Given an array and an integer k, find the maximum for each and every contiguous subarray of size k.

Examples:

Input :

arr[] = {1, 2, 3, 1, 4, 5, 2, 3, 6}

k = 3

Output :

3 3 4 5 5 5 6

Input :

arr[] = {8, 5, 10, 7, 9, 4, 15, 12, 90, 13}

k = 4

Output :

10 10 10 15 15 90 90

**Method 1 (Simple)**

Run two loops. In the outer loop, take all subarrays of size k. In the inner loop, get the maximum of the current subarray.

## C/C++

#include<stdio.h> void printKMax(int arr[], int n, int k) { int j, max; for (int i = 0; i <= n-k; i++) { max = arr[i]; for (j = 1; j < k; j++) { if (arr[i+j] > max) max = arr[i+j]; } printf("%d ", max); } } int main() { int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; int n = sizeof(arr)/sizeof(arr[0]); int k = 3; printKMax(arr, n, k); return 0; }

## Java

//Java Program to find the maximum for each and every contiguous subarray of size k. public class GFG { // Method to find the maximum for each and every contiguous subarray of size k. static void printKMax(int arr[], int n, int k) { int j, max; for (int i = 0; i <= n - k; i++) { max = arr[i]; for (j = 1; j < k; j++) { if (arr[i + j] > max) max = arr[i + j]; } System.out.print(max + " "); } } // Driver method public static void main(String args[]) { int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; int k = 3; printKMax(arr, arr.length, k); } } //This code is contributed by Sumit Ghosh

## Python3

# Python program to find the maximum for # each and every contiguous subarray of # size k # Method to find the maximum for each # and every contiguous subarray of s # of size k def printMax(arr, n, k): max = 0 for i in range(n - k + 1): max = arr[i] for j in range(1, k): if arr[i+j] > max: max = arr[i + j] print(str(max) + " ", end = "") # Driver method if __name__=="__main__": arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] n = len(arr) k = 3 printMax(arr, n, k) # This code is contributed by Shiv Shankar

Output :

3 4 5 6 7 8 9 10

Time Complexity: The outer loop runs n-k+1 times and the inner loop runs k times for every iteration of outer loop. So time complexity is O((n-k+1)*k) which can also be written as O(nk).

**Method 2 (Use Self-Balancing BST)**

1) Pick first k elements and create a Self-Balancing Binary Search Tree (BST) of size k.

2) Run a loop for i = 0 to n – k

…..a) Get the maximum element from the BST, and print it.

…..b) Search for arr[i] in the BST and delete it from the BST.

…..c) Insert arr[i+k] into the BST.

Time Complexity: Time Complexity of step 1 is O(kLogk). Time Complexity of steps 2(a), 2(b) and 2(c) is O(Logk). Since steps 2(a), 2(b) and 2(c) are in a loop that runs n-k+1 times, time complexity of the complete algorithm is O(kLogk + (n-k+1)*Logk) which can also be written as O(nLogk).

**Method 3 (A O(n) method: use Dequeue)**

We create a Dequeue, *Qi *of capacity k, that stores only useful elements of current window of k elements. An element is useful if it is in current window and is greater than all other elements on left side of it in current window. We process all array elements one by one and maintain *Qi *to contain useful elements of current window and these useful elements are maintained in sorted order. The element at front of the *Qi *is the largest and element at rear of *Qi *is the smallest of current window. Thanks to Aashish for suggesting this method.

Following is the implementation of this method.

## C++

#include <iostream> #include <deque> using namespace std; // A Dequeue (Double ended queue) based method for printing maixmum element of // all subarrays of size k void printKMax(int arr[], int n, int k) { // Create a Double Ended Queue, Qi that will store indexes of array elements // The queue will store indexes of useful elements in every window and it will // maintain decreasing order of values from front to rear in Qi, i.e., // arr[Qi.front[]] to arr[Qi.rear()] are sorted in decreasing order std::deque<int> Qi(k); /* Process first k (or first window) elements of array */ int i; for (i = 0; i < k; ++i) { // For very element, the previous smaller elements are useless so // remove them from Qi while ( (!Qi.empty()) && arr[i] >= arr[Qi.back()]) Qi.pop_back(); // Remove from rear // Add new element at rear of queue Qi.push_back(i); } // Process rest of the elements, i.e., from arr[k] to arr[n-1] for ( ; i < n; ++i) { // The element at the front of the queue is the largest element of // previous window, so print it cout << arr[Qi.front()] << " "; // Remove the elements which are out of this window while ( (!Qi.empty()) && Qi.front() <= i - k) Qi.pop_front(); // Remove from front of queue // Remove all elements smaller than the currently // being added element (remove useless elements) while ( (!Qi.empty()) && arr[i] >= arr[Qi.back()]) Qi.pop_back(); // Add current element at the rear of Qi Qi.push_back(i); } // Print the maximum element of last window cout << arr[Qi.front()]; } // Driver program to test above functions int main() { int arr[] = {12, 1, 78, 90, 57, 89, 56}; int n = sizeof(arr)/sizeof(arr[0]); int k = 3; printKMax(arr, n, k); return 0; }

