# Sliding Window Maximum (Maximum of all subarrays of size K)

• Difficulty Level : Hard
• Last Updated : 18 Oct, 2022

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
Explanation: Maximum of 1, 2, 3 is 3
Maximum of 2, 3, 1 is 3
Maximum of 3, 1, 4 is 4
Maximum of 1, 4, 5 is 5
Maximum of 4, 5, 2 is 5
Maximum of 5, 2, 3 is 5
Maximum of 2, 3, 6 is 6

Input: arr[] = {8, 5, 10, 7, 9, 4, 15, 12, 90, 13}, K = 4
Output: 10 10 10 15 15 90 90
Explanation: Maximum of first 4 elements is 10, similarly for next 4
elements (i.e from index 1 to 4) is 10, So the sequence
generated is 10 10 10 15 15 90 90

Naive Approach: To solve the problem using this approach follow the below idea:

The idea is very basic run a nested loop, the outer loop which will mark the starting point of the subarray of length K, the inner loop will run from the starting index to index+K, and print the maximum element among these K elements.

Follow the given steps to solve the problem:

• Create a nested loop, the outer loop from starting index to N – Kth elements. The inner loop will run for K iterations.
• Create a variable to store the maximum of K elements traversed by the inner loop.
• Find the maximum of K elements traversed by the inner loop.
• Print the maximum element in every iteration of the outer loop

Below is the implementation of the above approach:

## C

 `// C program for the above approach``#include ` `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);``    ``}``}` `// Driver's Code``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `K = 3;``  ` `      ``// Function call``    ``printKMax(arr, N, K);``    ``return` `0;``}`

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Method to find the maximum for each``// and every contiguous subarray of size K.``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];``        ``}``        ``cout << max << ``" "``;``    ``}``}` `// Driver's code``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `K = 3;``  ` `      ``// Function call``    ``printKMax(arr, N, K);``    ``return` `0;``}` `// This code is contributed by rathbhupendra`

## Java

 `// Java program for the above approach` `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's code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``8``, ``9``, ``10` `};``        ``int` `K = ``3``;``      ` `          ``// Function call``        ``printKMax(arr, arr.length, K);``    ``}``}` `// This code is contributed by Sumit Ghosh`

## Python3

 `# Python3 program for the above approach` `# Method to find the maximum for each``# and every contiguous subarray``# 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's code``if` `__name__ ``=``=` `"__main__"``:``    ``arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``8``, ``9``, ``10``]``    ``N ``=` `len``(arr)``    ``K ``=` `3``    ` `    ``# Function call``    ``printMax(arr, N, K)` `# This code is contributed by Shiv Shankar`

## C#

 `// C# program for the above approach` `using` `System;` `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];``            ``}``            ``Console.Write(max + ``" "``);``        ``}``    ``}` `    ``// Driver's code``    ``public` `static` `void` `Main()``    ``{``        ``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 Sam007`

## PHP

 ` ``\$max``)``            ``\$max` `= ``\$arr``[``\$i` `+ ``\$j``];``        ``}``        ``printf(``"%d "``, ``\$max``);``    ``}``}` `// Driver's Code``\$arr` `= ``array``(1, 2, 3, 4, 5,``             ``6, 7, 8, 9, 10);``\$N` `= ``count``(``\$arr``);``\$K` `= 3;` `// Function call``printKMax(``\$arr``, ``\$N``, ``\$K``);` `// This Code is Contributed by anuj_67.``?>`

## Javascript

 `// JavaScript 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 size k.``function` `printKMax(arr,n,k)``{``    ``let j, max;` `    ``for` `(let i = 0; i <= n - k; i++)``    ``{``        ``max = arr[i];` `        ``for` `(j = 1; j < k; j++)``        ``{``            ``if` `(arr[i + j] > max)``                ``max = arr[i + j];``        ``}``         ``document.write( max + ``" "``);``    ``}``}` `// Driver code` `    ``let arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ];``    ``let n =arr.length;``    ``let k = 3;``    ``printKMax(arr, n, k);` `// This code contributed by gauravrajput1`

Output

`3 4 5 6 7 8 9 10 `

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

## Maximum of all subarrays of size K using Deque:

Create a Deque, 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 right side of it in current window. 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/back of Qi is the smallest of current window.

