Given an array , if 2 adjacent number are equal then they can merge and their value is increased by one. Find the smallest possible number of element left in the array after this process.
Ex: [1,1,1,2,1] ->[1,2,2,1]-> [1,3,1]. Thus the answer is 3.
I have tried using a linked list to store the array then iterate through the whole thing until no equal, adjacent number is detected but this seem very insufficent.
Any hint or suggestion is very appriciated. Thank you for your time
Here is a recursive solution, but I don't know if it is optimal:
Start with the smallest number in the array. In the example [1 1 1 2 1], it is 1. The reason is that you will not get new 1's after merging other elements. So they are easy to work with.
Obviously, if you have an even number of consecutive 1s, merging them all is never subobtimal. So, we need to decide what to do with an odd number of consecutive elements. One of them needs to be left out (not merged), and once we choose that one, remaining parts have both an even number of 1s.
The important observation here is that once you choose the element to be left out, the array to the left of it and the array to the right of it constitute two independent problems. Since there will be a single 1 in the middle, you can't merge any number at the right side with the left side. So, for every possible choice, you can recursively solve the problem for the right- and left-sub-arrays, and then find the minimum result.
Algorithm
To summarize the method, these are the steps to be followed:
If the length of array is 0, return 0.
Find the minimum element in the array. Call it x.
Go over the array one more time, create a new array where even number of consecutive x's are all merged.
If you saw an odd number of x's anywhere in the array, do this:
Let the index of first element be i. For each j = i, i+2, i+4, ... that belongs to the sequence of x's, solve the problem for sub-arrays [0 .. j-1] and [j+1 .. end]. Call their results n1 and n2.
Return the minimum n1 + n2 + 1 from these possible splits.
If you didn't see an odd number of x's, then there are no x's left in the array. Go back to step 1.
Note that you can substitute x's with x+1's in the 4th step, and choose the sub-problem indices accordingly, to possibly save some work in the recursive calls.
Code
Here is a c++ code that does this:
#include <iostream>
#include <limits>
#include <vector>
// the range is [start, end)
int
solve(std::vector<int>& array, int start, int end)
{
if (start >= end)
return 0;
int length = end - start;
// find the minimum element
int min = array[start];
for (int i = start; i < end; i++)
if (array[i] < min)
min = array[i];
std::vector<int> newArray;
newArray.reserve(length + 1);
int minCount = 0; // number of consecutive elements that are equal to min
int firstOddNumber =
-1; // index of an odd number of consecutive min's in the new array
int oddNumbers = 0; // number of min's starting at firstOddNumber
for (int i = start; i <= end; i++) {
// iterate one last time with i == end to run the checks again.
// hence the special case. we pop this element after the loop.
int elem = i < end ? array[i] : min + 1;
if (elem == min) {
minCount++;
} else if (minCount != 0) {
// even number of min's
if (minCount % 2 == 0) {
// merge them
for (int j = 0; j < minCount / 2; j++)
newArray.push_back(min + 1);
} else {
// do not merge them but save their index in the new array
firstOddNumber = newArray.size();
oddNumbers = minCount;
for (int j = 0; j < minCount; j++)
newArray.push_back(min);
// ^^^ this part could be modified as I wrote in the note in my
// answer
}
minCount = 0;
newArray.push_back(elem);
} else
newArray.push_back(elem);
}
// remove the min+1 element pushed when i == end
newArray.pop_back();
if (firstOddNumber == -1)
// no odd number of consecutive min's, repeat the procedure
return solve(newArray, 0, newArray.size());
else {
int minResult = newArray.size();
// solve two subproblems for each possible split
for (int i = firstOddNumber; i <= firstOddNumber + oddNumbers; i += 2) {
int result = 1 + solve(newArray, 0, i) +
solve(newArray, i + 1, newArray.size());
if (result < minResult)
minResult = result;
// ^^^ this part could be modified as I wrote in the note in my
// answer
}
return minResult;
}
}
void
test(std::vector<int> v, int expected)
{
int result = solve(v, 0, v.size());
std::cout << result << '\n';
if (result == expected)
std::cout << "CORRECT\n" << std::endl;
else
std::cout << "EXPECTED: " << expected << '\n' << std::endl;
}
int
main()
{
test({ 1, 1, 1, 2, 1 }, 3);
test({ 1, 1, 1, 1, 1 }, 2);
test({ 1, 1, 1, 1, 1, 1 }, 2);
test({ 1, 2, 1, 1, 1 }, 3);
test({ 1, 2, 1, 2, 1 }, 5);
}
I am assuming that the question requires reading the input from the end, instead of the start. Because if it required reading from the start then your second iteration must have been : [2, 1, 2, 1]
Assuming the question requires reading the input from the end, here's the solution:
Here's the algorithm:
Add all the elements to stack 1. Stack1: [1, 2, 1, 1, 1]; result:[] and top = 1;
Checking if 1 and 2 are equal, if not equal adding it to the result stack. result: [2, 1]
Checking if 1 and 1 are equal, if they are, incrementing the element and pushing the element to stack 1, and also adding all the result stack elements to stack1. Stack1: [1, 2, 2, 1], result: [].
