As a homework, I have to find the missing number from 0 to n using a divide and conquer (DAC) algorithm.
As an input, I get n-1 numbers from [0, n] and n.
I can easily do this with a quicksort and then just see which number is missing, but that would mean the complexity of my algorithm will be O(n*log n).
I'm wondering if there is any way I can do lower than that.
I was thinking that I might get the sum of the input (somehow) using DAC, and then the number missing will be n - sum. This would be O(n) complexity.
Is there any other way to get a complexity lower than O(n) (without using any space) and also, is my idea a good one? If not, can you give me other ideas for this problem, please?
Thanks.
Edit:
I know I should post another question, but I can post only once every 90 minutes (as I recall) and I want to finish this problem now if possible.
How can I calculate the sum of an array using DAC?
int DAC(int low, int high, int a[], int& s)
{
if (low <= high)
{
int pivot = (low + high)/2;
s += DAC(low, pivot - 1, a, s);
s += DAC(pivot+1, high, a, s);
return a[pivot];
}
}
for this call
cout << DAC(0, n-1, a, s);
Input:
7
1 2 3 4 5 6 7
I get 4 and I don't understand why. I didn't expect it to return only 4.
Edit 2:
I was getting call because I had to cout<<s, not DAC, I'm sorry.
Now I get 52 for the following code, with input: n=7, a=1 2 3 4 5 6 7
#include <iostream>
#include <algorithm>
using namespace std;
void citire(int& n, int a[])
{
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> a[i];
}
}
int DAC(int low, int high, int a[], int& s)
{
if (low <= high)
{
int pivot = (low + high)/2;
s += DAC(low, pivot - 1, a, s);
s += DAC(pivot+1, high, a, s);
return a[pivot];
}
}
int main() {
int a[100], n, s = 0;
citire(n, a);
DAC(0, n-1, a, s);
cout << s;
return 0;
}
As sis modified internally, the function DAC doesn't have to return anything.
#include <iostream>
#include <algorithm>
void citire(int& n, int a[])
{
std::cin >> n;
for (int i = 0; i < n; i++)
{
std::cin >> a[i];
}
}
void DAC(int low, int high, int a[], int& s)
{
if (low <= high)
{
int pivot = (low + high)/2;
DAC(low, pivot - 1, a, s);
DAC(pivot+1, high, a, s);
s += a[pivot];
}
}
int main() {
int a[100], n, s = 0;
citire(n, a);
DAC(0, n-1, a, s);
std::cout << s << "\n";
return 0;
}
But std:accumulatewould be much simpler
I search everywhere on the internet for the best case time complexity of selection sort that is o(n^2). But i write and tested this below code of selection sort that can work in O(n) for best case (that is array is already sorted). Please find the mistake in this program
This is my code:
#include <bits/stdc++.h>
using namespace std;
/* Function to print an array */
void printArray(int arr[], int size)
{
int i;
for (i = 0; i < size; i++)
cout << arr[i] << " ";
cout << endl;
}
void swap(int *xp, int *yp)
{
int temp = *xp;
*xp = *yp;
*yp = temp;
}
void selectionSort(int arr[], int n)
{
int i, j, max_idx;
// One by one move boundary of unsorted subarray
for (i = 0; i < n - 1; i++)
{
cout << endl;
printArray(arr, n);
// Find the minimum element in unsorted array
max_idx = 0;
int count = 0;
for (j = 1; j < n - i; j++)
{
if (arr[j] >= arr[max_idx])
{
max_idx = j;
count++;
}
}
if (count != n - i - 1)
{ //swap only if not already sorted
// Swap the found minimum element with the first element
swap(&arr[max_idx], &arr[n - i - 1]);
}
else //already Sorted so returning
{
return;
}
//cout << "Sorted array: \n";
printArray(arr, n);
}
}
// Driver program to test above functions
int main()
{
int arr[] = {2, 1, 4, 3, 6, 5, 8, 7};
int n = sizeof(arr) / sizeof(arr[0]);
selectionSort(arr, n);
cout << "Sorted array: \n";
printArray(arr, n);
return 0;
}
// This is code is contributed by www.bhattiacademy.com
Yes, your algorithm has a best case running time of Θ(n), because if the array is already in ascending order then count will equal n - 1 on the first iteration of the outer loop, so the algorithm will terminate early.