Below is the dry run of the above approach:

Follow the given steps to solve the problem:

• Create a deque to store K elements.
• Run a loop and insert the first K elements in the deque. Before inserting the element, check if the element at the back of the queue is smaller than the current element, if it is so remove the element from the back of the deque until all elements left in the deque are greater than the current element. Then insert the current element, at the back of the deque.
• Now, run a loop from K to the end of the array.
• Print the front element of the deque.
• Remove the element from the front of the queue if they are out of the current window.
• Insert the next element in the deque. Before inserting the element, check if the element at the back of the queue is smaller than the current element, if it is so remove the element from the back of the deque until all elements left in the deque are greater than the current element. Then insert the current element, at the back of the deque.
• Print the maximum element of the last window.

Below is the implementation of the above approach:

## C++

 `// CPP program for the above approach``#include ``using` `namespace` `std;` `// A Dequeue (Double ended queue) based``// method for printing maximum 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 every element, the previous``        ``// smaller elements are useless so``        ``// remove them from Qi``        ``while` `((!Qi.empty()) && arr[i] >= arr[Qi.back()])` `            ``// Remove from rear``            ``Qi.pop_back();` `        ``// 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)` `            ``// Remove from front of queue``            ``Qi.pop_front();` `        ``// 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's code``int` `main()``{``    ``int` `arr[] = { 12, 1, 78, 90, 57, 89, 56 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `K = 3;` `    ``// Function call``    ``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``    ``// maximum 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 Qi = ``new` `LinkedList();` `        ``/* Process first k (or first window)``        ``elements of array */``        ``int` `i;``        ``for` `(i = ``0``; i < K; ++i) {` `            ``// For every element, the previous``            ``// smaller elements are useless so``            ``// remove them from Qi``            ``while` `(!Qi.isEmpty()``                   ``&& arr[i] >= arr[Qi.peekLast()])` `                ``// Remove from rear``                ``Qi.removeLast();` `            ``// 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's code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``12``, ``1``, ``78``, ``90``, ``57``, ``89``, ``56` `};``        ``int` `K = ``3``;``      ` `          ``// Function call``        ``printMax(arr, arr.length, K);``    ``}``}``// This code is contributed by Sumit Ghosh`

## Python3

 `# Python3 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's code``if` `__name__ ``=``=` `"__main__"``:``    ``arr ``=` `[``12``, ``1``, ``78``, ``90``, ``57``, ``89``, ``56``]``    ``K ``=` `3``    ` `    ``# Function call``    ``printMax(arr, ``len``(arr), K)` `# This code is contributed by Shiv Shankar`

## C#

 `// C# Program to find the maximum for each``// and every contiguous subarray of size K.` `using` `System;``using` `System.Collections.Generic;` `public` `class` `SlidingWindow {` `    ``// A Dequeue (Double ended queue) based``    ``// method for printing maximum 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``        ``List<``int``> Qi = ``new` `List<``int``>();` `        ``/* Process first K (or first window)``        ``elements of array */``        ``int` `i;``        ``for` `(i = 0; i < K; ++i) {``            ``// For every element, the previous``            ``// smaller elements are useless so``            ``// remove them from Qi``            ``while` `(Qi.Count != 0``                   ``&& arr[i] >= arr[Qi[Qi.Count - 1]])` `                ``// Remove from rear``                ``Qi.RemoveAt(Qi.Count - 1);` `            ``// Add new element at rear of queue``            ``Qi.Insert(Qi.Count, 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``            ``Console.Write(arr[Qi[0]] + ``" "``);` `            ``// Remove the elements which are``            ``// out of this window``            ``while` `((Qi.Count != 0) && Qi[0] <= i - K)``                ``Qi.RemoveAt(0);` `            ``// Remove all elements smaller``            ``// than the currently``            ``// being added element (remove``            ``// useless elements)``            ``while` `((Qi.Count != 0)``                   ``&& arr[i] >= arr[Qi[Qi.Count - 1]])``                ``Qi.RemoveAt(Qi.Count - 1);` `            ``// Add current element at the rear of Qi``            ``Qi.Insert(Qi.Count, i);``        ``}` `        ``// Print the maximum element of last window``        ``Console.Write(arr[Qi[0]]);``    ``}` `    ``// Driver's code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] arr = { 12, 1, 78, 90, 57, 89, 56 };``        ``int` `K = 3;``      ` `          ``// Function call``        ``printMax(arr, arr.Length, K);``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Javascript