Repeating the process until stack1 is empty.
class Solution {
public int mergeAdjacentSimilarElements(int[] arr) {
//stack 1
Stack stack = new Stack<>();
//stack 2
Stack result = new Stack<>();
//add all the elements to the stack, as stack follows LIFO, the last element would be at the top
for (int i = 0; i < arr.length; i++) {
stack.push(arr[i]);
}
while (!stack.isEmpty()) {
int top = !stack.isEmpty() ? stack.pop() : -1; // assign -1 in case stack is empty
int temp = !stack.isEmpty() ? stack.peek() : -1; // assign -1 in case stack is empty
//if top and temp are equal
if (top != -1 && temp != -1 && top == temp) {
stack.pop();
//increment the value of the top, and add it to stack
stack.push(++top);
//check if there are any elements in the result stack,
// as they have to be added to stack1, as stack1 is modified.
if (!result.isEmpty()) {
stack.push(result.pop());
}
} else {
//else simply add the element to the stack.
result.push(top);
}
}
//for testing
result.stream().forEach(System.out::println);
return result.size();
}
}
Given array : 8 3 5 2 10 6 7 9 5 2
So the o/p will be Yes.
as: {8,3,5} {10,6} {9,5,2} they all have same sum value i.e. 16.
But for this array : 1 4 9 6 2 12
o/p will be No.
as: No contiguous slide have same sum value
I was thinking to go with SubSetSum Algorithm / Kadane Maximum SubArray Algorithm but later I end up as all of the algorithms requires a target sum which is predefined.
But here we don't know the target sum
If desired sum is given, and all subarrays should be contiguous, then it's easily can be done in O(n).
Run a loop over array and maintain boundaries of slices (left and right indexes) and currentSum.
Start with first element as a 0. Boundaries will be [0, 0] (for simplicity we include right). Then in a loop you have three conditions.
If sum is less than desired, add right element to the sum and advance right index
If sum is greater than desired, remove left element from the sum and advance left index
If sum is equal to given, print the slice. To avoid this slice in next iteration, advance left index and adjust the sum.
Translated to code
public static void main(String[] args) {
int givenSum = 16;
int[] a = new int[] {8, 3, 5, 2, 10, 6, 7, 9, 5, 2};
// boundaries of slice
int left = 0; // defines position of slice
int right = 0; // exclusive
int currentSum = 0;
while (right < a.length) {
if (currentSum < givenSum) { // sum is not enough, add from the right
currentSum += a[right];
right++;
}
if (currentSum > givenSum) { // sum exceeds given, remove from the left
currentSum -= a[left];
left++;
}
if (currentSum == givenSum) { // boundaries of given sum found, print it
System.out.println(Arrays.toString(Arrays.copyOfRange(a, left, right)));
// remove the left element, so we can process next sums
currentSum -= a[left];
left++;
}
}
}
For your case it prints 4 slices which yields sum 16
[8, 3, 5]
[10, 6]
[7, 9]
[9, 5, 2]
EDIT:
As OP clarified, no given sum available, the goal is to check if there are at least two different contiguous subarrays present which yields equal sum.
The most straightforward algorithm is to generate all possible sums and check if there are duplicates
int[] a = new int[] {1, 4, 9, 6, 2, 12};
HashSet<Integer> sums = new HashSet<>();
int numOfSums = 0;
for (int left = 0; left < a.length - 1; left++) {
for (int right = left; right < a.length; right++) {
// sum from left to right
int sum = 0;
for (int k = left; k <= right; k++) {
sum += a[k];
}
numOfSums++;
sums.add(sum);
}
}
System.out.println(sums.size() == numOfSums);
Complexity of this is O(n^3), not a good one, but works.
Hint: One trick could be explored to boost it to O(n^2), you don't need to calculate sum for every pair of slices!
You can do it in the following way
You have the total sum = 48
Now the each subset would have a sum which would be equal to a factor of 48. The smaller the factor the more number of subsets you can break it into
For all factors of the sum, check if the answer is possible for that factor or not. This can be done in O(n) by simply traversing the array.
Time Complexity would be O(n * factors(sum))
Use dynamic programming to find all sub-sums of the array, then find the sub array with same sum. The complexity should be O(n2).
void subsum(int n, int* arr, int** sum) {
for (int i = 0; i < n; ++i) {
sum[i][i] = arr[i];
}
for (int l = 2; l <= n; ++l) {
for (int i = 0; i < n - l + 1; ++i) {
sum[i][i + l - 1] = sum[i][i + l - 2] + arr[i + l -1];
}
}
}
This question already has answers here:
Maximum sum sublist?
(13 answers)
Closed 8 years ago.
I had an interview question a while back that I never got a solution for. Apparently there is a "very efficient" algorithm to solve it.
The question: Given an array of random positive and negative numbers, find the continuous subset that has the greatest sum.
Example:
[1, -7, 4, 5, -1, 5]
The best subset here is {4, 5, -1, 5}
I can think of no solution but the brute-force method. What is the efficient method?
Iterate through the list, keeping track of the local sum of the list elements so far.
If the local sum is the highest sum so far, then keep a record of it.
If the local sum reaches 0 or below, then reset it and restart from the next element.
Theory
If the current subset sum is greater than zero it will contribute to future subset sums, so we keep it. On the other hand if the current subset sum is zero or below it will not contribute to future subset sums. So we throw it away and start fresh with a new subset sum. Then it's just a matter of keeping track of when the current subset sum is greater then any previous encountered.
Pseudocode
In-parameter is an array list of length N. The result is stored in best_start and best_end.
best_sum = -MAX
best_start = best_end = -1
local_start = local_sum = 0
for i from 0 to N-1 {
local_sum = local_sum + list[i]
if local_sum > best_sum {
best_sum = local_sum
best_start = local_start
best_end = i
}
if local_sum <= 0 {
local_sum = 0
local_start = i+1
}
}
Convert the list into a list of cumulative sums, [1,-7,4,5,-1,5] to [1, -6, -2, -3, 2]. Then walk through the list of cumulative sums, saving the smallest value so far and the maximum difference between what you see as the current value and what is currently the smallest value.
