftmost interval (i.e smallest i-A[i]).
For the current interval, do a binary search to see where the right end point of the interval (i.e. i+A[i]) will go (called the rank). Now you know that it intersects all the elements to the left.
Increment a counter with the rank and subtract current position (assuming one indexed) as we don't want to double count intervals and self intersections.
O(nlogn) time, O(n) space.
Python 100 / 100 (tested) on codility, with O(nlogn) time and O(n) space.
Here is #noisyboiler's python implementation of #Aryabhatta's method with comments and an example.
Full credit to original authors, any errors / poor wording are entirely my fault.
from bisect import bisect_right
def number_of_disc_intersections(A):
pairs = 0
# create an array of tuples, each containing the start and end indices of a disk
# some indices may be less than 0 or greater than len(A), this is fine!
# sort the array by the first entry of each tuple: the disk start indices
intervals = sorted( [(i-A[i], i+A[i]) for i in range(len(A))] )
# create an array of starting indices using tuples in intervals
starts = [i[0] for i in intervals]
# for each disk in order of the *starting* position of the disk, not the centre
for i in range(len(starts)):
# find the end position of that disk from the array of tuples
disk_end = intervals[i][1]
# find the index of the rightmost value less than or equal to the interval-end
# this finds the number of disks that have started before disk i ends
count = bisect_right(starts, disk_end )
# subtract current position to exclude previous matches
# this bit seemed 'magic' to me, so I think of it like this...
# for disk i, i disks that start to the left have already been dealt with
# subtract i from count to prevent double counting
# subtract one more to prevent counting the disk itsself
count -= (i+1)
pairs += count
if pairs > 10000000:
return -1
return pairs
Worked example: given [3, 0, 1, 6] the disk radii would look like this:
disk0 ------- start= -3, end= 3
disk1 . start= 1, end= 1
disk2 --- start= 1, end= 3
disk3 ------------- start= -3, end= 9
index 3210123456789 (digits left of zero are -ve)
intervals = [(-3, 3), (-3, 9), (1, 1), (1,3)]
starts = [-3, -3, 1, 1]
the loop order will be: disk0, disk3, disk1, disk2
0th loop:
by the end of disk0, 4 disks have started
one of which is disk0 itself
none of which could have already been counted
so add 3
1st loop:
by the end of disk3, 4 disks have started
one of which is disk3 itself
one of which has already started to the left so is either counted OR would not overlap
so add 2
2nd loop:
by the end of disk1, 4 disks have started
one of which is disk1 itself
two of which have already started to the left so are either counted OR would not overlap
so add 1
3rd loop:
by the end of disk2, 4 disks have started
one of which is disk2 itself
two of which have already started to the left so are either counted OR would not overlap
so add 0
pairs = 6
to check: these are (0,1), (0,2), (0,2), (1,2), (1,3), (2,3),
Well, I adapted Falk Hüffner's idea to c++, and made a change in the range.
Opposite to what is written above, there is no need to go beyond the scope of the array (no matter how large are the values in it).
On Codility this code received 100%.
Thank you Falk for your great idea!
int number_of_disc_intersections ( const vector<int> &A ) {
int sum=0;
vector<int> start(A.size(),0);
vector<int> end(A.size(),0);
for (unsigned int i=0;i<A.size();i++){
if ((int)i<A[i]) start[0]++;
else start[i-A[i]]++;
if (i+A[i]>=A.size()) end[A.size()-1]++;
else end[i+A[i]]++;
}
int active=0;
for (unsigned int i=0;i<A.size();i++){
sum+=active*start[i]+(start[i]*(start[i]-1))/2;
if (sum>10000000) return -1;
active+=start[i]-end[i];
}
return sum;
}
This can even be done in linear time [EDIT: this is not linear time, see comments]. In fact, it becomes easier if you ignore the fact that there is exactly one interval centered at each point, and just treat it as a set of start- and endpoints of intervals. You can then just scan it from the left (Python code for simplicity):
from collections import defaultdict
a = [1, 5, 2, 1, 4, 0]
start = defaultdict(int)
stop = defaultdict(int)
for i in range(len(a)):
start[i - a[i]] += 1
stop[i + a[i]] += 1
active = 0
intersections = 0
for i in range(-len(a), len(a)):
intersections += active * start[i] + (start[i] * (start[i] - 1)) / 2
active += start[i]
active -= stop[i]
print intersections
Here's a O(N) time, O(N) space algorithm requiring 3 runs across the array and no sorting, verified scoring 100%:
You're interested in pairs of discs. Each pair involves one side of one disc and the other side of the other disc. Therefore we won't have duplicate pairs if we handle one side of each disc. Let's call the sides right and left (I rotated the space while thinking about it).
An overlap is either due to a right side overlapping another disc directly at the center (so pairs equal to the radius with some care about the array length) or due to the number of left sides existing at the rightmost edge.
So we create an array that contains the number of left sides at each point and then it's a simple sum.
C code:
int solution(int A[], int N) {
int C[N];
int a, S=0, t=0;
// Mark left and middle of disks
for (int i=0; i<N; i++) {
C[i] = -1;
a = A[i];
if (a>=i) {
C[0]++;
} else {
C[i-a]++;
}
}
// Sum of left side of disks at location
for (int i=0; i<N; i++) {
t += C[i];
C[i] = t;
}
// Count pairs, right side only:
// 1. overlaps based on disk size
// 2. overlaps based on disks but not centers
for (int i=0; i<N; i++) {
a = A[i];
S += ((a<N-i) ? a: N-i-1);
if (i != N-1) {
S += C[((a<N-i) ? i+a: N-1)];
}
if (S>10000000) return -1;
}
return S;
}
I got 100 out of 100 with this C++ implementation:
#include <map>
#include <algorithm>
inline bool mySortFunction(pair<int,int> p1, pair<int,int> p2)
{
return ( p1.first < p2.first );
}
int number_of_disc_intersections ( const vector<int> &A ) {
int i, size = A.size();
if ( size <= 1 ) return 0;
// Compute lower boundary of all discs and sort them in ascending order
vector< pair<int,int> > lowBounds(size);
for(i=0; i<size; i++) lowBounds[i] = pair<int,int>(i-A[i],i+A[i]);
sort(lowBounds.begin(), lowBounds.end(), mySortFunction);
// Browse discs
int nbIntersect = 0;
for(i=0; i<size; i++)
{
int curBound = lowBounds[i].second;
for(int j=i+1; j<size && lowBounds[j].first<=curBound; j++)
{
nbIntersect++;
// Maximal number of intersections
if ( nbIntersect > 10000000 ) return -1;
}
}
return nbIntersect;
}
A Python answer
from bisect import bisect_right
def number_of_disc_intersections(li):
pairs = 0
# treat as a series of intervals on the y axis at x=0
intervals = sorted( [(i-li[i], i+li[i]) for i in range(len(li))] )
# do this by creating a list of start points of each interval
starts = [i[0] for i in intervals]
for i in range(len(starts)):
# find the index of the rightmost value less than or equal to the interval-end
count = bisect_right(starts, intervals[i][1])
# subtract current position to exclude previous matches, and subtract self
count -= (i+1)
pairs += count
if pairs > 10000000:
return -1
return pairs
100/100 c#
class Solution
{
class Interval
{
public long Left;
public long Right;
}
public int solution(int[] A)
{
if (A == null || A.Length < 1)
{
return 0;
}
var itervals = new Interval[A.Length];
for (int i = 0; i < A.Length; i++)
{
// use long to avoid data overflow (eg. int.MaxValue + 1)
long radius = A[i];
itervals[i] = new Interval()
{
Left = i - radius,
Right = i + radius
};
}
itervals = itervals.OrderBy(i => i.Left).ToArray();
int result = 0;
for (int i = 0; i < itervals.Length; i++)
{
var right = itervals[i].Right;
for (int j = i + 1; j < itervals.Length && itervals[j].Left <= right; j++)
{
result++;
if (result > 10000000)
{
return -1;
}
}
}
return result;
}
}
I'm offering one more solution because I did not find the counting principle of the previous solutions easy to follow. Though the results are the same, an explanation and more intuitive counting procedure seems worth presenting.
To begin, start by considering the O(N^2) solution that iterates over the discs in order of their center points, and counts the number of discs centered to the right of the current disc's that intersect the current disc, using the condition current_center + radius >= other_center - radius. Notice that we could get the same result counting discs centered to the left of the current disc using the condition current_center - radius <= other_center + radius.
def simple(A):
"""O(N^2) solution for validating more efficient solution."""
N = len(A)
unique_intersections = 0
# Iterate over discs in order of their center positions
for j in range(N):
# Iterate over discs whose center is to the right, to avoid double-counting.
for k in range(j+1, N):
# Increment cases where edge of current disk is at or right of the left edge of another disk.
if j + A[j] >= k - A[k]:
unique_intersections += 1
# Stop early if we have enough intersections.
# BUT: if the discs are small we still N^2 compare them all and time out.
if unique_intersections > 10000000:
return -1
return unique_intersections
We can go from O(N^2) to O(N) if we could only "look up" the number of discs to the right (or to the left!) that intersect the current disc. The key insight is to reinterpret the intersection condition as "the right edge of one disc overlaps the left edge of another disc", meaning (a ha!) the centers don't matter, only the edges.
The next insight is to try sorting the edges, taking O(N log N) time. Given a sorted array of the left edges and a sorted array of the right edges, as we scan our way from left to right along the number line, the number of left or right edges to the left of the current location point is simply the current index into left_edges and right_edges respectively: a constant-time deduction.
Finally, we use the "right edge > left edge" condition to deduce that the number of intersections between the current disc and discs that start only to the left of the current disc (to avoid duplicates) is the number of left edges to the left of the current edge, minus the number of right edges to the left of the current edge. That is, the number of discs starting to left of this one, minus the ones that closed already.