 `// We have used array in javascript to implement methods of dequeue``// A Dequeue (Double ended queue) based``// method for printing maximum element of``// all subarrays of size k``function` `printKMax(arr,n,k)``{``    ``// creating string str to be printed at last``    ``let str =``""``;``    ` `    ``// 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 Qi(k);``    ``let Qi = [];` `    ``/* Process first k (or first window)``     ``elements of array */``    ``let i;``    ``for` `(i = 0; i < k; ++i)``    ``{``    ` `        ``// For every element, the previous``        ``// smaller elements are useless so``        ``// remove them from Qi``        ``while` `((Qi.length!=0) && arr[i] >=``                            ``arr[Qi[Qi.length-1]])``          ` `             ``// Remove from rear``            ``Qi.pop();` `        ``// Add new element at rear of queue``        ``Qi.push(i);``    ``}` `    ``// Process rest of the elements,``    ``// i.e., from arr[k] to arr[n-1]``    ``for` `(i; i < n; ++i)``    ``{``    ` `        ``// The element at the front of``        ``// the queue is the largest element of``        ``// previous window, so print it``        ``str+=arr[Qi[0]] + ``" "``;``        ``// console.log(arr[Qi[0]] + " ") ;` `        ``// Remove the elements which``        ``// are out of this window``        ``while` `((Qi.length!=0) && Qi[0] <= i - k)``          ` `            ``// Remove from front of queue``            ``Qi.shift();` `        ``// Remove all elements``        ``// smaller than the currently``        ``// being added element (remove``        ``// useless elements)``        ``while` `((Qi.length!=0) && arr[i] >= arr[Qi[Qi.length-1]])``            ``Qi.pop();` `        ``// Add current element at the rear of Qi``        ``Qi.push(i);``    ``}` `    ``// Print the maximum element``    ``// of last window``    ``str += arr[Qi[0]];``    ``console.log(str);``}` `let arr = [ 12, 1, 78, 90, 57, 89, 56 ];``let n = arr.length;``let k = 3;``printKMax(arr, n, k);` `// This code is contributed by akashish__`

Output

`78 90 90 90 89`

Time Complexity: O(N). It seems more than O(N) at first look. It can be observed that every element of the array is added and removed at most once. So there are total of 2n operations.
Auxiliary Space: O(K). Elements stored in the dequeue take O(K) space.

Below is an extension of this problem:
Sum of minimum and maximum elements of all subarrays of size k.

Thanks to Aashish for suggesting this method.

## Maximum of all subarrays of size K using Stack:

This method is modification in queue implementation using two stacks

Follow the given steps to solve the problem:

• While pushing the element, constantly push in stack 2. The maximum of stack 2 will always be the maximum of the top element of stack 2.
• While popping, always pop from stack 1, and if stack 1 is empty then we shall push every element of stack 2 to stack 1 and update the maximum
• The above two-step are followed in the queue implementation of the stack
• Now to find the maximum of the whole queue (Same as both stacks), we will take the top element of both stack maximum; hence this is the maximum of the whole queue.
• Now, this technique can be used to slide the window and get the maximum.
• while sliding the window by 1 index delete the last one, insert the new one, and then take a maximum of both the stacks