Got it from here
You can answer this question from CLRS, which includes a tip:
Use the following ideas to develop a nonrecursive, linear-time algorithm for the
maximum-subarray problem.
Start at the left end of the array, and progress toward
the right, keeping track of the maximum subarray seen so far.
Knowing a maximum sub array of A[1..j], extend the answer to find a maximum subarray ending at index j+1 by using the following observation:
a maximum sub array of A[1..j+1] is either a maximum sub array of A[1..j] or a sub array A[i..j+1], for some 1 <= i <= j + 1.
Determine a maximum sub array of the form A[i..j+1] in constant time based on knowing a maximum subarray ending at index j.
max-sum = A[1]
current-sum = A[1]
left = right = 1
current-left = current-right = 1
for j = 2 to n
if A[j] > current-sum + A[j]
current-sum = A[j]
current-left = current-right = j
else
current-sum += A[j]
current-right = j
if current-sum > max-sum
max-sum = current-sum
left = current-left
right = current-right
return (max-sum, left, right)
Too bad Java does not have tuple return type. So, had to print the indices and sum in the method.
public class Kadane {
public static void main(String[] args) {
int[] intArr = {-1, 3, -5, 4, 6, -1, 2, -7, 13, -3};
findMaxSubArray(intArr);
}
public static void findMaxSubArray(int[] inputArray){
int maxStartIndex=0;
int maxEndIndex=0;
int maxSum = Integer.MIN_VALUE;
int sum= 0;
for (int currentIndex = 0; currentIndex < inputArray.length; currentIndex++) {
int eachArrayItem = inputArray[currentIndex];
sum+=eachArrayItem;
if( eachArrayItem > sum){
maxStartIndex = currentIndex;
sum = eachArrayItem;
}
if(sum>maxSum){
maxSum = sum;
maxEndIndex = currentIndex;
}
}
System.out.println("Max sum : "+maxSum);
System.out.println("Max start index : "+maxStartIndex);
System.out.println("Max end index : "+maxEndIndex);
}
}
And here is some shameless marketing : I managed to pull together a slide on how this works
Here is the java class which runs in linear time
public class MaxSumOfContinousSubset {
public static void main(String[] args) {
System.out.println(maxSum(1, -7, 4, 5, -1, 5));
}
private static int maxSum (int... nums) {
int maxsofar = 0;
int maxhere = 0;
for (int i = 0; i < nums.length; i++) {
maxhere = Math.max(maxhere + nums[i], 0);
maxsofar = Math.max(maxhere, maxsofar);
}
return maxsofar;
}
}
Given an array of size n and k, how do you find the maximum for every contiguous subarray of size k?
For example
arr = 1 5 2 6 3 1 24 7
k = 3
ans = 5 6 6 6 24 24
I was thinking of having an array of size k and each step evict the last element out and add the new element and find maximum among that. It leads to a running time of O(nk). Is there a better way to do this?
You have heard about doing it in O(n) using dequeue.
Well that is a well known algorithm for this question to do in O(n).
The method i am telling is quite simple and has time complexity O(n).
Your Sample Input:
n=10 , W = 3
10 3
1 -2 5 6 0 9 8 -1 2 0
Answer = 5 6 6 9 9 9 8 2
Concept: Dynamic Programming
Algorithm:
N is number of elements in an array and W is window size. So, Window number = N-W+1
Now divide array into blocks of W starting from index 1.
Here divide into blocks of size 'W'=3.
For your sample input:
We have divided into blocks because we will calculate maximum in 2 ways A.) by traversing from left to right B.) by traversing from right to left.
but how ??
Firstly, Traversing from Left to Right. For each element ai in block we will find maximum till that element ai starting from START of Block to END of that block.
So here,
Secondly, Traversing from Right to Left. For each element 'ai' in block we will find maximum till that element 'ai' starting from END of Block to START of that block.
So Here,
Now we have to find maximum for each subarray or window of size 'W'.
So, starting from index = 1 to index = N-W+1 .
max_val[index] = max(RL[index], LR[index+w-1]);
for index=1: max_val[1] = max(RL[1],LR[3]) = max(5,5)= 5
Simliarly, for all index i, (i<=(n-k+1)), value at RL[i] and LR[i+w-1]
are compared and maximum among those two is answer for that subarray.
So Final Answer : 5 6 6 9 9 9 8 2
Time Complexity: O(n)
Implementation code:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LIM 100001
using namespace std;
int arr[LIM]; // Input Array
int LR[LIM]; // maximum from Left to Right
int RL[LIM]; // maximum from Right to left
int max_val[LIM]; // number of subarrays(windows) will be n-k+1
int main(){
int n, w, i, k; // 'n' is number of elements in array
// 'w' is Window's Size
cin >> n >> w;
k = n - w + 1; // 'K' is number of Windows
for(i = 1; i <= n; i++)
cin >> arr[i];
for(i = 1; i <= n; i++){ // for maximum Left to Right
if(i % w == 1) // that means START of a block
LR[i] = arr[i];
else
LR[i] = max(LR[i - 1], arr[i]);
}
for(i = n; i >= 1; i--){ // for maximum Right to Left
if(i == n) // Maybe the last block is not of size 'W'.