## Java

//Java Program to find the maximum for each and every contiguous subarray of size k. import java.util.Deque; import java.util.LinkedList; public class SlidingWindow { // A Dequeue (Double ended queue) based method for printing maixmum element of // all subarrays of size k static void printMax(int arr[],int n, int k) { // Create a Double Ended Queue, Qi that will store indexes of array elements // The queue will store indexes of useful elements in every window and it will // maintain decreasing order of values from front to rear in Qi, i.e., // arr[Qi.front[]] to arr[Qi.rear()] are sorted in decreasing order Deque<Integer> Qi = new LinkedList<Integer>(); /* Process first k (or first window) elements of array */ int i; for(i = 0; i < k; ++i) { // For very element, the previous smaller elements are useless so // remove them from Qi while(!Qi.isEmpty() && arr[i] >= arr[Qi.peekLast()]) Qi.removeLast(); // Remove from rear // Add new element at rear of queue Qi.addLast(i); } // Process rest of the elements, i.e., from arr[k] to arr[n-1] for( ;i < n; ++i) { // The element at the front of the queue is the largest element of // previous window, so print it System.out.print(arr[Qi.peek()] + " "); // Remove the elements which are out of this window while((!Qi.isEmpty()) && Qi.peek() <= i-k) Qi.removeFirst(); // Remove all elements smaller than the currently // being added element (remove useless elements) while((!Qi.isEmpty()) && arr[i] >= arr[Qi.peekLast()]) Qi.removeLast(); // Add current element at the rear of Qi Qi.addLast(i); } // Print the maximum element of last window System.out.print(arr[Qi.peek()]); } // Driver program to test above functions public static void main(String[] args) { int arr[]={12, 1, 78, 90, 57, 89, 56}; int k=3; printMax(arr, arr.length,k); } } //This code is contributed by Sumit Ghosh

## Python3

# Python program to find the maximum for # each and every contiguous subarray of # size k from collections import deque # A Deque (Double ended queue) based # method for printing maximum element # of all subarrays of size k def printMax(arr, n, k): """ Create a Double Ended Queue, Qi that will store indexes of array elements. The queue will store indexes of useful elements in every window and it will maintain decreasing order of values from front to rear in Qi, i.e., arr[Qi.front[]] to arr[Qi.rear()] are sorted in decreasing order""" Qi = deque() # Process first k (or first window) # elements of array for i in range(k): # For every element, the previous # smaller elements are useless # so remove them from Qi while Qi and arr[i] >= arr[Qi[-1]] : Qi.pop() # Add new element at rear of queue Qi.append(i); # Process rest of the elements, i.e. # from arr[k] to arr[n-1] for i in range(k, n): # The element at the front of the # queue is the largest element of # previous window, so print it print(str(arr[Qi[0]]) + " ", end = "") # Remove the elements which are # out of this window while Qi and Qi[0] <= i-k: # remove from front of deque Qi.popleft() # Remove all elements smaller than # the currently being added element # (Remove useless elements) while Qi and arr[i] >= arr[Qi[-1]] : Qi.pop() # Add current element at the rear of Qi Qi.append(i) # Print the maximum element of last window print(str(arr[Qi[0]])) # Driver programm to test above fumctions if __name__=="__main__": arr = [12, 1, 78, 90, 57, 89, 56] k = 3 printMax(arr, len(arr), k) # This code is contributed by Shiv Shankar

Output:

78 90 90 90 89

Time Complexity: O(n). It seems more than O(n) at first look. If we take a closer look, we can observe that every element of array is added and removed at most once. So there are total 2n operations.

Auxiliary Space: O(k)

Below is an extension of this problem.

Sum of minimum and maximum elements of all subarrays of size k.

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.