Now for this code, tested 100% on Codility:
def solution(A):
"""O(N log N) due to sorting, with O(N) pass over sorted arrays"""
N = len(A)
# Left edges of the discs, in increasing order of position.
left_edges = sorted([(p-r) for (p,r) in enumerate(A)])
# Right edges of the discs, in increasing order of position.
right_edges = sorted([(p+r) for (p,r) in enumerate(A)])
#print("left edges:", left_edges[:10])
#print("right edges:", right_edges[:10])
intersections = 0
right_i = 0
# Iterate over the discs in order of their leftmost edge position.
for left_i in range(N):
# Find the first right_edge that's right of or equal to the current left_edge, naively:
# right_i = bisect.bisect_left(right_edges, left_edges[left_i])
# Just scan from previous index until right edge is at or beyond current left:
while right_edges[right_i] < left_edges[left_i]:
right_i += 1
# Count number of discs starting left of current, minus the ones that already closed.
intersections += left_i - right_i
# Return early if we find more than 10 million intersections.
if intersections > 10000000:
return -1
#print("correct:", simple(A))
return intersections
Java 2*100%.
result is declared as long for a case codility doesn't test, namely 50k*50k intersections at one point.
class Solution {
public int solution(int[] A) {
int[] westEnding = new int[A.length];
int[] eastEnding = new int[A.length];
for (int i=0; i<A.length; i++) {
if (i-A[i]>=0) eastEnding[i-A[i]]++; else eastEnding[0]++;
if ((long)i+A[i]<A.length) westEnding[i+A[i]]++; else westEnding[A.length-1]++;
}
long result = 0; //long to contain the case of 50k*50k. codility doesn't test for this.
int wests = 0;
int easts = 0;
for (int i=0; i<A.length; i++) {
int balance = easts*wests; //these are calculated elsewhere
wests++;
easts+=eastEnding[i];
result += (long) easts*wests - balance - 1; // 1 stands for the self-intersection
if (result>10000000) return -1;
easts--;
wests-= westEnding[i];
}
return (int) result;
}
}
Swift 4 Solution 100% (Codility do not check the worst case for this solution)
public func solution(_ A : inout [Int]) -> Int {
// write your code in Swift 4.2.1 (Linux)
var count = 0
let sortedA = A.sorted(by: >)
if sortedA.isEmpty{ return 0 }
let maxVal = sortedA[0]
for i in 0..<A.count{
let maxIndex = min(i + A[i] + maxVal + 1,A.count)
for j in i + 1..<maxIndex{
if j - A[j] <= i + A[i]{
count += 1
}
}
if count > 10_000_000{
return -1
}
}
return count
}
Here my JavaScript solution, based in other solutions in this thread but implemented in other languages.
function solution(A) {
let circleEndpoints = [];
for(const [index, num] of Object.entries(A)) {
circleEndpoints.push([parseInt(index)-num, true]);
circleEndpoints.push([parseInt(index)+num, false]);
}
circleEndpoints = circleEndpoints.sort(([a, openA], [b, openB]) => {
if(a == b) return openA ? -1 : 1;
return a - b;
});
let openCircles = 0;
let intersections = 0;
for(const [endpoint, opening] of circleEndpoints) {
if(opening) {
intersections += openCircles;
openCircles ++;
} else {
openCircles --;
}
if(intersections > 10000000) return -1;
}
return intersections;
}
count = 0
for (int i = 0; i < N; i++) {
for (int j = i+1; j < N; j++) {
if (i + A[i] >= j - A[j]) count++;
}
}
It is O(N^2) so pretty slow, but it works.
This is a ruby solution that scored 100/100 on codility. I'm posting it now because I'm finding it difficult to follow the already posted ruby answer.
def solution(a)
end_points = []
a.each_with_index do |ai, i|
end_points << [i - ai, i + ai]
end
end_points = end_points.sort_by { |points| points[0]}
intersecting_pairs = 0
end_points.each_with_index do |point, index|
lep, hep = point
pairs = bsearch(end_points, index, end_points.size - 1, hep)
return -1 if 10000000 - pairs + index < intersecting_pairs
intersecting_pairs += (pairs - index)
end
return intersecting_pairs
end
# This method returns the maximally appropriate position
# where the higher end-point may have been inserted.
def bsearch(a, l, u, x)
if l == u
if x >= a[u][0]
return u
else
return l - 1
end
end
mid = (l + u)/2
# Notice that we are searching in higher range
# even if we have found equality.
if a[mid][0] <= x
return bsearch(a, mid+1, u, x)
else
return bsearch(a, l, mid, x)
end
end
Probably extremely fast. O(N). But you need to check it out. 100% on Codility.
Main idea:
1. At any point of the table, there are number of circles "opened" till the right edge of the circle, lets say "o".
2. So there are (o-1-used) possible pairs for the circle in that point. "used" means circle that have been processed and pairs for them counted.
public int solution(int[] A) {
final int N = A.length;
final int M = N + 2;
int[] left = new int[M]; // values of nb of "left" edges of the circles in that point
int[] sleft = new int[M]; // prefix sum of left[]
int il, ir; // index of the "left" and of the "right" edge of the circle
for (int i = 0; i < N; i++) { // counting left edges
il = tl(i, A);
left[il]++;
}
sleft[0] = left[0];
for (int i = 1; i < M; i++) {// counting prefix sums for future use
sleft[i]=sleft[i-1]+left[i];
}
int o, pairs, total_p = 0, total_used=0;
for (int i = 0; i < N; i++) { // counting pairs
ir = tr(i, A, M);
o = sleft[ir]; // nb of open till right edge
pairs = o -1 - total_used;
total_used++;
total_p += pairs;
}
if(total_p > 10000000){
total_p = -1;
}
return total_p;
}
private int tl(int i, int[] A){
int tl = i - A[i]; // index of "begin" of the circle
if (tl < 0) {
tl = 0;
} else {
tl = i - A[i] + 1;
}
return tl;
}
int tr(int i, int[] A, int M){
int tr; // index of "end" of the circle
if (Integer.MAX_VALUE - i < A[i] || i + A[i] >= M - 1) {
tr = M - 1;
} else {
tr = i + A[i] + 1;
}
return tr;
}
There are a lot of great answers here already, including the great explanation from the accepted answer. However, I wanted to point out a small observation about implementation details in the Python language.
Originally, I've came up with the solution shown below. I was expecting to get O(N*log(N)) time complexity as soon as we have a single for-loop with N iterations, and each iteration performs a binary search that takes at most log(N).
def solution(a):
import bisect
if len(a) <= 1:
return 0
cuts = [(c - r, c + r) for c, r in enumerate(a)]
cuts.sort(key=lambda pair: pair[0])
lefts, rights = zip(*cuts)
n = len(cuts)
total = 0
for i in range(n):
r = rights[i]
pos = bisect.bisect_right(lefts[i+1:], r)
total += pos
if total > 10e6:
return -1
return total
However, I've get O(N**2) and a timeout failure. Do you see what is wrong here? Right, this line:
pos = bisect.bisect_right(lefts[i+1:], r)
In this line, you are actually taking a copy of the original list to pass it into binary search function, and it totally ruins the efficiency of the proposed solution! It makes your code just a bit more consice (i.e., you don't need to write pos - i - 1) but heavily undermies the performance. So, as it was shown above, the solution should be:
def solution(a):
import bisect
if len(a) <= 1:
return 0
cuts = [(c - r, c + r) for c, r in enumerate(a)]
cuts.sort(key=lambda pair: pair[0])
lefts, rights = zip(*cuts)
n = len(cuts)
total = 0
for i in range(n):
r = rights[i]
pos = bisect.bisect_right(lefts, r)
total += (pos - i - 1)
if total > 10e6:
return -1
return total
It seems that sometimes one could be too eager about making slices and copies because Python allows you to do it so easily :) Probably not a great insight, but for me it was a good lesson to pay more attention to these "technical" moments when converting ideas and algorithms into real-word solutions.
I know that this is an old questions but it is still active on codility.
private int solution(int[] A)
{
int openedCircles = 0;
int intersectCount = 0;
We need circles with their start and end values. For that purpose I have used Tuple.
True/False indicates if we are adding Circle Starting or Circle Ending value.
List<Tuple<decimal, bool>> circles = new List<Tuple<decimal, bool>>();
for(int i = 0; i < A.Length; i ++)
{
// Circle start value
circles.Add(new Tuple<decimal, bool>((decimal)i - (decimal)A[i], true));
// Circle end value
circles.Add(new Tuple<decimal, bool>((decimal)i + (decimal)A[i], false));
}
Order "circles" by their values.
If one circle is ending at same value where other circle is starting, it should be counted as intersect (because of that "opening" should be in front of "closing" if in same point)
circles = circles.OrderBy(x => x.Item1).ThenByDescending(x => x.Item2).ToList();
Counting and returning counter
foreach (var circle in circles)
{
// We are opening new circle (within existing circles)
if(circle.Item2 == true)
{
intersectCount += openedCircles;
if (intersectCount > 10000000)
{
return -1;
}
openedCircles++;
}
else
{
// We are closing circle
openedCircles--;
}
}
return intersectCount;
}
Javascript solution 100/100 based on this video https://www.youtube.com/watch?v=HV8tzIiidSw
function sortArray(A) {
return A.sort((a, b) => a - b)
}
function getDiskPoints(A) {
const diskStarPoint = []
const diskEndPoint = []
for(i = 0; i < A.length; i++) {
diskStarPoint.push(i - A[i])
diskEndPoint.push(i + A[i])
}
return {
diskStarPoint: sortArray(diskStarPoint),
diskEndPoint: sortArray(diskEndPoint)
};
}
function solution(A) {
const { diskStarPoint, diskEndPoint } = getDiskPoints(A)
let index = 0;
let openDisks = 0;
let intersections = 0;
for(i = 0; i < diskStarPoint.length; i++) {
while(diskStarPoint[i] > diskEndPoint[index]) {
openDisks--
index++
}
intersections += openDisks
openDisks++
}
return intersections > 10000000 ? -1 : intersections
}
so, I was doing this test in Scala and I would like to share here my example. My idea to solve is:
Extract the limits to the left and right of each position on the array.