RL[i] = arr[i];
else if(i % w == 0) // that means END of a block
RL[i] = arr[i];
else
RL[i] = max(RL[i+1], arr[i]);
}
for(i = 1; i <= k; i++) // maximum
max_val[i] = max(RL[i], LR[i + w - 1]);
for(i = 1; i <= k ; i++)
cout << max_val[i] << " ";
cout << endl;
return 0;
}
Running Code Link
I'll try to proof: (by #johnchen902)
If k % w != 1 (k is not the begin of a block)
Let k* = The begin of block containing k
ans[k] = max( arr[k], arr[k + 1], arr[k + 2], ..., arr[k + w - 1])
= max( max( arr[k], arr[k + 1], arr[k + 2], ..., arr[k*]),
max( arr[k*], arr[k* + 1], arr[k* + 2], ..., arr[k + w - 1]) )
= max( RL[k], LR[k+w-1] )
Otherwise (k is the begin of a block)
ans[k] = max( arr[k], arr[k + 1], arr[k + 2], ..., arr[k + w - 1])
= RL[k] = LR[k+w-1]
= max( RL[k], LR[k+w-1] )
Dynamic programming approach is very neatly explained by Shashank Jain. I would like to explain how to do the same using dequeue.
The key is to maintain the max element at the top of the queue(for a window ) and discarding the useless elements and we also need to discard the elements that are out of index of current window.
useless elements = If Current element is greater than the last element of queue than the last element of queue is useless .
Note : We are storing the index in queue not the element itself. It will be more clear from the code itself.
1. If Current element is greater than the last element of queue than the last element of queue is useless . We need to delete that last element.
(and keep deleting until the last element of queue is smaller than current element).
2. If if current_index - k >= q.front() that means we are going out of window so we need to delete the element from front of queue.
vector<int> max_sub_deque(vector<int> &A,int k)
{
deque<int> q;
for(int i=0;i<k;i++)
{
while(!q.empty() && A[i] >= A[q.back()])
q.pop_back();
q.push_back(i);
}
vector<int> res;
for(int i=k;i<A.size();i++)
{
res.push_back(A[q.front()]);
while(!q.empty() && A[i] >= A[q.back()] )
q.pop_back();
while(!q.empty() && q.front() <= i-k)
q.pop_front();
q.push_back(i);
}
res.push_back(A[q.front()]);
return res;
}
Since each element is enqueued and dequeued atmost 1 time to time complexity is O(n+n) = O(2n) = O(n).
And the size of queue can not exceed the limit k . so space complexity = O(k).
An O(n) time solution is possible by combining the two classic interview questions:
Make a stack data-structure (called MaxStack) which supports push, pop and max in O(1) time.
This can be done using two stacks, the second one contains the minimum seen so far.
Model a queue with a stack.
This can done using two stacks. Enqueues go into one stack, and dequeues come from the other.
For this problem, we basically need a queue, which supports enqueue, dequeue and max in O(1) (amortized) time.
We combine the above two, by modelling a queue with two MaxStacks.
To solve the question, we queue k elements, query the max, dequeue, enqueue k+1 th element, query the max etc. This will give you the max for every k sized sub-array.
I believe there are other solutions too.
1)
I believe the queue idea can be simplified. We maintain a queue and a max for every k. We enqueue a new element, and dequeu all elements which are not greater than the new element.
2) Maintain two new arrays which maintain the running max for each block of k, one array for one direction (left to right/right to left).
3) Use a hammer: Preprocess in O(n) time for range maximum queries.
The 1) solution above might be the most optimal.
You need a fast data structure that can add, remove and query for the max element in less than O(n) time (you can just use an array if O(n) or O(nlogn) is acceptable). You can use a heap, a balanced binary search tree, a skip list, or any other sorted data structure that performs these operations in O(log(n)).
The good news is that most popular languages have a sorted data structure implemented that supports these operations for you. C++ has std::set and std::multiset (you probably need the latter) and Java has PriorityQueue and TreeSet.
Here is the java implementation
public static Integer[] maxsInEveryWindows(int[] arr, int k) {
Deque<Integer> deque = new ArrayDeque<Integer>();
/* Process first k (or first window) elements of array */
for (int i = 0; i < k; i++) {
// For very element, the previous smaller elements are useless so
// remove them from deque
while (!deque.isEmpty() && arr[i] >= arr[deque.peekLast()]) {
deque.removeLast(); // Remove from rear
}
// Add new element at rear of queue
deque.addLast(i);
}
List<Integer> result = new ArrayList<Integer>();
// Process rest of the elements, i.e., from arr[k] to arr[n-1]
for (int i = k; i < arr.length; i++) {
// The element at the front of the queue is the largest element of
// previous window, so add to result.
result.add(arr[deque.getFirst()]);
// Remove all elements smaller than the currently
// being added element (remove useless elements)
while (!deque.isEmpty() && arr[i] >= arr[deque.peekLast()]) {
deque.removeLast();
}
// Remove the elements which are out of this window
while (!deque.isEmpty() && deque.getFirst() <= i - k) {
deque.removeFirst();
}
// Add current element at the rear of deque
deque.addLast(i);
}
// Print the maximum element of last window
result.add(arr[deque.getFirst()]);
return result.toArray(new Integer[0]);
}
Here is the corresponding test case
#Test
public void maxsInWindowsOfSizeKTest() {
Integer[] result = ArrayUtils.maxsInEveryWindows(new int[]{1, 2, 3, 1, 4, 5, 2, 3, 6}, 3);
assertThat(result, equalTo(new Integer[]{3, 3, 4, 5, 5, 5, 6}));
result = ArrayUtils.maxsInEveryWindows(new int[]{8, 5, 10, 7, 9, 4, 15, 12, 90, 13}, 4);
assertThat(result, equalTo(new Integer[]{10, 10, 10, 15, 15, 90, 90}));
}
Using a heap (or tree), you should be able to do it in O(n * log(k)). I'm not sure if this would be indeed better.