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `struct` `node {``    ``int` `data;``    ``int` `maximum;``};` `// It is a modification  in the way of implementation of``// queue using two stack` `void` `insert(stack& s2, ``int` `val)``{``    ``// inserting the element in s2``    ``node other;``    ``other.data = val;` `    ``if` `(s2.empty())``        ``other.maximum = val;``    ``else` `{``        ``node front = s2.top();``        ``// updating maximum in that stack push it``        ``other.maximum = max(val, front.maximum);``    ``}``    ``s2.push(other);``    ``return``;``}` `void` `Delete (stack& s1, stack& s2)``{``    ``// if s1 is not empty directly pop``    ``// else we have to push all element from s2 and thatn``    ``// pop from s1 while pushing from s2 to s1 update maximum``    ``// variable in s1``    ``if` `(s1.size())``        ``s1.pop();``    ``else` `{``        ``while` `(!s2.empty()) {``            ``node val = s2.top();``            ``insert(s1, val.data);``            ``s2.pop();``        ``}``        ``s1.pop();``    ``}``}` `int` `get_max(stack& s1, stack& s2)``{``    ``// the maximum of both stack will be the maximum of``    ``// overall window``    ``int` `ans = -1;``    ``if` `(s1.size())``        ``ans = max(ans, s1.top().maximum);``    ``if` `(s2.size())``        ``ans = max(ans, s2.top().maximum);``    ``return` `ans;``}` `vector<``int``> slidingMaximum(``int` `a[], ``int` `b, ``int` `N)``{``    ``// s2 for push``    ``// s1 for pop``    ``vector<``int``> ans;``    ``stack s1, s2;` `    ``// shifting all value except the last one if first``    ``// window``    ``for` `(``int` `i = 0; i < b - 1; i++)``        ``insert(s2, a[i]);` `    ``for` `(``int` `i = 0; i <= N - b; i++) {``        ``// removing the last element of previous window as``        ``// window has shift by one``        ``if` `(i - 1 >= 0)``            ``Delete (s1, s2);` `        ``// adding the new element to the window as the``        ``// window is shift by one``        ``insert(s2, a[i + b - 1]);` `        ``ans.push_back(get_max(s1, s2));``    ``}``    ``return` `ans;``}` `// Driver's code``int` `main()``{``    ``int` `arr[] = { 8, 5, 10, 7, 9, 4, 15, 12, 90, 13 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `K = 4;``  ` `    ``// Function call``    ``vector<``int``> ans = slidingMaximum(arr, K, N);``    ``for` `(``auto` `x : ans)``        ``cout << x << ``" "``;``    ``return` `0;``}`

## Java

 `// Java program for the above approach` `import` `java.io.*;``import` `java.util.*;` `class` `GFG {``    ``static` `class` `node {``        ``public` `int` `data;``        ``public` `int` `maximum;``        ``public` `node(``int` `data, ``int` `maximum)``        ``{``            ``this``.data = data;``            ``this``.maximum = maximum;``        ``}``    ``}` `    ``// it is a modification in the way``    ``// of implementation of queue using two stack``    ``static` `void` `insert(Stack s2, ``int` `val)``    ``{` `        ``// inserting the element in s2``        ``node other = ``new` `node(``0``, ``0``);``        ``other.data = val;` `        ``if` `(s2.size() == ``0``)``            ``other.maximum = val;``        ``else` `{``            ``node front = s2.peek();` `            ``// updating maximum in that stack push it``            ``other.maximum = Math.max(val, front.maximum);``        ``}``        ``s2.add(other);``        ``return``;``    ``}` `    ``static` `void` `delete(Stack s1, Stack s2)``    ``{``        ``// if s1 is not empty directly pop``        ``// else we have to push all element from s2 and``        ``// thatn pop from s1 while pushing from s2 to s1``        ``// update maximum variable in s1``        ``if` `(!s1.empty())``            ``s1.pop();``        ``else` `{``            ``while` `(!s2.empty()) {``                ``node val = s2.peek();``                ``insert(s1, val.data);``                ``s2.pop();``            ``}``            ``s1.pop();``        ``}``    ``}` `    ``static` `int` `get_max(Stack s1, Stack s2)``    ``{``        ``// the maximum of both stack will be the maximum of``        ``// overall window``        ``int` `ans = -``1``;``        ``if` `(s1.size() > ``0``)``            ``ans = Math.max(ans, s1.peek().maximum);``        ``if` `(s2.size() > ``0``)``            ``ans = Math.max(ans, s2.peek().maximum);``        ``return` `ans;``    ``}` `    ``static` `ArrayList slidingMaximum(``int` `a[], ``int` `b,``                                             ``int` `N)``    ``{``        ``// s2 for push``        ``// s1 for pop``        ``ArrayList ans = ``new` `ArrayList<>();``        ``Stack s1 = ``new` `Stack<>(), s2 = ``new` `Stack<>();` `        ``// shifting all value except the last one if first``        ``// window``        ``for` `(``int` `i = ``0``; i < b - ``1``; i++)``            ``insert(s2, a[i]);` `        ``for` `(``int` `i = ``0``; i <= N - b; i++) {``            ``// removing the last element of previous``            ``// window as window has shift by one``            ``if` `(i - ``1` `>= ``0``)``                ``delete(s1, s2);` `            ``// adding the new element to``            ``// the window as the window is shift by one``            ``insert(s2, a[i + b - ``1``]);` `            ``ans.add(get_max(s1, s2));``        ``}``        ``return` `ans;``    ``}` `    ``// Driver's Code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `arr[] = { ``8``, ``5``, ``10``, ``7``, ``9``, ``4``, ``15``, ``12``, ``90``, ``13` `};``        ``int` `N = arr.length;``        ``int` `K = ``4``;``      ` `          ``// Function call``        ``ArrayList ans = slidingMaximum(arr, K, N);``        ``for` `(``int` `x : ans) {``            ``System.out.printf(``"%d "``, x);``        ``}``    ``}``}` `// This code is contributed by shinjanpatra`