A[0] = 1 --> (0-1, 0+1) = A0(-1, 1)
A[1] = 5 --> (1-5, 1+5) = A1(-4, 6)
A[2] = 2 --> (2-2, 2+2) = A2(0, 4)
A[3] = 1 --> (3-1, 3+1) = A3(2, 4)
A[4] = 4 --> (4-4, 4+4) = A4(0, 8)
A[5] = 0 --> (5-0, 5+0) = A5(5, 5)
Check if there is intersections between any two positions
(A0_0 >= A1_0 AND A0_0 <= A1_1) OR // intersection
(A0_1 >= A1_0 AND A0_1 <= A1_1) OR // intersection
(A0_0 <= A1_0 AND A0_1 >= A1_1) // one circle contain inside the other
if any of these two checks is true count one intersection.
object NumberOfDiscIntersections {
def solution(a: Array[Int]): Int = {
var count: Long = 0
for (posI: Long <- 0L until a.size) {
for (posJ <- (posI + 1) until a.size) {
val tupleI = (posI - a(posI.toInt), posI + a(posI.toInt))
val tupleJ = (posJ - a(posJ.toInt), posJ + a(posJ.toInt))
if ((tupleI._1 >= tupleJ._1 && tupleI._1 <= tupleJ._2) ||
(tupleI._2 >= tupleJ._1 && tupleI._2 <= tupleJ._2) ||
(tupleI._1 <= tupleJ._1 && tupleI._2 >= tupleJ._2)) {
count += 1
}
}
}
count.toInt
}
}
This got 100/100 in c#
class CodilityDemo3
{
public static int GetIntersections(int[] A)
{
if (A == null)
{
return 0;
}
int size = A.Length;
if (size <= 1)
{
return 0;
}
List<Line> lines = new List<Line>();
for (int i = 0; i < size; i++)
{
if (A[i] >= 0)
{
lines.Add(new Line(i - A[i], i + A[i]));
}
}
lines.Sort(Line.CompareLines);
size = lines.Count;
int intersects = 0;
for (int i = 0; i < size; i++)
{
Line ln1 = lines[i];
for (int j = i + 1; j < size; j++)
{
Line ln2 = lines[j];
if (ln2.YStart <= ln1.YEnd)
{
intersects += 1;
if (intersects > 10000000)
{
return -1;
}
}
else
{
break;
}
}
}
return intersects;
}
}
public class Line
{
public Line(double ystart, double yend)
{
YStart = ystart;
YEnd = yend;
}
public double YStart { get; set; }
public double YEnd { get; set; }
public static int CompareLines(Line line1, Line line2)
{
return (line1.YStart.CompareTo(line2.YStart));
}
}
}
Thanks to Falk for the great idea! Here is a ruby implementation that takes advantage of sparseness.
def int(a)
event = Hash.new{|h,k| h[k] = {:start => 0, :stop => 0}}
a.each_index {|i|
event[i - a[i]][:start] += 1
event[i + a[i]][:stop ] += 1
}
sorted_events = (event.sort_by {|index, value| index}).map! {|n| n[1]}
past_start = 0
intersect = 0
sorted_events.each {|e|
intersect += e[:start] * (e[:start]-1) / 2 +
e[:start] * past_start
past_start += e[:start]
past_start -= e[:stop]
}
return intersect
end
puts int [1,1]
puts int [1,5,2,1,4,0]
#include <stdio.h>
#include <stdlib.h>
void sortPairs(int bounds[], int len){
int i,j, temp;
for(i=0;i<(len-1);i++){
for(j=i+1;j<len;j++){
if(bounds[i] > bounds[j]){
temp = bounds[i];
bounds[i] = bounds[j];
bounds[j] = temp;
temp = bounds[i+len];
bounds[i+len] = bounds[j+len];
bounds[j+len] = temp;
}
}
}
}
int adjacentPointPairsCount(int a[], int len){
int count=0,i,j;
int *bounds;
if(len<2) {
goto toend;
}
bounds = malloc(sizeof(int)*len *2);
for(i=0; i< len; i++){
bounds[i] = i-a[i];
bounds[i+len] = i+a[i];
}
sortPairs(bounds, len);
for(i=0;i<len;i++){
int currentBound = bounds[i+len];
for(j=i+1;a[j]<=currentBound;j++){
if(count>100000){
count=-1;
goto toend;
}
count++;
}
}
toend:
free(bounds);
return count;
}
An Implementation of Idea stated above in Java:
public class DiscIntersectionCount {
public int number_of_disc_intersections(int[] A) {
int[] leftPoints = new int[A.length];
for (int i = 0; i < A.length; i++) {
leftPoints[i] = i - A[i];
}
Arrays.sort(leftPoints);
// System.out.println(Arrays.toString(leftPoints));
int count = 0;
for (int i = 0; i < A.length - 1; i++) {
int rpoint = A[i] + i;
int rrank = getRank(leftPoints, rpoint);
//if disk has sifnificant radius, exclude own self
if (rpoint > i) rrank -= 1;
int rank = rrank;
// System.out.println(rpoint+" : "+rank);
rank -= i;
count += rank;
}
return count;
}
public int getRank(int A[], int num) {
if (A==null || A.length == 0) return -1;
int mid = A.length/2;
while ((mid >= 0) && (mid < A.length)) {
if (A[mid] == num) return mid;
if ((mid == 0) && (A[mid] > num)) return -1;
if ((mid == (A.length - 1)) && (A[mid] < num)) return A.length;
if (A[mid] < num && A[mid + 1] >= num) return mid + 1;
if (A[mid] > num && A[mid - 1] <= num) return mid - 1;
if (A[mid] < num) mid = (mid + A.length)/2;
else mid = (mid)/2;
}
return -1;
}
public static void main(String[] args) {
DiscIntersectionCount d = new DiscIntersectionCount();
int[] A =
//{1,5,2,1,4,0}
//{0,0,0,0,0,0}
// {1,1,2}
{3}
;
int count = d.number_of_disc_intersections(A);
System.out.println(count);
}
}
Here is the PHP code that scored 100 on codility:
$sum=0;
//One way of cloning the A:
$start = array();
$end = array();
foreach ($A as $key=>$value)
{
$start[]=0;
$end[]=0;
}
for ($i=0; $i<count($A); $i++)
{
if ($i<$A[$i])
$start[0]++;
else
$start[$i-$A[$i]]++;
if ($i+$A[$i] >= count($A))
$end[count($A)-1]++;
else
$end[$i+$A[$i]]++;
}
$active=0;
for ($i=0; $i<count($A);$i++)
{
$sum += $active*$start[$i]+($start[$i]*($start[$i]-1))/2;
if ($sum>10000000) return -1;
$active += $start[$i]-$end[$i];
}
return $sum;
However I dont understand the logic. This is just transformed C++ code from above. Folks, can you elaborate on what you were doing here, please?
A 100/100 C# implementation as described by Aryabhatta (the binary search solution).
using System;
class Solution {
public int solution(int[] A)
{
return IntersectingDiscs.Execute(A);
}
}
class IntersectingDiscs
{
public static int Execute(int[] data)
{
int counter = 0;
var intervals = Interval.GetIntervals(data);
Array.Sort(intervals); // sort by Left value
for (int i = 0; i < intervals.Length; i++)
{
counter += GetCoverage(intervals, i);
if(counter > 10000000)
{
return -1;
}
}
return counter;
}
private static int GetCoverage(Interval[] intervals, int i)
{
var currentInterval = intervals[i];
// search for an interval starting at currentInterval.Right
int j = Array.BinarySearch(intervals, new Interval { Left = currentInterval.Right });
if(j < 0)
{
// item not found
j = ~j; // bitwise complement (see Array.BinarySearch documentation)
// now j == index of the next item larger than the searched one
j = j - 1; // set index to the previous element
}
while(j + 1 < intervals.Length && intervals[j].Left == intervals[j + 1].Left)
{
j++; // get the rightmost interval starting from currentInterval.Righ
}
return j - i; // reduce already processed intervals (the left side from currentInterval)
}
}
class Interval : IComparable
{
public long Left { get; set; }
public long Right { get; set; }
// Implementation of IComparable interface
// which is used by Array.Sort().
public int CompareTo(object obj)
{
// elements will be sorted by Left value
var another = obj as Interval;
if (this.Left < another.Left)
{
return -1;
}
if (this.Left > another.Left)
{
return 1;
}
return 0;
}
/// <summary>
/// Transform array items into Intervals (eg. {1, 2, 4} -> {[-1,1], [-1,3], [-2,6]}).
/// </summary>
public static Interval[] GetIntervals(int[] data)
{
var intervals = new Interval[data.Length];
for (int i = 0; i < data.Length; i++)
{
// use long to avoid data overflow (eg. int.MaxValue + 1)
long radius = data[i];
intervals[i] = new Interval
{
Left = i - radius,
Right = i + radius
};
}
return intervals;
}
}
100% score in Codility.
Here is an adaptation to C# of Толя solution:
public int solution(int[] A)
{
long result = 0;
Dictionary<long, int> dps = new Dictionary<long, int>();
Dictionary<long, int> dpe = new Dictionary<long, int>();
for (int i = 0; i < A.Length; i++)
{
Inc(dps, Math.Max(0, i - A[i]));
Inc(dpe, Math.Min(A.Length - 1, i + A[i]));
}
long t = 0;
for (int i = 0; i < A.Length; i++)
{
int value;
if (dps.TryGetValue(i, out value))
{
result += t * value;
result += value * (value - 1) / 2;
t += value;
if (result > 10000000)
return -1;
}
dpe.TryGetValue(i, out value);
t -= value;
}
return (int)result;
}
private static void Inc(Dictionary<long, int> values, long index)
{
int value;
values.TryGetValue(index, out value);
values[index] = ++value;
}
Here's a two-pass C++ solution that doesn't require any libraries, binary searching, sorting, etc.
int solution(vector<int> &A) {
#define countmax 10000000
int count = 0;
// init lower edge array
vector<int> E(A.size());
for (int i = 0; i < (int) E.size(); i++)
E[i] = 0;
// first pass
// count all lower numbered discs inside this one
// mark lower edge of each disc
for (int i = 0; i < (int) A.size(); i++)
{
// if disc overlaps zero
if (i - A[i] <= 0)
count += i;
// doesn't overlap zero
else {
count += A[i];
E[i - A[i]]++;
}
if (count > countmax)
return -1;
}
// second pass
// count higher numbered discs with edge inside this one
for (int i = 0; i < (int) A.size(); i++)
{
// loop up inside this disc until top of vector
int jend = ((int) E.size() < (long long) i + A[i] + 1 ?
(int) E.size() : i + A[i] + 1);
// count all discs with edge inside this disc
// note: if higher disc is so big that edge is at or below
// this disc center, would count intersection in first pass
for (int j = i + 1; j < jend; j++)
count += E[j];
if (count > countmax)
return -1;
}
return count;
}
My answer in Swift; gets a 100% score.
import Glibc
struct Interval {
let start: Int
let end: Int
}
func bisectRight(intervals: [Interval], end: Int) -> Int {
var pos = -1
var startpos = 0
var endpos = intervals.count - 1
if intervals.count == 1 {
if intervals[0].start < end {
return 1
} else {
return 0
}
}
while true {
let currentLength = endpos - startpos
if currentLength == 1 {
pos = startpos
pos += 1
if intervals[pos].start <= end {
pos += 1
}
break
} else {
let middle = Int(ceil( Double((endpos - startpos)) / 2.0 ))
let middlepos = startpos + middle
if intervals[middlepos].start <= end {
startpos = middlepos
} else {
endpos = middlepos
}
}
}
return pos
}
public func solution(inout A: [Int]) -> Int {
let N = A.count
var nIntersections = 0
// Create array of intervals
var unsortedIntervals: [Interval] = []
for i in 0 ..< N {
let interval = Interval(start: i-A[i], end: i+A[i])
unsortedIntervals.append(interval)
}
// Sort array
let intervals = unsortedIntervals.sort {
$0.start < $1.start
}
for i in 0 ..< intervals.count {
let end = intervals[i].end
var count = bisectRight(intervals, end: end)
count -= (i + 1)
nIntersections += count
if nIntersections > Int(10E6) {
return -1
}
}
return nIntersections
}
C# solution 100/100
using System.Linq;
class Solution
{
private struct Interval
{
public Interval(long #from, long to)
{
From = #from;
To = to;
}
public long From { get; }
public long To { get; }
}
public int solution(int[] A)
{
int result = 0;
Interval[] intervals = A.Select((value, i) =>
{
long iL = i;
return new Interval(iL - value, iL + value);
})
.OrderBy(x => x.From)
.ToArray();
for (int i = 0; i < intervals.Length; i++)
{
for (int j = i + 1; j < intervals.Length && intervals[j].From <= intervals[i].To; j++)
result++;
if (result > 10000000)
return -1;
}
return result;
}
}
I was asked this question in a job interview, and I'd like to know how others would solve it. I'm most comfortable with Java, but solutions in other languages are welcome.