here is the Python implementation in O(1)...Thanks to #Shahshank Jain in advance..
from sys import stdin,stdout
from operator import *
n,w=map(int , stdin.readline().strip().split())
Arr=list(map(int , stdin.readline().strip().split()))
k=n-w+1 # window size = k
leftA=[0]*n
rightA=[0]*n
result=[0]*k
for i in range(n):
if i%w==0:
leftA[i]=Arr[i]
else:
leftA[i]=max(Arr[i],leftA[i-1])
for i in range(n-1,-1,-1):
if i%w==(w-1) or i==n-1:
rightA[i]=Arr[i]
else:
rightA[i]=max(Arr[i],rightA[i+1])
for i in range(k):
result[i]=max(rightA[i],leftA[i+w-1])
print(*result,sep=' ')
Method 1: O(n) time, O(k) space
We use a deque (it is like a list but with constant-time insertion and deletion from both ends) to store the index of useful elements.
The index of the current max is kept at the leftmost element of deque. The rightmost element of deque is the smallest.
In the following, for easier explanation we say an element from the array is in the deque, while in fact the index of that element is in the deque.
Let's say {5, 3, 2} are already in the deque (again, if fact their indexes are).
If the next element we read from the array is bigger than 5 (remember, the leftmost element of deque holds the max), say 7: We delete the deque and create a new one with only 7 in it (we do this because the current elements are useless, we have found a new max).
If the next element is less than 2 (which is the smallest element of deque), say 1: We add it to the right ({5, 3, 2, 1})
If the next element is bigger than 2 but less than 5, say 4: We remove elements from right that are smaller than the element and then add the element from right ({5, 4}).
Also we keep elements of the current window only (we can do this in constant time because we are storing the indexes instead of elements).
from collections import deque
def max_subarray(array, k):
deq = deque()
for index, item in enumerate(array):
if len(deq) == 0:
deq.append(index)
elif index - deq[0] >= k: # the max element is out of the window
deq.popleft()
elif item > array[deq[0]]: # found a new max
deq = deque()
deq.append(index)
elif item < array[deq[-1]]: # the array item is smaller than all the deque elements
deq.append(index)
elif item > array[deq[-1]] and item < array[deq[0]]:
while item > array[deq[-1]]:
deq.pop()
deq.append(index)
if index >= k - 1: # start printing when the first window is filled
print(array[deq[0]])
Proof of O(n) time: The only part we need to check is the while loop. In the whole runtime of the code, the while loop can perform at most O(n) operations in total. The reason is that the while loop pops elements from the deque, and since in other parts of the code, we do at most O(n) insertions into the deque, the while loop cannot exceed O(n) operations in total. So the total runtime is O(n) + O(n) = O(n)
Method 2: O(n) time, O(n) space
This is the explanation of the method suggested by S Jain (as mentioned in the comments of his post, this method doesn't work with data streams, which most sliding window questions are designed for).
The reason that method works is explained using the following example:
array = [5, 6, 2, 3, 1, 4, 2, 3]
k = 4
[5, 6, 2, 3 1, 4, 2, 3 ]
LR: 5 6 6 6 1 4 4 4
RL: 6 6 3 3 4 4 3 3
6 6 4 4 4
To get the max for the window [2, 3, 1, 4],
we can get the max of [2, 3] and max of [1, 4], and return the bigger of the two.
Max of [2, 3] is calculated in the RL pass and max of [1, 4] is calculated in LR pass.
Using Fibonacci heap, you can do it in O(n + (n-k) log k), which is equal to O(n log k) for small k, for k close to n this becomes O(n).
The algorithm: in fact, you need:
n inserts to the heap
n-k deletions
n-k findmax's
How much these operations cost in Fibonacci heaps? Insert and findmax is O(1) amortized, deletion is O(log n) amortized. So, we have
O(n + (n-k) log k + (n-k)) = O(n + (n-k) log k)
Sorry, this should have been a comment but I am not allowed to comment for now.
#leo and #Clay Goddard
You can save yourselves from re-computing the maximum by storing both maximum and 2nd maximum of the window in the beginning
(2nd maximum will be the maximum only if there are two maximums in the initial window). If the maximum slides out of the window you still have the next best candidate to compare with the new entry. So you get O(n) , otherwise if you allowed the whole re-computation again the worst case order would be O(nk), k is the window size.
class MaxFinder
{
// finds the max and its index
static int[] findMaxByIteration(int arr[], int start, int end)
{
int max, max_ndx;
max = arr[start];
max_ndx = start;
for (int i=start; i<end; i++)
{
if (arr[i] > max)
{
max = arr[i];
max_ndx = i;
}
}
int result[] = {max, max_ndx};
return result;
}
// optimized to skip iteration, when previous windows max element
// is present in current window
static void optimizedPrintKMax(int arr[], int n, int k)
{
int i, j, max, max_ndx;
// for first window - find by iteration.