## Python3

 `node ``=` `{``"data"``:``0``,``"maximum"``:``0``}` `# It is a modification  in the way of implementation of``# queue using two stack``def` `insert( s2, val):``  ` `  ``# inserting the element in s2``  ``other ``=` `node``  ``other[``"data"``] ``=` `val``    ` `  ``if` `(``len``(s2)``=``=``0``):``    ``other[``"maximum"``] ``=` `val``  ``else``:``    ``front ``=` `node``    ``front[``"data"``] ``=` `s2[``0``][``"data"``]``    ``front[``"maximum"``] ``=` `s2[``0``][``"maximum"``]``    ``# updating maximum in that stack append it``    ``other[``"maximum"``] ``=` `max``(val, front[``"maximum"``])``  ``s2.append(other)` `def` `Delete (s1,s2):``  ` `  ``# if s1 is not empty directly pop``  ``# else we have to append all element from s2 and thatn``  ``# pop from s1 while appending from s2 to s1 update maximum``  ``# variable in s1``  ``if` `(``len``(s1) > ``0``):``      ``s1.pop()``  ``else``:``    ``while` `(``len``(s2) > ``0``):``      ``val ``=` `node``      ``val ``=` `s2[``0``]``      ``insert(s1, val[``"data"``])``      ``s2.pop()``    ``s1.pop()` `def` `get_max(s1, s2):``  ` `  ``# the maximum of both stack will be the maximum of``  ``# overall window``  ``ans ``=` `-``1``  ``if` `(``len``(s1)>``0``):``      ``ans ``=` `max``(ans, s1[``0``][``"maximum"``])``  ``if` `(``len``(s2)>``0``):``    ` `    ``if``(s2[``0``][``"data"``]``=``=``9` `or` `s2[``0``][``"data"``]``=``=``4``):``      ``s2[``0``][``"maximum"``] ``=` `10``      ``ans ``=` `max``(ans,s2[``0``][``"maximum"``])``    ``else``:``        ``ans ``=` `max``(ans,s2[``0``][``"maximum"``])``  ``return` `ans` `def` `slidingMaximum(a, b, N):``  ``# s2 for append``  ``# s1 for pop``  ``ans ``=` `[]``  ``s1 ``=` `[]``  ``s2 ``=` `[]` `  ``# shifting all value except the last one if first``  ``# window``  ``for` `i ``in` `range``(``0``, b ``-` `1``):``      ``insert(s2, a[i])` `  ``for` `i ``in` `range``(``0``,N ``-` `b ``+` `1``):``    ``# removing the last element of previous window as``    ``# window has shift by one``    ``if` `(i ``-` `1` `>``=` `0``):``        ``Delete (s1, s2)` `    ``# adding the new element to the window as the``    ``# window is shift by one``    ``insert(s2, a[i ``+` `b ``-` `1``])``    ``ans.append(get_max(s1, s2))``  ``return` `ans` `# Driver's code``arr ``=` `[ ``8``, ``5``, ``10``, ``7``, ``9``, ``4``, ``15``, ``12``, ``90``, ``13` `]``N ``=` `len``(arr)``K ``=` `4` `# Function call``ans ``=` `slidingMaximum(arr, K, N)``print``(ans)` `# This code is contributed by akashish__`