Given an array of numbers, nums, return an array of numbers products, where products[i] is the product of all nums[j], j != i.
Input : [1, 2, 3, 4, 5]
Output: [(2*3*4*5), (1*3*4*5), (1*2*4*5), (1*2*3*5), (1*2*3*4)]
= [120, 60, 40, 30, 24]
You must do this in O(N) without using division.
An explanation of polygenelubricants method is:
The trick is to construct the arrays (in the case for 4 elements):
{ 1, a[0], a[0]*a[1], a[0]*a[1]*a[2], }
{ a[1]*a[2]*a[3], a[2]*a[3], a[3], 1, }
Both of which can be done in O(n) by starting at the left and right edges respectively.
Then, multiplying the two arrays element-by-element gives the required result.
My code would look something like this:
int a[N] // This is the input
int products_below[N];
int p = 1;
for (int i = 0; i < N; ++i) {
products_below[i] = p;
p *= a[i];
}
int products_above[N];
p = 1;
for (int i = N - 1; i >= 0; --i) {
products_above[i] = p;
p *= a[i];
}
int products[N]; // This is the result
for (int i = 0; i < N; ++i) {
products[i] = products_below[i] * products_above[i];
}
If you need the solution be O(1) in space as well, you can do this (which is less clear in my opinion):
int a[N] // This is the input
int products[N];
// Get the products below the current index
int p = 1;
for (int i = 0; i < N; ++i) {
products[i] = p;
p *= a[i];
}
// Get the products above the current index
p = 1;
for (int i = N - 1; i >= 0; --i) {
products[i] *= p;
p *= a[i];
}
Here is a small recursive function (in C++) to do the modification in-place. It requires O(n) extra space (on stack) though. Assuming the array is in a and N holds the array length, we have:
int multiply(int *a, int fwdProduct, int indx) {
int revProduct = 1;
if (indx < N) {
revProduct = multiply(a, fwdProduct*a[indx], indx+1);
int cur = a[indx];
a[indx] = fwdProduct * revProduct;
revProduct *= cur;
}
return revProduct;
}
Here's my attempt to solve it in Java. Apologies for the non-standard formatting, but the code has a lot of duplication, and this is the best I can do to make it readable.
import java.util.Arrays;
public class Products {
static int[] products(int... nums) {
final int N = nums.length;
int[] prods = new int[N];
Arrays.fill(prods, 1);
for (int
i = 0, pi = 1 , j = N-1, pj = 1 ;
(i < N) && (j >= 0) ;
pi *= nums[i++] , pj *= nums[j--] )
{
prods[i] *= pi ; prods[j] *= pj ;
}
return prods;
}
public static void main(String[] args) {
System.out.println(
Arrays.toString(products(1, 2, 3, 4, 5))
); // prints "[120, 60, 40, 30, 24]"
}
}
The loop invariants are pi = nums[0] * nums[1] *.. nums[i-1] and pj = nums[N-1] * nums[N-2] *.. nums[j+1]. The i part on the left is the "prefix" logic, and the j part on the right is the "suffix" logic.
Recursive one-liner
Jasmeet gave a (beautiful!) recursive solution; I've turned it into this (hideous!) Java one-liner. It does in-place modification, with O(N) temporary space in the stack.
static int multiply(int[] nums, int p, int n) {
return (n == nums.length) ? 1
: nums[n] * (p = multiply(nums, nums[n] * (nums[n] = p), n + 1))
+ 0*(nums[n] *= p);
}
int[] arr = {1,2,3,4,5};
multiply(arr, 1, 0);
System.out.println(Arrays.toString(arr));
// prints "[120, 60, 40, 30, 24]"
Translating Michael Anderson's solution into Haskell:
otherProducts xs = zipWith (*) below above
where below = scanl (*) 1 $ init xs
above = tail $ scanr (*) 1 xs
Sneakily circumventing the "no divisions" rule:
sum = 0.0
for i in range(a):
sum += log(a[i])
for i in range(a):
output[i] = exp(sum - log(a[i]))
Here you go, simple and clean solution with O(N) complexity:
int[] a = {1,2,3,4,5};
int[] r = new int[a.length];
int x = 1;
r[0] = 1;
for (int i=1;i<a.length;i++){
r[i]=r[i-1]*a[i-1];
}
for (int i=a.length-1;i>0;i--){
x=x*a[i];
r[i-1]=x*r[i-1];
}
for (int i=0;i<r.length;i++){
System.out.println(r[i]);
}
Travel Left->Right and keep saving product. Call it Past. -> O(n)
Travel Right -> left keep the product. Call it Future. -> O(n)
Result[i] = Past[i-1] * future[i+1] -> O(n)
Past[-1] = 1; and Future[n+1]=1;
O(n)
C++, O(n):
long long prod = accumulate(in.begin(), in.end(), 1LL, multiplies<int>());
transform(in.begin(), in.end(), back_inserter(res),
bind1st(divides<long long>(), prod));
Here is my solution in modern C++. It makes use of std::transform and is pretty easy to remember.
Online code (wandbox).
#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
vector<int>& multiply_up(vector<int>& v){
v.insert(v.begin(),1);
transform(v.begin()+1, v.end()
,v.begin()
,v.begin()+1
,[](auto const& a, auto const& b) { return b*a; }
);
v.pop_back();
return v;
}
int main() {
vector<int> v = {1,2,3,4,5};
auto vr = v;
reverse(vr.begin(),vr.end());
multiply_up(v);
multiply_up(vr);
reverse(vr.begin(),vr.end());
transform(v.begin(),v.end()
,vr.begin()
,v.begin()
,[](auto const& a, auto const& b) { return b*a; }
);
for(auto& i: v) cout << i << " ";
}
Precalculate the product of the numbers to the left and to the right of each element.
For every element the desired value is the product of it's neigbors's products.
#include <stdio.h>
unsigned array[5] = { 1,2,3,4,5};
int main(void)
{
unsigned idx;
unsigned left[5]
, right[5];
left[0] = 1;
right[4] = 1;
/* calculate products of numbers to the left of [idx] */
for (idx=1; idx < 5; idx++) {
left[idx] = left[idx-1] * array[idx-1];
}
/* calculate products of numbers to the right of [idx] */
for (idx=4; idx-- > 0; ) {
right[idx] = right[idx+1] * array[idx+1];
}
for (idx=0; idx <5 ; idx++) {
printf("[%u] Product(%u*%u) = %u\n"
, idx, left[idx] , right[idx] , left[idx] * right[idx] );
}
return 0;
}
Result:
$ ./a.out
[0] Product(1*120) = 120
[1] Product(1*60) = 60
[2] Product(2*20) = 40
[3] Product(6*5) = 30
[4] Product(24*1) = 24
(UPDATE: now I look closer, this uses the same method as Michael Anderson, Daniel Migowski and polygenelubricants above)
Tricky:
Use the following:
public int[] calc(int[] params) {
int[] left = new int[n-1]
in[] right = new int[n-1]
int fac1 = 1;
int fac2 = 1;
for( int i=0; i<n; i++ ) {
fac1 = fac1 * params[i];
fac2 = fac2 * params[n-i];
left[i] = fac1;
right[i] = fac2;
}
fac = 1;
int[] results = new int[n];
for( int i=0; i<n; i++ ) {
results[i] = left[i] * right[i];
}
Yes, I am sure i missed some i-1 instead of i, but thats the way to solve it.
This is O(n^2) but f# is soooo beautiful:
List.fold (fun seed i -> List.mapi (fun j x -> if i=j+1 then x else x*i) seed)
[1;1;1;1;1]
[1..5]
There also is a O(N^(3/2)) non-optimal solution. It is quite interesting, though.
First preprocess each partial multiplications of size N^0.5(this is done in O(N) time complexity). Then, calculation for each number's other-values'-multiple can be done in 2*O(N^0.5) time(why? because you only need to multiple the last elements of other ((N^0.5) - 1) numbers, and multiply the result with ((N^0.5) - 1) numbers that belong to the group of the current number). Doing this for each number, one can get O(N^(3/2)) time.
Example:
4 6 7 2 3 1 9 5 8
partial results:
4*6*7 = 168
2*3*1 = 6
9*5*8 = 360
To calculate the value of 3, one needs to multiply the other groups' values 168*360, and then with 2*1.
public static void main(String[] args) {
int[] arr = { 1, 2, 3, 4, 5 };
int[] result = { 1, 1, 1, 1, 1 };
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < i; j++) {
result[i] *= arr[j];
}
for (int k = arr.length - 1; k > i; k--) {
result[i] *= arr[k];
}
}
for (int i : result) {
System.out.println(i);
}
}
This solution i came up with and i found it so clear what do you think!?
Based on Billz answer--sorry I can't comment, but here is a scala version that correctly handles duplicate items in the list, and is probably O(n):
val list1 = List(1, 7, 3, 3, 4, 4)
val view = list1.view.zipWithIndex map { x => list1.view.patch(x._2, Nil, 1).reduceLeft(_*_)}
view.force
returns:
List(1008, 144, 336, 336, 252, 252)
Adding my javascript solution here as I didn't find anyone suggesting this.