int result[] = findMaxByIteration(arr, 0, k);
System.out.printf("%d ", result[0]);
max = result[0];
max_ndx = result[1];
for (j=1; j <= (n-k); j++)
{
// if previous max has fallen out of current window, iterate and find
if (max_ndx < j)
{
result = findMaxByIteration(arr, j, j+k);
max = result[0];
max_ndx = result[1];
}
// optimized path, just compare max with new_elem that has come into the window
else
{
int new_elem_ndx = j + (k-1);
if (arr[new_elem_ndx] > max)
{
max = arr[new_elem_ndx];
max_ndx = new_elem_ndx;
}
}
System.out.printf("%d ", max);
}
}
public static void main(String[] args)
{
int arr[] = {10, 9, 8, 7, 6, 5, 4, 3, 2, 1};
//int arr[] = {1,5,2,6,3,1,24,7};
int n = arr.length;
int k = 3;
optimizedPrintKMax(arr, n, k);
}
}
package com;
public class SlidingWindow {
public static void main(String[] args) {
int[] array = { 1, 5, 2, 6, 3, 1, 24, 7 };
int slide = 3;//say
List<Integer> result = new ArrayList<Integer>();
for (int i = 0; i < array.length - (slide-1); i++) {
result.add(getMax(array, i, slide));
}
System.out.println("MaxList->>>>" + result.toString());
}
private static Integer getMax(int[] array, int i, int slide) {
List<Integer> intermediate = new ArrayList<Integer>();
System.out.println("Initial::" + intermediate.size());
while (intermediate.size() < slide) {
intermediate.add(array[i]);
i++;
}
Collections.sort(intermediate);
return intermediate.get(slide - 1);
}
}
Here is the solution in O(n) time complexity with auxiliary deque
public class TestSlidingWindow {
public static void main(String[] args) {
int[] arr = { 1, 5, 7, 2, 1, 3, 4 };
int k = 3;
printMaxInSlidingWindow(arr, k);
}
public static void printMaxInSlidingWindow(int[] arr, int k) {
Deque<Integer> queue = new ArrayDeque<Integer>();
Deque<Integer> auxQueue = new ArrayDeque<Integer>();
int[] resultArr = new int[(arr.length - k) + 1];
int maxElement = 0;
int j = 0;
for (int i = 0; i < arr.length; i++) {
queue.add(arr[i]);
if (arr[i] > maxElement) {
maxElement = arr[i];
}
/** we need to maintain the auxiliary deque to maintain max element in case max element is removed.
We add the element to deque straight away if subsequent element is less than the last element
(as there is a probability if last element is removed this element can be max element) otherwise
remove all lesser element then insert current element **/
if (auxQueue.size() > 0) {
if (arr[i] < auxQueue.peek()) {
auxQueue.push(arr[i]);
} else {
while (auxQueue.size() > 0 && (arr[i] > auxQueue.peek())) {
auxQueue.pollLast();
}
auxQueue.push(arr[i]);
}
}else {
auxQueue.push(arr[i]);
}
if (queue.size() > 3) {
int removedEl = queue.removeFirst();
if (maxElement == removedEl) {
maxElement = auxQueue.pollFirst();
}
}
if (queue.size() == 3) {
resultArr[j++] = maxElement;
}
}
for (int i = 0; i < resultArr.length; i++) {
System.out.println(resultArr[i]);
}
}
}
static void countDistinct(int arr[], int n, int k)
{
System.out.print("\nMaximum integer in the window : ");
// Traverse through every window
for (int i = 0; i <= n - k; i++) {
System.out.print(findMaximuminAllWindow(Arrays.copyOfRange(arr, i, arr.length), k)+ " ");
}
}
private static int findMaximuminAllWindow(int[] win, int k) {
// TODO Auto-generated method stub
int max= Integer.MIN_VALUE;
for(int i=0; i<k;i++) {
if(win[i]>max)
max=win[i];
}
return max;
}
arr = 1 5 2 6 3 1 24 7
We have to find the maximum of subarray, Right?
So, What is meant by subarray?
SubArray = Partial set and it should be in order and contiguous.
From the above array
{1,5,2} {6,3,1} {1,24,7} all are the subarray examples
n = 8 // Array length
k = 3 // window size
For finding the maximum, we have to iterate through the array, and find the maximum.
From the window size k,
{1,5,2} = 5 is the maximum
{5,2,6} = 6 is the maximum
{2,6,3} = 6 is the maximum
and so on..
ans = 5 6 6 6 24 24
It can be evaluated as the n-k+1
Hence, 8-3+1 = 6
And the length of an answer is 6 as we seen.
How can we solve this now?
When the data is moving from the pipe, the first thought for the data structure came in mind is the Queue
But, rather we are not discussing much here, we directly jump on the deque
Thinking Would be:
Window is fixed and data is in and out
Data is fixed and window is sliding
EX: Time series database
While (Queue is not empty and arr[Queue.back() < arr[i]] {
Queue.pop_back();
Queue.push_back();
For the rest:
Print the front of queue
// purged expired element
While (queue not empty and queue.front() <= I-k) {
Queue.pop_front();
While (Queue is not empty and arr[Queue.back() < arr[i]] {
Queue.pop_back();
Queue.push_back();
}
}
arr = [1, 2, 3, 1, 4, 5, 2, 3, 6]
k = 3
for i in range(len(arr)-k):
k=k+1
print (max(arr[i:k]),end=' ') #3 3 4 5 5 5 6
Two approaches.
Segment Tree O(nlog(n-k))
Build a maximum segment-tree.
Query between [i, i+k)
Something like..
public static void printMaximums(int[] a, int k) {
int n = a.length;
SegmentTree tree = new SegmentTree(a);
for (int i=0; i<=n-k; i++) System.out.print(tree.query(i, i+k));
}
Deque O(n)
If the next element is greater than the rear element, remove the rear element.
If the element in the front of the deque is out of the window, remove the front element.
public static void printMaximums(int[] a, int k) {
int n = a.length;
Deque<int[]> deck = new ArrayDeque<>();
List<Integer> result = new ArrayList<>();
for (int i=0; i<n; i++) {
while (!deck.isEmpty() && a[i] >= deck.peekLast()[0]) deck.pollLast();
deck.offer(new int[] {a[i], i});
while (!deck.isEmpty() && deck.peekFirst()[1] <= i - k) deck.pollFirst();
if (i >= k - 1) result.add(deck.peekFirst()[0]);
}
System.out.println(result);
}
Here is an optimized version of the naive (conditional) nested loop approach I came up with which is much faster and doesn't require any auxiliary storage or data structure.
