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

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

Examples :Â

Input: arr[] = {1, 2, 3, 1, 4, 5, 2, 3, 6}, K = 3Â
Output: 3 3 4 5 5 5 6
Explanation: Maximum of 1, 2, 3 is 3
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Maximum of 2, 3, 1 is 3
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Maximum of 3, 1, 4 is 4
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Maximum of 1, 4, 5 is 5
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Maximum of 4, 5, 2 is 5Â
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Maximum of 5, 2, 3 is 5
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Maximum of 2, 3, 6 is 6

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

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### Naive Approach:

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.Â

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 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 codeint 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

## 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 Codeint 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;}

## 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 codeif __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

## 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

## 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 callprintKMax(\$arr, \$N, \$K); // This Code is Contributed by anuj_67.?>

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 Max-Heap:Â

• Initialize an empty priority queue heap to store elements in decreasing order of their values, along with their indices.
• Push the first k elements of the input array arr into the priority queue heap along with their respective indices.
• The maximum element in the first window is obtained by accessing the top element of the priority queue heap. Push this maximum element into the answer vector ans.
• Process the remaining elements of arr starting from index k:
• Add the current element along with its index to the priority queue heap.
• Remove elements from the priority queue heap that are outside the current window. This is done by comparing the index of the top element in the heap with the index i – k. If the index of the top element is less than or equal to i – k, it means the element is outside the current window and should be removed.
• The maximum element in the current window is obtained by accessing the top element of the priority queue heap. Push this maximum element into the answer vector ans.
• Finally, return the answer vector ans containing the maximum elements in each sliding window.

Below is the implementation of the above approach:

## C++

 #include using namespace std; // Function to find the maximum element in each sliding// window of size kvector maxSlidingWindow(vector& arr, int k){    vector ans;    priority_queue > heap;     // Initialize the heap with the first k elements    for (int i = 0; i < k; i++)        heap.push({ arr[i], i });     // The maximum element in the first window    ans.push_back(heap.top().first);     // Process the remaining elements    for (int i = k; i < arr.size(); i++) {         // Add the current element to the heap        heap.push({ arr[i], i });         // Remove elements that are outside the current        // window        while (heap.top().second <= i - k)            heap.pop();         // The maximum element in the current window        ans.push_back(heap.top().first);    }     return ans;} int main(){    vector arr = { 2, 3, 7, 9, 5, 1, 6, 4, 3 };    int k = 3;     // Find the maximum element in each sliding window of    // size k    vector result = maxSlidingWindow(arr, k);     // Print the results    for (auto i : result)        cout << i << " ";     return 0;}

## Java

 import java.util.ArrayList;import java.util.List;import java.util.PriorityQueue; public class Main {    public static List maxSlidingWindow(int[] arr,                                                 int k)    {        List ans = new ArrayList<>();        PriorityQueue heap = new PriorityQueue<>(            (a, b) -> b.value - a.value);         // Initialize the heap with the first k elements        for (int i = 0; i < k; i++) {            heap.offer(new Pair(arr[i], i));        }         // The maximum element in the first window        ans.add(heap.peek().value);         // Process the remaining elements        for (int i = k; i < arr.length; i++) {            heap.offer(new Pair(arr[i], i));             // Remove elements that are outside the current            // window            while (heap.peek().index <= i - k) {                heap.poll();            }             // The maximum element in the current window            ans.add(heap.peek().value);        }         return ans;    }     static class Pair {        int value;        int index;         public Pair(int value, int index)        {            this.value = value;            this.index = index;        }    }     public static void main(String[] args)    {        int[] arr = { 2, 3, 7, 9, 5, 1, 6, 4, 3 };        int k = 3;         // Find the maximum element in each sliding window        // of size k        List result = maxSlidingWindow(arr, k);         // Print the results        for (int num : result) {            System.out.print(num + " ");        }    }}