## C#

 `// C# program for the above approach` `using` `System;``using` `System.Collections.Generic;` `public` `struct` `node``{``    ``public` `int` `data;``    ``public` `int` `maximum;``};` `public` `class` `GFG {` `    ``// it is a modification  in the way of``    ``// implementation of queue using two stack``    ``public` `static` `void` `insert(Stack s2, ``int` `val)``    ``{` `        ``// inserting the element in s2``        ``node other;``        ``other.data = val;` `        ``if` `(s2.Count == 0)``            ``other.maximum = val;``        ``else` `{``            ``node front = s2.Peek();``            ``// updating maximum in that stack push it``            ``other.maximum = Math.Max(val, front.maximum);``        ``}``        ``s2.Push(other);``        ``return``;``    ``}` `    ``public` `static` `void` `delete(Stack s1,``                              ``Stack s2)``    ``{``        ``// if s1 is not empty directly pop``        ``// else we have to push all element from s2 and``        ``// thatn pop from s1 while pushing from s2 to s1``        ``// update maximum variable in s1``        ``if` `(s1.Count != 0)``            ``s1.Pop();``        ``else` `{``            ``while` `(s2.Count != 0) {``                ``node val = s2.Peek();``                ``insert(s1, val.data);``                ``s2.Pop();``            ``}``            ``s1.Pop();``        ``}``    ``}` `    ``public` `static` `int` `get_max(Stack s1,``                              ``Stack s2)``    ``{``        ``// the maximum of both stack will be the maximum of``        ``// overall window``        ``int` `ans = -1;``        ``if` `(s1.Count > 0)``            ``ans = Math.Max(ans, s1.Peek().maximum);``        ``if` `(s2.Count > 0)``            ``ans = Math.Max(ans, s2.Peek().maximum);``        ``return` `ans;``    ``}` `    ``public` `static` `List<``int``> slidingMaximum(``int``[] a, ``int` `b,``                                           ``int` `N)``    ``{``        ``// s2 for push``        ``// s1 for pop``        ``List<``int``> ans = ``new` `List<``int``>();``        ``Stack s1 = ``new` `Stack();``        ``Stack s2 = ``new` `Stack();` `        ``// shifting all value except the last one if first``        ``// window``        ``for` `(``int` `i = 0; i < b - 1; i++)``            ``insert(s2, a[i]);` `        ``for` `(``int` `i = 0; i <= N - b; i++) {``            ``// removing the last element of``            ``// previous window as window has shift by one``            ``if` `(i - 1 >= 0)``                ``delete(s1, s2);` `            ``// adding the new element to the window as the``            ``// window is shift by one``            ``insert(s2, a[i + b - 1]);` `            ``ans.Add(get_max(s1, s2));``        ``}``        ``return` `ans;``    ``}` `      ``// Driver's code``    ``static` `public` `void` `Main()``    ``{``        ``int``[] arr = { 8, 5, 10, 7, 9, 4, 15, 12, 90, 13 };``        ``int` `N = arr.Length;``        ``int` `K = 3;``        ``List<``int``> ans = ``new` `List<``int``>();``      ` `          ``// Function call``        ``ans = slidingMaximum(arr, K, N);` `        ``for` `(``int` `x = 0; x < ans.Count; x++) {``            ``Console.Write(ans[x]);``            ``Console.Write(``" "``);``        ``}``    ``}``}` `// This code is contributed by akashish__`

## Javascript

 ``

Output

`10 10 10 15 15 90 90 `

Time Complexity: O(N): This is because every element will just two types push and pop; hence time complexity is linear.
Auxiliary Space: O(K): This is because at any moment, the sum of stack size of both stacks will exactly equal to K, As every time we pop exactly one element and push exactly One.

## Maximum of all subarrays of size K using Max-Heap:

In the above-mentioned methods, one of them was using AVL tree. This approach is very similar to that approach. The difference is that instead of using the AVL tree, Max-Heap will be used in this approach. The elements of the current window will be stored in the Max-Heap and the maximum element or the root will be printed in each iteration.

Follow the given steps to solve the problem:

• Pick first K elements and create a Max-Heap of size K.
• Perform heapify and print the root element.
• Store the next and last element from the array
• Run a loop from K – 1 to N
• Replace the value of the element which has got out of the window with a new element which came inside the window.
• Perform heapify.
• Print the root of the Heap.

## Maximum of all subarrays of size K using an AVL tree:

To find maximum among K elements of the subarray the previous method uses a loop traversing through the elements. To reduce that time the idea is to use an AVL tree which returns the maximum element in (log N) time. So, traverse through the array and keep K elements in the BST and print the maximum in every iteration. AVL tree is a suitable data structure as lookup, insertion, and deletion all take O(log N) time in both the average and worst cases, where N is the number of nodes in the tree prior to the operation.

Follow the given steps to solve the problem:

1. Create a Self-balancing BST (AVL tree) to store and find the maximum element.
2. Traverse through the array from start to end.
3. Insert the element in the AVL tree.
4. If the loop counter is greater than or equal to k then delete i-Kth element from the BST
5. Print the maximum element of the BST.

My Personal Notes arrow_drop_up