Your algorithm is different to the standard selection sort algorithm, which looks like this:
for(int i = 0; i < n - 1; i++) {
int min_idx = i;
for(int j = i + 1; j < n; j++) {
if(arr[j] < arr[min_idx]) {
min_idx = j;
}
}
swap(&arr[i], &arr[min_idx]);
}
The selection sort algorithm iteratively searches for the minimum remaining element and swaps it into place. This doesn't create an opportunity to detect that the array is already in increasing order, so there's no opportunity to terminate early, and selection sort's best case running time is therefore Θ(n2).
Selection Sort: Idea Given an array of n items
1.Find the largest item x, in the range of [0…n−1]
2.Swap x with the (n−1)th item
3.Reduce n by 1 and go to Step 1
Selection sort function you can use following algorithm has hint to write the code:
Hello everyone i have a question. It's my task which one is below:
Let A[] be a natural numbers array of length N, which is partially sorted, i.e. there exists such index i(0 < i < N-1), that the subaray A[0],...,A[i] is incrementally sorted and also the subarray A[i+1],...,A[N] is incrementally sorted. Design the algorithm, which sorts the whole array A[] and works in place (so has space complexity O(1)) and the result must be stored in the same array A[]. Describe the algorithm, its correctness and its time complexity approximation.
For this question which approaching is better? Bubble sorting or Insertion sort? Or is there more effective solution? I prefered bubble sorting for this task but i am open to other opinions
static void bubbleSort(int arr[], int n)
{
int i, j, temp;
boolean swapped;
for (i = 0; i < n - 1; i++)
{
swapped = false;
for (j = 0; j < n - i - 1; j++)
{
if (arr[j] > arr[j + 1])
{
// swap arr[j] and arr[j+1]
temp = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = temp;
swapped = true;
}
}
if (swapped == false)
break;
}
}
static void printArray(int arr[], int size)
{
int i;
for (i = 0; i < size; i++)
System.out.print(arr[i] + " ");
System.out.println();
}
public static void main(String args[])
{
int arr[] = { 1, 8, 45, 12, 22, 11, 90 };
int n = arr.length;
bubbleSort(arr, n);
System.out.println("Sorted array: ");
printArray(arr, n);
}
}
Bubble sort algorithm complexity is O(n^2). Even using if (swapped == false) break; this will not help to reduce the complexity (try for {2,3,4,5,1}, you will find out).
Since there exists such index i(0 < i < N-1), that the subaray A[0],...,A[i] is incrementally sorted and also the subarray A[i+1],...,A[N] is incrementally sorted.This problem can be solve in O(n) run time complexity. If we can find the index i where A[0:i] and A[i+1:n] are sorted, then we can think this problem as merging two sorted array into one array which can be done in O(n) time. Algorithm is given below:
void sortPartialSortedArray(int arr[], int n)
{
int pos = 0;
// find the position for which arr[0:pos] and arr[pos+1:n] is sorted
for(int i=0; i+1<n; i++) {
if(arr[i]>arr[i+1]) {
pos = i;
}
}
int i = pos, j= n-1;
// sort it from last position
while(i>=0 && j>=0) {
if(arr[i] > arr[j]) {
swap(arr[i],arr[j]);
}
j--;
if(i==j) {
i--;
}
}
}
This question already has answers here:
Quick sort Worst case
(6 answers)
What is the worst case scenario for quicksort?
(6 answers)
Closed 4 years ago.
#include <iostream>
#include<stdio.h>
#include<fstream>
using namespace std;
void swap(int* a, int* b)
{
int t = *a;
*a = *b;
*b = t;
}
int partition (int arr[], int low, int high)
{
int pivot = arr[high];
int i = (low - 1);
for (int j = low; j <= high- 1; j++)
{
if (arr[j] <= pivot)
{
i++;
swap(&arr[i], &arr[j]);
}
}
swap(&arr[i + 1], &arr[high]);
return (i + 1);
}
void quickSort(int arr[], int low, int high)
{
if (low < high)
{
int pi = partition(arr, low, high);
quickSort(arr, low, pi - 1);
quickSort(arr, pi + 1, high);
}
}
int main()
{
int arr[100000];
int i;
ifstream fin;
int n = 20000;
fin.open("reverse20k.txt");
if(fin.is_open())
{
for(i=0;i<n;i++)
fin>>arr[i];
}
quickSort(arr, 0, n-1);
return 0;
}
It takes this about 1.25 seconds to sort a 20k purely descending array, while it takes merge sort only 0.05. Is quick sort just extremely inefficient when sorting descending arrays, or is there just something wrong with the algorithm?