As the program moves from window to window, the start index and end index moves forward by 1. In other words, two consecutive windows have adjacent start and end indices.
For the first window of size W , the inner loop finds the maximum of elements with index (0 to W-1). (Hence i == 0 in the if in 4th line of the code).
Now instead of computing for the second window which only has one new element, since we have already computed the maximum for elements of indices 0 to W-1, we only need to compare this maximum to the only new element in the new window with the index W.
But if the element at 0 was the maximum which is the only element not part of the new window, we need to compute the maximum using the inner loop from 1 to W again using the inner loop (hence the second condition maxm == arr[i-1] in the if in line 4), otherwise just compare the maximum of the previous window and the only new element in the new window.
void print_max_for_each_subarray(int arr[], int n, int k)
{
int maxm;
for(int i = 0; i < n - k + 1 ; i++)
{
if(i == 0 || maxm == arr[i-1]) {
maxm = arr[i];
for(int j = i+1; j < i+k; j++)
if(maxm < arr[j]) maxm = arr[j];
}
else {
maxm = maxm < arr[i+k-1] ? arr[i+k-1] : maxm;
}
cout << maxm << ' ';
}
cout << '\n';
}
You can use Deque data structure to implement this. Deque has an unique facility that you can insert and remove elements from both the ends of the queue unlike the traditional queue where you can only insert from one end and remove from other.
Following is the code for the above problem.
public int[] maxSlidingWindow(int[] nums, int k) {
int n = nums.length;
int[] maxInWindow = new int[n - k + 1];
Deque<Integer> dq = new LinkedList<Integer>();
int i = 0;
for(; i<k; i++){
while(!dq.isEmpty() && nums[dq.peekLast()] <= nums[i]){
dq.removeLast();
}
dq.addLast(i);
}
for(; i <n; i++){
maxInWindow[i - k] = nums[dq.peekFirst()];
while(!dq.isEmpty() && dq.peekFirst() <= i - k){
dq.removeFirst();
}
while(!dq.isEmpty() && nums[dq.peekLast()] <= nums[i]){
dq.removeLast();
}
dq.addLast(i);
}
maxInWindow[i - k] = nums[dq.peekFirst()];
return maxInWindow;
}
the resultant array will have n - k + 1 elements where n is length of the given array, k is the given window size.
We can solve it using the Python , applying the slicing.
def sliding_window(a,k,n):
max_val =[]
val =[]
val1=[]
for i in range(n-k-1):
if i==0:
val = a[0:k+1]
print("The value in val variable",val)
val1 = max(val)
max_val.append(val1)
else:
val = a[i:i*k+1]
val1 =max(val)
max_val.append(val1)
return max_val
Driver Code
a = [15,2,3,4,5,6,2,4,9,1,5]
n = len(a)
k = 3
sl=s liding_window(a,k,n)
print(sl)
Create a TreeMap of size k. Put first k elements as keys in it and assign any value like 1(doesn't matter). TreeMap has the property to sort the elements based on key so now, first element in map will be min and last element will be max element. Then remove 1 element from the map whose index in the arr is i-k. Here, I have considered that Input elements are taken in array arr and from that array we are filling the map of size k. Since, we can't do anything with sorting happening inside TreeMap, therefore this approach will also take O(n) time.
100% working Tested (Swift)
func maxOfSubArray(arr:[Int],n:Int,k:Int)->[Int]{
var lenght = arr.count
var resultArray = [Int]()
for i in 0..<arr.count{
if lenght+1 > k{
let tempArray = Array(arr[i..<k+i])
resultArray.append(tempArray.max()!)
}
lenght = lenght - 1
}
print(resultArray)
return resultArray
}
This way we can use:
maxOfSubArray(arr: [1,2,3,1,4,5,2,3,6], n: 9, k: 3)
Result:
[3, 3, 4, 5, 5, 5, 6]
Just notice that you only have to find in the new window if:
* The new element in the window is smaller than the previous one (if it's bigger, it's for sure this one).
OR
* The element that just popped out of the window was the current bigger.
In this case, re-scan the window.
for how big k? for reasonable-sized k. you can create k k-sized buffers and just iterate over the array keeping track of max element pointers in the buffers - needs no data structures and is O(n) k^2 pre-allocation.
A complete working solution in Amortised Constant O(1) Complexity.
https://github.com/varoonverma/code-challenge.git
Compare the first k elements and find the max, this is your first number
then compare the next element to the previous max. If the next element is bigger, that is your max of the next subarray, if its equal or smaller, the max for that sub array is the same
then move on to the next number
max(1 5 2) = 5
max(5 6) = 6
max(6 6) = 6
... and so on
max(3 24) = 24
max(24 7) = 24
It's only slightly better than your answer
For example, given an integer array and its two consecutive sequence 's beginning position which are 'b1' and 'b2', furthermore provided with the position 'last' which indicates the second sequence's ending position. From array[b1] to array [b2-1] and from array [b2] to array[last] are both in order separately, how to merge them in place using O(n) time and O(1) space cost?
Kronrod's merge was the first published algorithm to do that. It goes roughly like this:
Split both parts of the array into blocks of size k=sqrt(n). Sort the blocks using their first elements as the basis for comparison. This can be done in sqrt(n)^2=O(n) by selection sort. The key property of selection sort here is that it has constant moves per block, so only #comparisons is square.