What is to divide, except to count the number of times you can extract a number from another number? I went through calculating the product of the whole array, and then iterate over each element, and substracting the current element until zero:
//No division operation allowed
// keep substracting divisor from dividend, until dividend is zero or less than divisor
function calculateProducsExceptCurrent_NoDivision(input){
var res = [];
var totalProduct = 1;
//calculate the total product
for(var i = 0; i < input.length; i++){
totalProduct = totalProduct * input[i];
}
//populate the result array by "dividing" each value
for(var i = 0; i < input.length; i++){
var timesSubstracted = 0;
var divisor = input[i];
var dividend = totalProduct;
while(divisor <= dividend){
dividend = dividend - divisor;
timesSubstracted++;
}
res.push(timesSubstracted);
}
return res;
}
Just 2 passes up and down. Job done in O(N)
private static int[] multiply(int[] numbers) {
int[] multiplied = new int[numbers.length];
int total = 1;
multiplied[0] = 1;
for (int i = 1; i < numbers.length; i++) {
multiplied[i] = numbers[i - 1] * multiplied[i - 1];
}
for (int j = numbers.length - 2; j >= 0; j--) {
total *= numbers[j + 1];
multiplied[j] = total * multiplied[j];
}
return multiplied;
}
def productify(arr, prod, i):
if i < len(arr):
prod.append(arr[i - 1] * prod[i - 1]) if i > 0 else prod.append(1)
retval = productify(arr, prod, i + 1)
prod[i] *= retval
return retval * arr[i]
return 1
if __name__ == "__main__":
arr = [1, 2, 3, 4, 5]
prod = []
productify(arr, prod, 0)
print(prod)
Well,this solution can be considered that of C/C++.
Lets say we have an array "a" containing n elements
like a[n],then the pseudo code would be as below.
for(j=0;j<n;j++)
{
prod[j]=1;
for (i=0;i<n;i++)
{
if(i==j)
continue;
else
prod[j]=prod[j]*a[i];
}
One more solution, Using division. with twice traversal.
Multiply all the elements and then start dividing it by each element.
{-
Recursive solution using sqrt(n) subsets. Runs in O(n).
Recursively computes the solution on sqrt(n) subsets of size sqrt(n).
Then recurses on the product sum of each subset.
Then for each element in each subset, it computes the product with
the product sum of all other products.
Then flattens all subsets.
Recurrence on the run time is T(n) = sqrt(n)*T(sqrt(n)) + T(sqrt(n)) + n
Suppose that T(n) ≤ cn in O(n).
T(n) = sqrt(n)*T(sqrt(n)) + T(sqrt(n)) + n
≤ sqrt(n)*c*sqrt(n) + c*sqrt(n) + n
≤ c*n + c*sqrt(n) + n
≤ (2c+1)*n
∈ O(n)
Note that ceiling(sqrt(n)) can be computed using a binary search
and O(logn) iterations, if the sqrt instruction is not permitted.
-}
otherProducts [] = []
otherProducts [x] = [1]
otherProducts [x,y] = [y,x]
otherProducts a = foldl' (++) [] $ zipWith (\s p -> map (*p) s) solvedSubsets subsetOtherProducts
where
n = length a
-- Subset size. Require that 1 < s < n.
s = ceiling $ sqrt $ fromIntegral n
solvedSubsets = map otherProducts subsets
subsetOtherProducts = otherProducts $ map product subsets
subsets = reverse $ loop a []
where loop [] acc = acc
loop a acc = loop (drop s a) ((take s a):acc)
Here is my code:
int multiply(int a[],int n,int nextproduct,int i)
{
int prevproduct=1;
if(i>=n)
return prevproduct;
prevproduct=multiply(a,n,nextproduct*a[i],i+1);
printf(" i=%d > %d\n",i,prevproduct*nextproduct);
return prevproduct*a[i];
}
int main()
{
int a[]={2,4,1,3,5};
multiply(a,5,1,0);
return 0;
}
Here's a slightly functional example, using C#:
Func<long>[] backwards = new Func<long>[input.Length];
Func<long>[] forwards = new Func<long>[input.Length];
for (int i = 0; i < input.Length; ++i)
{
var localIndex = i;
backwards[i] = () => (localIndex > 0 ? backwards[localIndex - 1]() : 1) * input[localIndex];
forwards[i] = () => (localIndex < input.Length - 1 ? forwards[localIndex + 1]() : 1) * input[localIndex];
}
var output = new long[input.Length];
for (int i = 0; i < input.Length; ++i)
{
if (0 == i)
{
output[i] = forwards[i + 1]();
}
else if (input.Length - 1 == i)
{
output[i] = backwards[i - 1]();
}
else
{
output[i] = forwards[i + 1]() * backwards[i - 1]();
}
}
I'm not entirely certain that this is O(n), due to the semi-recursion of the created Funcs, but my tests seem to indicate that it's O(n) in time.
To be complete here is the code in Scala:
val list1 = List(1, 2, 3, 4, 5)
for (elem <- list1) println(list1.filter(_ != elem) reduceLeft(_*_))
This will print out the following:
120
60
40
30
24
The program will filter out the current elem (_ != elem); and multiply the new list with reduceLeft method. I think this will be O(n) if you use scala view or Iterator for lazy eval.
// This is the recursive solution in Java
// Called as following from main product(a,1,0);
public static double product(double[] a, double fwdprod, int index){
double revprod = 1;
if (index < a.length){
revprod = product2(a, fwdprod*a[index], index+1);
double cur = a[index];
a[index] = fwdprod * revprod;
revprod *= cur;
}
return revprod;
}
A neat solution with O(n) runtime:
For each element calculate the product of all the elements that occur before that and it store in an array "pre".
For each element calculate the product of all the elements that occur after that element and store it in an array "post"
Create a final array "result", for an element i,
result[i] = pre[i-1]*post[i+1];
Here is the ptyhon version
# This solution use O(n) time and O(n) space
def productExceptSelf(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
N = len(nums)
if N == 0: return
# Initialzie list of 1, size N
l_prods, r_prods = [1]*N, [1]*N
for i in range(1, N):
l_prods[i] = l_prods[i-1] * nums[i-1]
for i in reversed(range(N-1)):
r_prods[i] = r_prods[i+1] * nums[i+1]
result = [x*y for x,y in zip(l_prods,r_prods)]
return result
# This solution use O(n) time and O(1) space
def productExceptSelfSpaceOptimized(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
N = len(nums)
if N == 0: return
# Initialzie list of 1, size N
result = [1]*N
for i in range(1, N):
result[i] = result[i-1] * nums[i-1]
r_prod = 1
for i in reversed(range(N)):
result[i] *= r_prod
r_prod *= nums[i]
return result
I'm use to C#:
public int[] ProductExceptSelf(int[] nums)
{
int[] returnArray = new int[nums.Length];
List<int> auxList = new List<int>();
int multTotal = 0;
// If no zeros are contained in the array you only have to calculate it once
if(!nums.Contains(0))
{
multTotal = nums.ToList().Aggregate((a, b) => a * b);
for (int i = 0; i < nums.Length; i++)
{
returnArray[i] = multTotal / nums[i];
}
}
else
{
for (int i = 0; i < nums.Length; i++)
{
auxList = nums.ToList();
auxList.RemoveAt(i);
if (!auxList.Contains(0))
{
returnArray[i] = auxList.Aggregate((a, b) => a * b);
}
else
{
returnArray[i] = 0;
}
}
}
return returnArray;
}
Here is simple Scala version in Linear O(n) time:
def getProductEff(in:Seq[Int]):Seq[Int] = {
//create a list which has product of every element to the left of this element
val fromLeft = in.foldLeft((1, Seq.empty[Int]))((ac, i) => (i * ac._1, ac._2 :+ ac._1))._2
//create a list which has product of every element to the right of this element, which is the same as the previous step but in reverse
val fromRight = in.reverse.foldLeft((1,Seq.empty[Int]))((ac,i) => (i * ac._1,ac._2 :+ ac._1))._2.reverse
//merge the two list by product at index
in.indices.map(i => fromLeft(i) * fromRight(i))
}
This works because essentially the answer is an array which has product of all elements to the left and to the right.
import java.util.Arrays;
public class Pratik
{
public static void main(String[] args)
{
int[] array = {2, 3, 4, 5, 6}; // OUTPUT: 360 240 180 144 120
int[] products = new int[array.length];
arrayProduct(array, products);
System.out.println(Arrays.toString(products));
}
public static void arrayProduct(int array[], int products[])
{
double sum = 0, EPSILON = 1e-9;
for(int i = 0; i < array.length; i++)
sum += Math.log(array[i]);
for(int i = 0; i < array.length; i++)
products[i] = (int) (EPSILON + Math.exp(sum - Math.log(array[i])));
}
}
OUTPUT:
[360, 240, 180, 144, 120]
Time complexity : O(n)
Space complexity: O(1)
Given an array A of N integers we draw N discs in a 2D plane, such that i-th disc has center in (0,i) and a radius A[i]. We say that k-th disc and j-th disc intersect, if k-th and j-th discs have at least one common point.
Write a function
int number_of_disc_intersections(int[] A);
which given an array A describing N discs as explained above, returns the number of pairs of intersecting discs. For example, given N=6 and
A[0] = 1
A[1] = 5
A[2] = 2
A[3] = 1
A[4] = 4
A[5] = 0
there are 11 pairs of intersecting discs:
0th and 1st
0th and 2nd
0th and 4th
1st and 2nd
1st and 3rd
1st and 4th
1st and 5th
2nd and 3rd
2nd and 4th
3rd and 4th
4th and 5th
so the function should return 11.
The function should return -1 if the number of intersecting pairs exceeds 10,000,000. The function may assume that N does not exceed 10,000,000.
O(N) complexity and O(N) memory solution.
private static int Intersections(int[] a)
{
int result = 0;
int[] dps = new int[a.length];
int[] dpe = new int[a.length];
for (int i = 0, t = a.length - 1; i < a.length; i++)
{
int s = i > a[i]? i - a[i]: 0;
int e = t - i > a[i]? i + a[i]: t;
dps[s]++;
dpe[e]++;
}
int t = 0;
for (int i = 0; i < a.length; i++)
{
if (dps[i] > 0)
{
result += t * dps[i];
result += dps[i] * (dps[i] - 1) / 2;
if (10000000 < result) return -1;
t += dps[i];
}
t -= dpe[i];
}
return result;
}
So you want to find the number of intersections of the intervals [i-A[i], i+A[i]].
Maintain a sorted array (call it X) containing the i-A[i] (also have some extra space which has the value i+A[i] in there).
Now walk the array X, starting at the leftmost interval (i.e smallest i-A[i]).
For the current interval, do a binary search to see where the right end point of the interval (i.e. i+A[i]) will go (called the rank). Now you know that it intersects all the elements to the left.
Increment a counter with the rank and subtract current position (assuming one indexed) as we don't want to double count intervals and self intersections.
O(nlogn) time, O(n) space.
Python 100 / 100 (tested) on codility, with O(nlogn) time and O(n) space.
Here is #noisyboiler's python implementation of #Aryabhatta's method with comments and an example.