Actually it is the problem #10 of chapter 8 of Programming Pearls 2nd edition. It asked two questions: given an array A[] of integers(positive and nonpositive), how can you find a continuous subarray of A[] whose sum is closest to 0? Or closest to a certain value t?
I can think of a way to solve the problem closest to 0. Calculate the prefix sum array S[], where S[i] = A[0]+A[1]+...+A[i]. And then sort this S according to the element value, along with its original index information kept, to find subarray sum closest to 0, just iterate the S array and do the diff of the two neighboring values and update the minimum absolute diff.
Question is, what is the best way so solve second problem? Closest to a certain value t? Can anyone give a code or at least an algorithm? (If anyone has better solution to closest to zero problem, answers are welcome too)
To solve this problem, you can build an interval-tree by your own,
or balanced binary search tree, or even beneficial from STL map, in O(nlogn).
Following is use STL map, with lower_bound().
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
int A[] = {10,20,30,30,20,10,10,20};
// return (i, j) s.t. A[i] + ... + A[j] is nearest to value c
pair<int, int> nearest_to_c(int c, int n, int A[]) {
map<int, int> bst;
bst[0] = -1;
// barriers
bst[-int(1e9)] = -2;
bst[int(1e9)] = n;
int sum = 0, start, end, ret = c;
for (int i=0; i<n; ++i) {
sum += A[i];
// it->first >= sum-c, and with the minimal value in bst
map<int, int>::iterator it = bst.lower_bound(sum - c);
int tmp = -(sum - c - it->first);
if (tmp < ret) {
ret = tmp;
start = it->second + 1;
end = i;
}
--it;
// it->first < sum-c, and with the maximal value in bst
tmp = sum - c - it->first;
if (tmp < ret) {
ret = tmp;
start = it->second + 1;
end = i;
}
bst[sum] = i;
}
return make_pair(start, end);
}
// demo
int main() {
int c;
cin >> c;
pair<int, int> ans = nearest_to_c(c, 8, A);
cout << ans.first << ' ' << ans.second << endl;
return 0;
}
You can adapt your method. Assuming you have an array S of prefix sums, as you wrote, and already sorted in increasing order of sum value. The key concept is to not only examine consecutive prefix sums, but instead use two pointers to indicate two positions in the array S. Written in a (slightly pythonic) pseudocode:
left = 0 # Initialize window of length 0 ...
right = 0 # ... at the beginning of the array
best = ∞ # Keep track of best solution so far
while right < length(S): # Iterate until window reaches the end of the array
diff = S[right] - S[left]
if diff < t: # Window is getting too small
if t - diff < best: # We have a new best subarray
best = t - diff
# remember left and right as well
right = right + 1 # Make window bigger
else: # Window getting too big
if diff - t < best # We have a new best subarray
best = diff - t
# remember left and right as well
left = left + 1 # Make window smaller
The complexity is bound by the sorting. The above search will take at most 2n=O(n) iterations of the loop, each with computation time bound by a constant. Note that the above code was conceived for positive t.
The code was conceived for positive elements in S, and positive t. If any negative integers crop up, you might end up with a situation where the original index of right is smaller than that of left. So you'd end up with a sub sequence sum of -t. You can check this condition in the if … < best checks, but if you only suppress such cases there, I believe that you might be missing some relevant cases. Bottom line is: take this idea, think it through, but you'll have to adapt it for negative numbers.
Note: I think that this is the same general idea which Boris Strandjev wanted to express in his solution. However, I found that solution somewhat hard to read and harder to understand, so I'm offering my own formulation of this.
Your solution for the 0 case seems ok to me. Here is my solution to the second case:
You again calculate the prefix sums and sort.
You initialize to indices start to 0 (first index in the sorted prefix array) end to last (last index of the prefix array)
you start iterating over start 0...last and for each you find the corresponding end - the last index in which the prefix sum is such that prefix[start] + prefix[end] > t. When you find that end the best solution for start is either prefix[start] + prefix[end] or prefix[start] + prefix[end - 1] (the latter taken only if end > 0)
the most important thing is that you do not search for end for each start from scratch - prefix[start] increases in value when iterating over all possible values for start, which means that in each iteration you are interested only in values <= the previous value of end.
you can stop iterating when start > end
you take the best of all values obtained for all start positions.
It can easily be proved that this will give you complexity of O(n logn) for the entire algorithm.