After this phase, for each element A[i] in the array there are at most k-1 elements "wrongly sorted" below it, that is elements at positions j<i such that A[j]>A[i]. These are (possibly) in the closest block below it that comes from the other merged part. Note that the first element of the block (and all other blocks below it) are already properly sorted relative to A[i] because of the blocks being sorted on their first elements. This is why the second phase works, i.e. achieves the fully sorted array:
Now merge the first block with the second, then second with the third, etc., using the last 2 blocks as temporary space for the output of the merge. This will scramble the contents of the last two blocks but in the last phase they (together with the preceding block) can be sorted by selection sort in sqrt(n)^2=O(n) time.
This is by no means a simple problem It is possible, but rarely done in practice because it's so much more complicated than a standard merge using N-scratch space. Huang and Langston's paper has been around since the late 80's, though practical implementations didn't really surface until later. Earlier, L. Trabb-Prado's paper in 1977 predates Huang and Langston significantly, but I'm challanged to find the exact text that paper; only references abound.
An excellent later publication, Asymptotically efficient in-place merging (1995) by Geert, Katajainenb, and Pasanen is a good coverage of multiple algorithms, and references Trabb-Prado's contributions to the subject.
There are such things as true in-place merges, but they are not straightforward enough that anybody is going to independently reinvent them in the middle of an interview - there have been papers describing a succession of pretty complex algorithms for this for years. One is Practical In-Place Merging, by Huang and Langston, CACM March 1988. The starting idea for this is to divide the data of length n into blocks of size sqrt(n), and use one block, filled with the largest elements of the data, to provide buffer space used in merging the others. The introduction to that paper says
"Given two sorted lists whose lengths sum to n, the obvious methods for merging in O(n) steps require a linear amount of extra memory as well. On the other hand, it is easy to merge in place using only a constant amount of additional space by heap-sorting, but at a cost of O(n log n) time"
Hence I claim that true merging in place can be done but is non-obvious.
Though it is not possible entirely in O(n) time, I have a proposition to do it faster than O(n^2). I use only O(1) space which is temp in my code. I am sure it should run better than O(n^2).
private static int[] mergeSortedArrays(int[] a1, int[] a2) {
int i = 0, j = 0;
while (a1[i] != Integer.MIN_VALUE) {
if (a1[i] > a2[j]) {
int temp = a1[i];
a1[i] = a2[j];
a2[j] = temp;
for (int k = 1; k < a2.length; k++) {
if (a2[k - 1] > a2[k]) {
temp = a2[k - 1];
a2[k - 1] = a2[k];
a2[k] = temp;
}
}
}
i++;
}
while(j < a2.length){
a1[i++] = a2[j++];
}
return a1;
}
Here is O(n-1) Memory (n+1)
/**
* Created by deian on 2016-12-22.
* We just need track the two smallest numbers
*/
public class Merge {
public static void swap(int[] a, int i1, int i2) {
int t = a[i1];
a[i1] = a[i2];
a[i2] = t;
}
public static void merge(int[] a) {
// i1 and i2 - always point to the smallest known numbers
// it would works as well with two m and n sized arrays
int i1 = 0;
int i2 = a.length / 2;
System.out.printf(" %s, i(%d,%d) \n", Arrays.toString(a), i1, i2);
for (int di = 0; di < a.length - 1; di++) {
int ni;
int oi1 = i1; int oi2 = i2;
if (a[i1] > a[i2]) {
ni = i2; i2++;
if (i2 >= a.length) { i2--; }
} else {
ni = i1; i1++;
if (i1 >= i2) { i1 = di; }
}
if (di == i1) { i1 = ni; }
swap(a, di, ni);
System.out.printf("#%d: %s, i(%d,%d)s(%d>%d)i(%d,%d) \n", di + 1, Arrays.toString(a), oi1, oi2, ni, di, i1, i2);
}
System.out.printf(" %s\n", Arrays.toString(a));
}
public static void main(String[] args) {
// int[] a = new int[]{1, 3, 6, 8, -5, -2, 3, 8};
// int[] a = new int[]{1, 3, 6, 8, -5, 2, 3, 8};
// int[] a = new int[]{1, 5, 6, 8, -5, 2, 3, 4};
// int[] a = new int[]{1, 5, 6, 8, -5, -2, -1, 4};
// int[] a = new int[]{ 1, 2, 3, 4, 5, 6, 7, 8};
// int[] a = new int[]{5, 6, 7, 8, 1, 2, 3, 4};
int[] a = new int[]{1, 3, 5, 7, 2, 4, 6, 8};
merge(a);
}
}
I had an interview (with a very important company) a couple of hours ago and I was asked that.
There is the answer in Java
public static void main(String[] args) {
int A[] = { 1, 3, 5, 6, 9 };
int B[] = new int[12];
B[0] = 3;
B[1] = 6;
B[2] = 8;
B[3] = 10;
B[4] = 11;
B[5] = 13;
B[6] = 15;
mergeInB(A, B, 7);
for (int n : B)
System.out.print(n + " ");
}
/**
* #param a
* #param b - it will be modified
* #param j = length of b
*/
public static void mergeInB(int[] a, int[] b, int j) {
int i = a.length - 1, k;
j --;
for (k = b.length-1; k >= 0; k--) {
if (i >= 0 && j >= 0) {
if (a[i] > b[j]) {
b[k] = a[i];
i --;
}
else
{
b[k] = b[j];
j --;
}
}
else break;
}
while(i>=0 && k >=0) {
b[k] = a[i];
k --;
i --;
}
while(j>= 0 && k >=0) {
b[k] = b[j];
j--;
k--;
}
}