Full credit to original authors, any errors / poor wording are entirely my fault.
from bisect import bisect_right
def number_of_disc_intersections(A):
pairs = 0
# create an array of tuples, each containing the start and end indices of a disk
# some indices may be less than 0 or greater than len(A), this is fine!
# sort the array by the first entry of each tuple: the disk start indices
intervals = sorted( [(i-A[i], i+A[i]) for i in range(len(A))] )
# create an array of starting indices using tuples in intervals
starts = [i[0] for i in intervals]
# for each disk in order of the *starting* position of the disk, not the centre
for i in range(len(starts)):
# find the end position of that disk from the array of tuples
disk_end = intervals[i][1]
# find the index of the rightmost value less than or equal to the interval-end
# this finds the number of disks that have started before disk i ends
count = bisect_right(starts, disk_end )
# subtract current position to exclude previous matches
# this bit seemed 'magic' to me, so I think of it like this...
# for disk i, i disks that start to the left have already been dealt with
# subtract i from count to prevent double counting
# subtract one more to prevent counting the disk itsself
count -= (i+1)
pairs += count
if pairs > 10000000:
return -1
return pairs
Worked example: given [3, 0, 1, 6] the disk radii would look like this:
disk0 ------- start= -3, end= 3
disk1 . start= 1, end= 1
disk2 --- start= 1, end= 3
disk3 ------------- start= -3, end= 9
index 3210123456789 (digits left of zero are -ve)
intervals = [(-3, 3), (-3, 9), (1, 1), (1,3)]
starts = [-3, -3, 1, 1]
the loop order will be: disk0, disk3, disk1, disk2
0th loop:
by the end of disk0, 4 disks have started
one of which is disk0 itself
none of which could have already been counted
so add 3
1st loop:
by the end of disk3, 4 disks have started
one of which is disk3 itself
one of which has already started to the left so is either counted OR would not overlap
so add 2
2nd loop:
by the end of disk1, 4 disks have started
one of which is disk1 itself
two of which have already started to the left so are either counted OR would not overlap
so add 1
3rd loop:
by the end of disk2, 4 disks have started
one of which is disk2 itself
two of which have already started to the left so are either counted OR would not overlap
so add 0
pairs = 6
to check: these are (0,1), (0,2), (0,2), (1,2), (1,3), (2,3),
Well, I adapted Falk Hüffner's idea to c++, and made a change in the range.
Opposite to what is written above, there is no need to go beyond the scope of the array (no matter how large are the values in it).
On Codility this code received 100%.
Thank you Falk for your great idea!
int number_of_disc_intersections ( const vector<int> &A ) {
int sum=0;
vector<int> start(A.size(),0);
vector<int> end(A.size(),0);
for (unsigned int i=0;i<A.size();i++){
if ((int)i<A[i]) start[0]++;
else start[i-A[i]]++;
if (i+A[i]>=A.size()) end[A.size()-1]++;
else end[i+A[i]]++;
}
int active=0;
for (unsigned int i=0;i<A.size();i++){
sum+=active*start[i]+(start[i]*(start[i]-1))/2;
if (sum>10000000) return -1;
active+=start[i]-end[i];
}
return sum;
}
This can even be done in linear time [EDIT: this is not linear time, see comments]. In fact, it becomes easier if you ignore the fact that there is exactly one interval centered at each point, and just treat it as a set of start- and endpoints of intervals. You can then just scan it from the left (Python code for simplicity):
from collections import defaultdict
a = [1, 5, 2, 1, 4, 0]
start = defaultdict(int)
stop = defaultdict(int)
for i in range(len(a)):
start[i - a[i]] += 1
stop[i + a[i]] += 1
active = 0
intersections = 0
for i in range(-len(a), len(a)):
intersections += active * start[i] + (start[i] * (start[i] - 1)) / 2
active += start[i]
active -= stop[i]
print intersections
Here's a O(N) time, O(N) space algorithm requiring 3 runs across the array and no sorting, verified scoring 100%:
You're interested in pairs of discs. Each pair involves one side of one disc and the other side of the other disc. Therefore we won't have duplicate pairs if we handle one side of each disc. Let's call the sides right and left (I rotated the space while thinking about it).
An overlap is either due to a right side overlapping another disc directly at the center (so pairs equal to the radius with some care about the array length) or due to the number of left sides existing at the rightmost edge.
So we create an array that contains the number of left sides at each point and then it's a simple sum.
C code:
int solution(int A[], int N) {
int C[N];
int a, S=0, t=0;
// Mark left and middle of disks
for (int i=0; i<N; i++) {
C[i] = -1;
a = A[i];
if (a>=i) {
C[0]++;
} else {
C[i-a]++;
}
}
// Sum of left side of disks at location
for (int i=0; i<N; i++) {
t += C[i];
C[i] = t;
}
// Count pairs, right side only:
// 1. overlaps based on disk size
// 2. overlaps based on disks but not centers
for (int i=0; i<N; i++) {
a = A[i];
S += ((a<N-i) ? a: N-i-1);
if (i != N-1) {
S += C[((a<N-i) ? i+a: N-1)];
}
if (S>10000000) return -1;
}
return S;
}
I got 100 out of 100 with this C++ implementation:
#include <map>
#include <algorithm>
inline bool mySortFunction(pair<int,int> p1, pair<int,int> p2)
{
return ( p1.first < p2.first );
}
int number_of_disc_intersections ( const vector<int> &A ) {
int i, size = A.size();
if ( size <= 1 ) return 0;
// Compute lower boundary of all discs and sort them in ascending order
vector< pair<int,int> > lowBounds(size);
for(i=0; i<size; i++) lowBounds[i] = pair<int,int>(i-A[i],i+A[i]);
sort(lowBounds.begin(), lowBounds.end(), mySortFunction);
// Browse discs
int nbIntersect = 0;
for(i=0; i<size; i++)
{
int curBound = lowBounds[i].second;
for(int j=i+1; j<size && lowBounds[j].first<=curBound; j++)
{
nbIntersect++;
// Maximal number of intersections
if ( nbIntersect > 10000000 ) return -1;
}
}
return nbIntersect;
}
A Python answer
from bisect import bisect_right
def number_of_disc_intersections(li):
pairs = 0
# treat as a series of intervals on the y axis at x=0
intervals = sorted( [(i-li[i], i+li[i]) for i in range(len(li))] )
# do this by creating a list of start points of each interval
starts = [i[0] for i in intervals]
for i in range(len(starts)):
# find the index of the rightmost value less than or equal to the interval-end
count = bisect_right(starts, intervals[i][1])
# subtract current position to exclude previous matches, and subtract self
count -= (i+1)
pairs += count
if pairs > 10000000:
return -1
return pairs
100/100 c#
class Solution
{
class Interval
{
public long Left;
public long Right;
}
public int solution(int[] A)
{
if (A == null || A.Length < 1)
{
return 0;
}
var itervals = new Interval[A.Length];
for (int i = 0; i < A.Length; i++)
{
// use long to avoid data overflow (eg. int.MaxValue + 1)
long radius = A[i];
itervals[i] = new Interval()
{
Left = i - radius,
Right = i + radius
};
}
itervals = itervals.OrderBy(i => i.Left).ToArray();
int result = 0;
for (int i = 0; i < itervals.Length; i++)
{
var right = itervals[i].Right;
for (int j = i + 1; j < itervals.Length && itervals[j].Left <= right; j++)
{
result++;
if (result > 10000000)
{
return -1;
}
}
}
return result;
}
}
I'm offering one more solution because I did not find the counting principle of the previous solutions easy to follow. Though the results are the same, an explanation and more intuitive counting procedure seems worth presenting.
To begin, start by considering the O(N^2) solution that iterates over the discs in order of their center points, and counts the number of discs centered to the right of the current disc's that intersect the current disc, using the condition current_center + radius >= other_center - radius. Notice that we could get the same result counting discs centered to the left of the current disc using the condition current_center - radius <= other_center + radius.
def simple(A):
"""O(N^2) solution for validating more efficient solution."""
N = len(A)
unique_intersections = 0
# Iterate over discs in order of their center positions
for j in range(N):
# Iterate over discs whose center is to the right, to avoid double-counting.
for k in range(j+1, N):
# Increment cases where edge of current disk is at or right of the left edge of another disk.
if j + A[j] >= k - A[k]:
unique_intersections += 1
# Stop early if we have enough intersections.
# BUT: if the discs are small we still N^2 compare them all and time out.
if unique_intersections > 10000000:
return -1
return unique_intersections
We can go from O(N^2) to O(N) if we could only "look up" the number of discs to the right (or to the left!) that intersect the current disc. The key insight is to reinterpret the intersection condition as "the right edge of one disc overlaps the left edge of another disc", meaning (a ha!) the centers don't matter, only the edges.
The next insight is to try sorting the edges, taking O(N log N) time. Given a sorted array of the left edges and a sorted array of the right edges, as we scan our way from left to right along the number line, the number of left or right edges to the left of the current location point is simply the current index into left_edges and right_edges respectively: a constant-time deduction.
Finally, we use the "right edge > left edge" condition to deduce that the number of intersections between the current disc and discs that start only to the left of the current disc (to avoid duplicates) is the number of left edges to the left of the current edge, minus the number of right edges to the left of the current edge. That is, the number of discs starting to left of this one, minus the ones that closed already.
Now for this code, tested 100% on Codility:
def solution(A):
"""O(N log N) due to sorting, with O(N) pass over sorted arrays"""
N = len(A)
# Left edges of the discs, in increasing order of position.
left_edges = sorted([(p-r) for (p,r) in enumerate(A)])
# Right edges of the discs, in increasing order of position.
right_edges = sorted([(p+r) for (p,r) in enumerate(A)])
#print("left edges:", left_edges[:10])
#print("right edges:", right_edges[:10])
intersections = 0
right_i = 0
# Iterate over the discs in order of their leftmost edge position.
for left_i in range(N):
# Find the first right_edge that's right of or equal to the current left_edge, naively:
# right_i = bisect.bisect_left(right_edges, left_edges[left_i])
# Just scan from previous index until right edge is at or beyond current left:
while right_edges[right_i] < left_edges[left_i]:
right_i += 1
# Count number of discs starting left of current, minus the ones that already closed.
intersections += left_i - right_i
# Return early if we find more than 10 million intersections.
if intersections > 10000000:
return -1
#print("correct:", simple(A))
return intersections
Java 2*100%.
result is declared as long for a case codility doesn't test, namely 50k*50k intersections at one point.
class Solution {
public int solution(int[] A) {
int[] westEnding = new int[A.length];
int[] eastEnding = new int[A.length];
for (int i=0; i<A.length; i++) {
if (i-A[i]>=0) eastEnding[i-A[i]]++; else eastEnding[0]++;
if ((long)i+A[i]<A.length) westEnding[i+A[i]]++; else westEnding[A.length-1]++;
}
long result = 0; //long to contain the case of 50k*50k. codility doesn't test for this.