I found this question by accident. Although it's been a while, I just post it. O(nlogn) time, O(n) space algorithm. This is running Java code. Hope this help people.
import java.util.*;
public class FindSubarrayClosestToZero {
void findSubarrayClosestToZero(int[] A) {
int curSum = 0;
List<Pair> list = new ArrayList<Pair>();
// 1. create prefix array: curSum array
for(int i = 0; i < A.length; i++) {
curSum += A[i];
Pair pair = new Pair(curSum, i);
list.add(pair);
}
// 2. sort the prefix array by value
Collections.sort(list, valueComparator);
// printPairList(list);
System.out.println();
// 3. compute pair-wise value diff: Triple< diff, i, i+1>
List<Triple> tList = new ArrayList<Triple>();
for(int i=0; i < A.length-1; i++) {
Pair p1 = list.get(i);
Pair p2 = list.get(i+1);
int valueDiff = p2.value - p1.value;
Triple Triple = new Triple(valueDiff, p1.index, p2.index);
tList.add(Triple);
}
// printTripleList(tList);
System.out.println();
// 4. Sort by min diff
Collections.sort(tList, valueDiffComparator);
// printTripleList(tList);
Triple res = tList.get(0);
int startIndex = Math.min(res.index1 + 1, res.index2);
int endIndex = Math.max(res.index1 + 1, res.index2);
System.out.println("\n\nThe subarray whose sum is closest to 0 is: ");
for(int i= startIndex; i<=endIndex; i++) {
System.out.print(" " + A[i]);
}
}
class Pair {
int value;
int index;
public Pair(int value, int index) {
this.value = value;
this.index = index;
}
}
class Triple {
int valueDiff;
int index1;
int index2;
public Triple(int valueDiff, int index1, int index2) {
this.valueDiff = valueDiff;
this.index1 = index1;
this.index2 = index2;
}
}
public static Comparator<Pair> valueComparator = new Comparator<Pair>() {
public int compare(Pair p1, Pair p2) {
return p1.value - p2.value;
}
};
public static Comparator<Triple> valueDiffComparator = new Comparator<Triple>() {
public int compare(Triple t1, Triple t2) {
return t1.valueDiff - t2.valueDiff;
}
};
void printPairList(List<Pair> list) {
for(Pair pair : list) {
System.out.println("<" + pair.value + " : " + pair.index + ">");
}
}
void printTripleList(List<Triple> list) {
for(Triple t : list) {
System.out.println("<" + t.valueDiff + " : " + t.index1 + " , " + t.index2 + ">");
}
}
public static void main(String[] args) {
int A1[] = {8, -3, 2, 1, -4, 10, -5}; // -3, 2, 1
int A2[] = {-3, 2, 4, -6, -8, 10, 11}; // 2, 4, 6
int A3[] = {10, -2, -7}; // 10, -2, -7
FindSubarrayClosestToZero f = new FindSubarrayClosestToZero();
f.findSubarrayClosestToZero(A1);
f.findSubarrayClosestToZero(A2);
f.findSubarrayClosestToZero(A3);
}
}
Solution time complexity : O(NlogN)
Solution space complexity : O(N)
[Note this problem can't be solved in O(N) as some have claimed]
Algorithm:-
Compute cumulative array(here,cum[]) of given array [Line 10]
Sort the cumulative array [Line 11]
Answer is minimum amongst C[i]-C[i+1] , $\forall$ i∈[1,n-1] (1-based index) [Line 12]
C++ Code:-
#include<bits/stdc++.h>
#define M 1000010
#define REP(i,n) for (int i=1;i<=n;i++)
using namespace std;
typedef long long ll;
ll a[M],n,cum[M],ans=numeric_limits<ll>::max(); //cum->cumulative array
int main() {
ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
cin>>n; REP(i,n) cin>>a[i],cum[i]=cum[i-1]+a[i];
sort(cum+1,cum+n+1);
REP(i,n-1) ans=min(ans,cum[i+1]-cum[i]);
cout<<ans; //min +ve difference from 0 we can get
}
After more thinking on this problem, I found that #frankyym's solution is the right solution. I have made some refinements on the original solution, here is my code:
#include <map>
#include <stdio.h>
#include <algorithm>
#include <limits.h>
using namespace std;
#define IDX_LOW_BOUND -2
// Return [i..j] range of A
pair<int, int> nearest_to_c(int A[], int n, int t) {
map<int, int> bst;
int presum, subsum, closest, i, j, start, end;
bool unset;
map<int, int>::iterator it;
bst[0] = -1;
// Barriers. Assume that no prefix sum is equal to INT_MAX or INT_MIN.