int wests = 0;
int easts = 0;
for (int i=0; i<A.length; i++) {
int balance = easts*wests; //these are calculated elsewhere
wests++;
easts+=eastEnding[i];
result += (long) easts*wests - balance - 1; // 1 stands for the self-intersection
if (result>10000000) return -1;
easts--;
wests-= westEnding[i];
}
return (int) result;
}
}
Swift 4 Solution 100% (Codility do not check the worst case for this solution)
public func solution(_ A : inout [Int]) -> Int {
// write your code in Swift 4.2.1 (Linux)
var count = 0
let sortedA = A.sorted(by: >)
if sortedA.isEmpty{ return 0 }
let maxVal = sortedA[0]
for i in 0..<A.count{
let maxIndex = min(i + A[i] + maxVal + 1,A.count)
for j in i + 1..<maxIndex{
if j - A[j] <= i + A[i]{
count += 1
}
}
if count > 10_000_000{
return -1
}
}
return count
}
Here my JavaScript solution, based in other solutions in this thread but implemented in other languages.
function solution(A) {
let circleEndpoints = [];
for(const [index, num] of Object.entries(A)) {
circleEndpoints.push([parseInt(index)-num, true]);
circleEndpoints.push([parseInt(index)+num, false]);
}
circleEndpoints = circleEndpoints.sort(([a, openA], [b, openB]) => {
if(a == b) return openA ? -1 : 1;
return a - b;
});
let openCircles = 0;
let intersections = 0;
for(const [endpoint, opening] of circleEndpoints) {
if(opening) {
intersections += openCircles;
openCircles ++;
} else {
openCircles --;
}
if(intersections > 10000000) return -1;
}
return intersections;
}
count = 0
for (int i = 0; i < N; i++) {
for (int j = i+1; j < N; j++) {
if (i + A[i] >= j - A[j]) count++;
}
}
It is O(N^2) so pretty slow, but it works.
This is a ruby solution that scored 100/100 on codility. I'm posting it now because I'm finding it difficult to follow the already posted ruby answer.
def solution(a)
end_points = []
a.each_with_index do |ai, i|
end_points << [i - ai, i + ai]
end
end_points = end_points.sort_by { |points| points[0]}
intersecting_pairs = 0
end_points.each_with_index do |point, index|
lep, hep = point
pairs = bsearch(end_points, index, end_points.size - 1, hep)
return -1 if 10000000 - pairs + index < intersecting_pairs
intersecting_pairs += (pairs - index)
end
return intersecting_pairs
end
# This method returns the maximally appropriate position
# where the higher end-point may have been inserted.
def bsearch(a, l, u, x)
if l == u
if x >= a[u][0]
return u
else
return l - 1
end
end
mid = (l + u)/2
# Notice that we are searching in higher range
# even if we have found equality.
if a[mid][0] <= x
return bsearch(a, mid+1, u, x)
else
return bsearch(a, l, mid, x)
end
end
Probably extremely fast. O(N). But you need to check it out. 100% on Codility.
Main idea:
1. At any point of the table, there are number of circles "opened" till the right edge of the circle, lets say "o".
2. So there are (o-1-used) possible pairs for the circle in that point. "used" means circle that have been processed and pairs for them counted.
public int solution(int[] A) {
final int N = A.length;
final int M = N + 2;
int[] left = new int[M]; // values of nb of "left" edges of the circles in that point
int[] sleft = new int[M]; // prefix sum of left[]
int il, ir; // index of the "left" and of the "right" edge of the circle
for (int i = 0; i < N; i++) { // counting left edges
il = tl(i, A);
left[il]++;
}
sleft[0] = left[0];
for (int i = 1; i < M; i++) {// counting prefix sums for future use
sleft[i]=sleft[i-1]+left[i];
}
int o, pairs, total_p = 0, total_used=0;
for (int i = 0; i < N; i++) { // counting pairs
ir = tr(i, A, M);
o = sleft[ir]; // nb of open till right edge
pairs = o -1 - total_used;
total_used++;
total_p += pairs;
}
if(total_p > 10000000){
total_p = -1;
}
return total_p;
}
private int tl(int i, int[] A){
int tl = i - A[i]; // index of "begin" of the circle
if (tl < 0) {
tl = 0;
} else {
tl = i - A[i] + 1;
}
return tl;
}
int tr(int i, int[] A, int M){
int tr; // index of "end" of the circle
if (Integer.MAX_VALUE - i < A[i] || i + A[i] >= M - 1) {
tr = M - 1;
} else {
tr = i + A[i] + 1;
}
return tr;
}
There are a lot of great answers here already, including the great explanation from the accepted answer. However, I wanted to point out a small observation about implementation details in the Python language.
Originally, I've came up with the solution shown below. I was expecting to get O(N*log(N)) time complexity as soon as we have a single for-loop with N iterations, and each iteration performs a binary search that takes at most log(N).
def solution(a):
import bisect
if len(a) <= 1:
return 0
cuts = [(c - r, c + r) for c, r in enumerate(a)]
cuts.sort(key=lambda pair: pair[0])
lefts, rights = zip(*cuts)
n = len(cuts)
total = 0
for i in range(n):
r = rights[i]
pos = bisect.bisect_right(lefts[i+1:], r)
total += pos
if total > 10e6:
return -1
return total
However, I've get O(N**2) and a timeout failure. Do you see what is wrong here? Right, this line:
pos = bisect.bisect_right(lefts[i+1:], r)
In this line, you are actually taking a copy of the original list to pass it into binary search function, and it totally ruins the efficiency of the proposed solution! It makes your code just a bit more consice (i.e., you don't need to write pos - i - 1) but heavily undermies the performance. So, as it was shown above, the solution should be:
def solution(a):
import bisect
if len(a) <= 1:
return 0
cuts = [(c - r, c + r) for c, r in enumerate(a)]
cuts.sort(key=lambda pair: pair[0])
lefts, rights = zip(*cuts)
n = len(cuts)
total = 0
for i in range(n):
r = rights[i]
pos = bisect.bisect_right(lefts, r)
total += (pos - i - 1)
if total > 10e6:
return -1
return total
It seems that sometimes one could be too eager about making slices and copies because Python allows you to do it so easily :) Probably not a great insight, but for me it was a good lesson to pay more attention to these "technical" moments when converting ideas and algorithms into real-word solutions.
I know that this is an old questions but it is still active on codility.
private int solution(int[] A)
{
int openedCircles = 0;
int intersectCount = 0;
We need circles with their start and end values. For that purpose I have used Tuple.
True/False indicates if we are adding Circle Starting or Circle Ending value.
List<Tuple<decimal, bool>> circles = new List<Tuple<decimal, bool>>();
for(int i = 0; i < A.Length; i ++)
{
// Circle start value
circles.Add(new Tuple<decimal, bool>((decimal)i - (decimal)A[i], true));
// Circle end value
circles.Add(new Tuple<decimal, bool>((decimal)i + (decimal)A[i], false));
}
Order "circles" by their values.
If one circle is ending at same value where other circle is starting, it should be counted as intersect (because of that "opening" should be in front of "closing" if in same point)
circles = circles.OrderBy(x => x.Item1).ThenByDescending(x => x.Item2).ToList();
Counting and returning counter
foreach (var circle in circles)
{
// We are opening new circle (within existing circles)
if(circle.Item2 == true)
{
intersectCount += openedCircles;
if (intersectCount > 10000000)
{
return -1;
}
openedCircles++;
}
else
{
// We are closing circle
openedCircles--;
}
}
return intersectCount;
}
Javascript solution 100/100 based on this video https://www.youtube.com/watch?v=HV8tzIiidSw
function sortArray(A) {
return A.sort((a, b) => a - b)
}
function getDiskPoints(A) {
const diskStarPoint = []
const diskEndPoint = []
for(i = 0; i < A.length; i++) {
diskStarPoint.push(i - A[i])
diskEndPoint.push(i + A[i])
}
return {
diskStarPoint: sortArray(diskStarPoint),
diskEndPoint: sortArray(diskEndPoint)
};
}
function solution(A) {
const { diskStarPoint, diskEndPoint } = getDiskPoints(A)
let index = 0;
let openDisks = 0;
let intersections = 0;
for(i = 0; i < diskStarPoint.length; i++) {
while(diskStarPoint[i] > diskEndPoint[index]) {
openDisks--
index++
}
intersections += openDisks
openDisks++
}
return intersections > 10000000 ? -1 : intersections
}
so, I was doing this test in Scala and I would like to share here my example. My idea to solve is:
Extract the limits to the left and right of each position on the array.
A[0] = 1 --> (0-1, 0+1) = A0(-1, 1)
A[1] = 5 --> (1-5, 1+5) = A1(-4, 6)
A[2] = 2 --> (2-2, 2+2) = A2(0, 4)
A[3] = 1 --> (3-1, 3+1) = A3(2, 4)
A[4] = 4 --> (4-4, 4+4) = A4(0, 8)
A[5] = 0 --> (5-0, 5+0) = A5(5, 5)
Check if there is intersections between any two positions
(A0_0 >= A1_0 AND A0_0 <= A1_1) OR // intersection
(A0_1 >= A1_0 AND A0_1 <= A1_1) OR // intersection
(A0_0 <= A1_0 AND A0_1 >= A1_1) // one circle contain inside the other
if any of these two checks is true count one intersection.
object NumberOfDiscIntersections {
def solution(a: Array[Int]): Int = {
var count: Long = 0
for (posI: Long <- 0L until a.size) {
for (posJ <- (posI + 1) until a.size) {
val tupleI = (posI - a(posI.toInt), posI + a(posI.toInt))
val tupleJ = (posJ - a(posJ.toInt), posJ + a(posJ.toInt))
if ((tupleI._1 >= tupleJ._1 && tupleI._1 <= tupleJ._2) ||
(tupleI._2 >= tupleJ._1 && tupleI._2 <= tupleJ._2) ||
(tupleI._1 <= tupleJ._1 && tupleI._2 >= tupleJ._2)) {
count += 1
}
}
}
count.toInt
}
}
This got 100/100 in c#
class CodilityDemo3
{
public static int GetIntersections(int[] A)
{
if (A == null)
{
return 0;
}
int size = A.Length;
if (size <= 1)
{
return 0;
}
List<Line> lines = new List<Line>();
for (int i = 0; i < size; i++)
{
if (A[i] >= 0)
{
lines.Add(new Line(i - A[i], i + A[i]));
}
}
lines.Sort(Line.CompareLines);
size = lines.Count;
int intersects = 0;
for (int i = 0; i < size; i++)
{
Line ln1 = lines[i];
for (int j = i + 1; j < size; j++)
{
Line ln2 = lines[j];
if (ln2.YStart <= ln1.YEnd)
{
intersects += 1;
if (intersects > 10000000)
{
return -1;
}
}
else
{
break;
}
}
}
return intersects;
}
}
public class Line
{
public Line(double ystart, double yend)
{
YStart = ystart;
YEnd = yend;
}
public double YStart { get; set; }
public double YEnd { get; set; }
public static int CompareLines(Line line1, Line line2)
{
return (line1.YStart.CompareTo(line2.YStart));
}
}
}
Thanks to Falk for the great idea! Here is a ruby implementation that takes advantage of sparseness.