bst[INT_MIN] = IDX_LOW_BOUND;
bst[INT_MAX] = n;
unset = true;
// This initial value is always overwritten afterwards.
closest = 0;
presum = 0;
for (i = 0; i < n; ++i) {
presum += A[i];
for (it = bst.lower_bound(presum - t), j = 0; j < 2; --it, j++) {
if (it->first == INT_MAX || it->first == INT_MIN)
continue;
subsum = presum - it->first;
if (unset || abs(closest - t) > abs(subsum - t)) {
closest = subsum;
start = it->second + 1;
end = i;
if (closest - t == 0)
goto ret;
unset = false;
}
}
bst[presum] = i;
}
ret:
return make_pair(start, end);
}
int main() {
int A[] = {10, 20, 30, 30, 20, 10, 10, 20};
int t;
scanf("%d", &t);
pair<int, int> ans = nearest_to_c(A, 8, t);
printf("[%d:%d]\n", ans.first, ans.second);
return 0;
}
As a side note: I agree with the algorithms provided by other threads here. There is another algorithm on top of my head recently. Make up another copy of A[], which is B[]. Inside B[], each element is A[i]-t/n, which means B[0]=A[0]-t/n, B[1]=A[1]-t/n ... B[n-1]=A[n-1]-t/n. Then the second problem is actually transformed to the first problem, once the smallest subarray of B[] closest to 0 is found, the subarray of A[] closest to t is found at the same time. (It is kinda tricky if t is not divisible by n, nevertheless, the precision has to be chosen appropriate. Also the runtime is O(n))
I think there is a little bug concerning the closest to 0 solution. At the last step we should not only inspect the difference between neighbor elements but also elements not near to each other if one of them is bigger than 0 and the other one is smaller than 0.
Sorry, I thought I am supposed to get all answers for the problem. Didn't see it only requires one.
Cant we use dynamic programming to solve this question similar to kadane's algorithm.Here is my solution to this problem.Please comment if this approach is wrong.
#include <bits/stdc++.h>
using namespace std;
int main() {
//code
int test;
cin>>test;
while(test--){
int n;
cin>>n;
vector<int> A(n);
for(int i=0;i<n;i++)
cin>>A[i];
int closest_so_far=A[0];
int closest_end_here=A[0];
int start=0;
int end=0;
int lstart=0;
int lend=0;
for(int i=1;i<n;i++){
if(abs(A[i]-0)<abs(A[i]+closest_end_here-0)){
closest_end_here=A[i]-0;
lstart=i;
lend=i;
}
else{
closest_end_here=A[i]+closest_end_here-0;
lend=i;
}
if(abs(closest_end_here-0)<abs(closest_so_far-0)){
closest_so_far=closest_end_here;
start=lstart;
end=lend;
}
}
for(int i=start;i<=end;i++)
cout<<A[i]<<" ";
cout<<endl;
cout<<closest_so_far<<endl;
}
return 0;
}
Here is a code implementation by java:
public class Solution {
/**
* #param nums: A list of integers
* #return: A list of integers includes the index of the first number
* and the index of the last number
*/
public ArrayList<Integer> subarraySumClosest(int[] nums) {
// write your code here
int len = nums.length;
ArrayList<Integer> result = new ArrayList<Integer>();
int[] sum = new int[len];
HashMap<Integer,Integer> mapHelper = new HashMap<Integer,Integer>();
int min = Integer.MAX_VALUE;
int curr1 = 0;
int curr2 = 0;
sum[0] = nums[0];
if(nums == null || len < 2){
result.add(0);
result.add(0);
return result;
}
for(int i = 1;i < len;i++){
sum[i] = sum[i-1] + nums[i];
}
for(int i = 0;i < len;i++){
if(mapHelper.containsKey(sum[i])){
result.add(mapHelper.get(sum[i])+1);
result.add(i);
return result;
}
else{
mapHelper.put(sum[i],i);
}
}
Arrays.sort(sum);
for(int i = 0;i < len-1;i++){
if(Math.abs(sum[i] - sum[i+1]) < min){
min = Math.abs(sum[i] - sum[i+1]);
curr1 = sum[i];
curr2 = sum[i+1];
}
}
if(mapHelper.get(curr1) < mapHelper.get(curr2)){
result.add(mapHelper.get(curr1)+1);
result.add(mapHelper.get(curr2));
}
else{
result.add(mapHelper.get(curr2)+1);
result.add(mapHelper.get(curr1));
}
return result;
}
}