def int(a)
event = Hash.new{|h,k| h[k] = {:start => 0, :stop => 0}}
a.each_index {|i|
event[i - a[i]][:start] += 1
event[i + a[i]][:stop ] += 1
}
sorted_events = (event.sort_by {|index, value| index}).map! {|n| n[1]}
past_start = 0
intersect = 0
sorted_events.each {|e|
intersect += e[:start] * (e[:start]-1) / 2 +
e[:start] * past_start
past_start += e[:start]
past_start -= e[:stop]
}
return intersect
end
puts int [1,1]
puts int [1,5,2,1,4,0]
#include <stdio.h>
#include <stdlib.h>
void sortPairs(int bounds[], int len){
int i,j, temp;
for(i=0;i<(len-1);i++){
for(j=i+1;j<len;j++){
if(bounds[i] > bounds[j]){
temp = bounds[i];
bounds[i] = bounds[j];
bounds[j] = temp;
temp = bounds[i+len];
bounds[i+len] = bounds[j+len];
bounds[j+len] = temp;
}
}
}
}
int adjacentPointPairsCount(int a[], int len){
int count=0,i,j;
int *bounds;
if(len<2) {
goto toend;
}
bounds = malloc(sizeof(int)*len *2);
for(i=0; i< len; i++){
bounds[i] = i-a[i];
bounds[i+len] = i+a[i];
}
sortPairs(bounds, len);
for(i=0;i<len;i++){
int currentBound = bounds[i+len];
for(j=i+1;a[j]<=currentBound;j++){
if(count>100000){
count=-1;
goto toend;
}
count++;
}
}
toend:
free(bounds);
return count;
}
An Implementation of Idea stated above in Java:
public class DiscIntersectionCount {
public int number_of_disc_intersections(int[] A) {
int[] leftPoints = new int[A.length];
for (int i = 0; i < A.length; i++) {
leftPoints[i] = i - A[i];
}
Arrays.sort(leftPoints);
// System.out.println(Arrays.toString(leftPoints));
int count = 0;
for (int i = 0; i < A.length - 1; i++) {
int rpoint = A[i] + i;
int rrank = getRank(leftPoints, rpoint);
//if disk has sifnificant radius, exclude own self
if (rpoint > i) rrank -= 1;
int rank = rrank;
// System.out.println(rpoint+" : "+rank);
rank -= i;
count += rank;
}
return count;
}
public int getRank(int A[], int num) {
if (A==null || A.length == 0) return -1;
int mid = A.length/2;
while ((mid >= 0) && (mid < A.length)) {
if (A[mid] == num) return mid;
if ((mid == 0) && (A[mid] > num)) return -1;
if ((mid == (A.length - 1)) && (A[mid] < num)) return A.length;
if (A[mid] < num && A[mid + 1] >= num) return mid + 1;
if (A[mid] > num && A[mid - 1] <= num) return mid - 1;
if (A[mid] < num) mid = (mid + A.length)/2;
else mid = (mid)/2;
}
return -1;
}
public static void main(String[] args) {
DiscIntersectionCount d = new DiscIntersectionCount();
int[] A =
//{1,5,2,1,4,0}
//{0,0,0,0,0,0}
// {1,1,2}
{3}
;
int count = d.number_of_disc_intersections(A);
System.out.println(count);
}
}
Here is the PHP code that scored 100 on codility:
$sum=0;
//One way of cloning the A:
$start = array();
$end = array();
foreach ($A as $key=>$value)
{
$start[]=0;
$end[]=0;
}
for ($i=0; $i<count($A); $i++)
{
if ($i<$A[$i])
$start[0]++;
else
$start[$i-$A[$i]]++;
if ($i+$A[$i] >= count($A))
$end[count($A)-1]++;
else
$end[$i+$A[$i]]++;
}
$active=0;
for ($i=0; $i<count($A);$i++)
{
$sum += $active*$start[$i]+($start[$i]*($start[$i]-1))/2;
if ($sum>10000000) return -1;
$active += $start[$i]-$end[$i];
}
return $sum;
However I dont understand the logic. This is just transformed C++ code from above. Folks, can you elaborate on what you were doing here, please?
A 100/100 C# implementation as described by Aryabhatta (the binary search solution).
using System;
class Solution {
public int solution(int[] A)
{
return IntersectingDiscs.Execute(A);
}
}
class IntersectingDiscs
{
public static int Execute(int[] data)
{
int counter = 0;
var intervals = Interval.GetIntervals(data);
Array.Sort(intervals); // sort by Left value
for (int i = 0; i < intervals.Length; i++)
{
counter += GetCoverage(intervals, i);
if(counter > 10000000)
{
return -1;
}
}
return counter;
}
private static int GetCoverage(Interval[] intervals, int i)
{
var currentInterval = intervals[i];
// search for an interval starting at currentInterval.Right
int j = Array.BinarySearch(intervals, new Interval { Left = currentInterval.Right });
if(j < 0)
{
// item not found
j = ~j; // bitwise complement (see Array.BinarySearch documentation)
// now j == index of the next item larger than the searched one
j = j - 1; // set index to the previous element
}
while(j + 1 < intervals.Length && intervals[j].Left == intervals[j + 1].Left)
{
j++; // get the rightmost interval starting from currentInterval.Righ
}
return j - i; // reduce already processed intervals (the left side from currentInterval)
}
}
class Interval : IComparable
{
public long Left { get; set; }
public long Right { get; set; }
// Implementation of IComparable interface
// which is used by Array.Sort().
public int CompareTo(object obj)
{
// elements will be sorted by Left value
var another = obj as Interval;
if (this.Left < another.Left)
{
return -1;
}
if (this.Left > another.Left)
{
return 1;
}
return 0;
}
/// <summary>
/// Transform array items into Intervals (eg. {1, 2, 4} -> {[-1,1], [-1,3], [-2,6]}).
/// </summary>
public static Interval[] GetIntervals(int[] data)
{
var intervals = new Interval[data.Length];
for (int i = 0; i < data.Length; i++)
{
// use long to avoid data overflow (eg. int.MaxValue + 1)
long radius = data[i];
intervals[i] = new Interval
{
Left = i - radius,
Right = i + radius
};
}
return intervals;
}
}
100% score in Codility.
Here is an adaptation to C# of Толя solution:
public int solution(int[] A)
{
long result = 0;
Dictionary<long, int> dps = new Dictionary<long, int>();
Dictionary<long, int> dpe = new Dictionary<long, int>();
for (int i = 0; i < A.Length; i++)
{
Inc(dps, Math.Max(0, i - A[i]));
Inc(dpe, Math.Min(A.Length - 1, i + A[i]));
}
long t = 0;
for (int i = 0; i < A.Length; i++)
{
int value;
if (dps.TryGetValue(i, out value))
{
result += t * value;
result += value * (value - 1) / 2;
t += value;
if (result > 10000000)
return -1;
}
dpe.TryGetValue(i, out value);
t -= value;
}
return (int)result;
}
private static void Inc(Dictionary<long, int> values, long index)
{
int value;
values.TryGetValue(index, out value);
values[index] = ++value;
}
Here's a two-pass C++ solution that doesn't require any libraries, binary searching, sorting, etc.
int solution(vector<int> &A) {
#define countmax 10000000
int count = 0;
// init lower edge array
vector<int> E(A.size());
for (int i = 0; i < (int) E.size(); i++)
E[i] = 0;
// first pass
// count all lower numbered discs inside this one
// mark lower edge of each disc
for (int i = 0; i < (int) A.size(); i++)
{
// if disc overlaps zero
if (i - A[i] <= 0)
count += i;
// doesn't overlap zero
else {
count += A[i];
E[i - A[i]]++;
}
if (count > countmax)
return -1;
}
// second pass
// count higher numbered discs with edge inside this one
for (int i = 0; i < (int) A.size(); i++)
{
// loop up inside this disc until top of vector
int jend = ((int) E.size() < (long long) i + A[i] + 1 ?
(int) E.size() : i + A[i] + 1);
// count all discs with edge inside this disc
// note: if higher disc is so big that edge is at or below
// this disc center, would count intersection in first pass
for (int j = i + 1; j < jend; j++)
count += E[j];
if (count > countmax)
return -1;
}
return count;
}
My answer in Swift; gets a 100% score.
import Glibc
struct Interval {
let start: Int
let end: Int
}
func bisectRight(intervals: [Interval], end: Int) -> Int {
var pos = -1
var startpos = 0
var endpos = intervals.count - 1
if intervals.count == 1 {
if intervals[0].start < end {
return 1
} else {
return 0
}
}
while true {
let currentLength = endpos - startpos
if currentLength == 1 {
pos = startpos
pos += 1
if intervals[pos].start <= end {
pos += 1
}
break
} else {
let middle = Int(ceil( Double((endpos - startpos)) / 2.0 ))
let middlepos = startpos + middle
if intervals[middlepos].start <= end {
startpos = middlepos
} else {
endpos = middlepos
}
}
}
return pos
}
public func solution(inout A: [Int]) -> Int {
let N = A.count
var nIntersections = 0
// Create array of intervals
var unsortedIntervals: [Interval] = []
for i in 0 ..< N {
let interval = Interval(start: i-A[i], end: i+A[i])
unsortedIntervals.append(interval)
}
// Sort array
let intervals = unsortedIntervals.sort {
$0.start < $1.start
}
for i in 0 ..< intervals.count {
let end = intervals[i].end
var count = bisectRight(intervals, end: end)
count -= (i + 1)
nIntersections += count
if nIntersections > Int(10E6) {
return -1
}
}
return nIntersections
}
C# solution 100/100
using System.Linq;
class Solution
{
private struct Interval
{
public Interval(long #from, long to)
{
From = #from;
To = to;
}
public long From { get; }
public long To { get; }
}
public int solution(int[] A)
{
int result = 0;
Interval[] intervals = A.Select((value, i) =>
{
long iL = i;
return new Interval(iL - value, iL + value);
})
.OrderBy(x => x.From)
.ToArray();
for (int i = 0; i < intervals.Length; i++)
{
for (int j = i + 1; j < intervals.Length && intervals[j].From <= intervals[i].To; j++)
result++;
if (result > 10000000)
return -1;
}
return result;
}
}