## Python3

 import heapq  def max_sliding_window(arr, k):    ans = []    heap = []     # Initialize the heap with the first k elements    for i in range(k):        heapq.heappush(heap, (-arr[i], i))     # The maximum element in the first window    ans.append(-heap[0][0])     # Process the remaining elements    for i in range(k, len(arr)):        heapq.heappush(heap, (-arr[i], i))         # Remove elements that are outside the current window        while heap[0][1] <= i - k:            heapq.heappop(heap)         # The maximum element in the current window        ans.append(-heap[0][0])     return ans  arr = [2, 3, 7, 9, 5, 1, 6, 4, 3]k = 3 # Find the maximum element in each sliding window of size kresult = max_sliding_window(arr, k) # Print the resultsfor num in result:    print(num, end=" ")

## C#

 using System;using System.Collections.Generic; class GFG{    static List MaxSlidingWindow(List arr, int k)    {        List ans = new List();        Queue deque = new Queue();         for (int i = 0; i < arr.Count; i++)        {            // Remove elements that are outside the current window           // from the front of the deque            while (deque.Count > 0 && deque.Peek() < i - k + 1)                deque.Dequeue();             // Remove elements that are less than the current element           // from the back of the deque            while (deque.Count > 0 && arr[deque.Peek()] < arr[i])                deque.Dequeue();             // Add the current element's index to the back of the deque            deque.Enqueue(i);             // If the current index is equal to or greater than k - 1,           // then we can start adding elements to the result list            if (i >= k - 1)                ans.Add(arr[deque.Peek()]);        }         return ans;    }     static void Main()    {        List arr = new List { 2, 3, 7, 9, 5, 1, 6, 4, 3 };        int k = 3;         // Find the maximum element in each sliding window of size k        List result = MaxSlidingWindow(arr, k);         // Print the results        foreach (int i in result)            Console.Write(i + " ");         Console.ReadLine();    }}

## Javascript

 class Pair {    constructor(value, index) {        this.value = value;        this.index = index;    }} function maxSlidingWindow(arr, k) {    const ans = [];    const heap = [];     // Initialize the heap with the first k elements    for (let i = 0; i < k; i++) {        heap.push(new Pair(arr[i], i));    }    heap.sort((a, b) => b.value - a.value);     // The maximum element in the first window    ans.push(heap[0].value);     // Process the remaining elements    for (let i = k; i < arr.length; i++) {        heap.push(new Pair(arr[i], i));         // Remove elements that are outside the current window        while (heap[0].index <= i - k) {            heap.shift();        }        heap.sort((a, b) => b.value - a.value);         // The maximum element in the current window        ans.push(heap[0].value);    }     return ans;} const arr = [2, 3, 7, 9, 5, 1, 6, 4, 3];const k = 3; // Find the maximum element in each sliding window of size kconst result = maxSlidingWindow(arr, k); // Print the resultsfor (const num of result) {    console.log(num + " ");}

Output
7 9 9 9 6 6 6

Time Complexity: O(NlogN), Where N is the size of the array.
Auxiliary Space: O(N), where N is the size of the array, this method requires O(N) space in the worst case when the input array is an increasing array

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

To reduce the auxilary space to O(K), Set Data Structure can be used which allows deletion of any element in O(logn N) time. Thus the size of set will never exceed K, if we delete delete the (i-Kt)h element from the Set while traversing.

Follow the given steps to solve the problem:

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

Below is the implementation of the above approach:

## C++

 #include using namespace std; // Function to find the maximum element in each sliding// window of size kvector maxSlidingWindow(vector& arr, int k){    vector ans;    set, greater>> st;     // Initialize the set with the first k elements    for (int i = 0; i < k; i++)        st.insert({ arr[i], i });     // The maximum element in the first window    ans.push_back((st.begin())->first);     // Process the remaining elements    for (int i = k; i < arr.size(); i++) {         // Add the current element to the set        st.insert({ arr[i], i });         // Remove the (i-k)th element from the window        st.erase({ arr[i - k], (i - k) });         // The maximum element in the current window        ans.push_back(st.begin()->first);    }     return ans;} int main(){    vector arr = { 2, 3, 7, 9, 5, 1, 6, 4, 3 };    int k = 3;     // Find the maximum element in each sliding window of    // size k    vector result = maxSlidingWindow(arr, k);     // Print the results    for (auto i : result)        cout << i << " ";     return 0;}

## Java

 import java.util.ArrayDeque;import java.util.Deque; public class MaxSlidingWindow {    // Function to find the maximum element in each sliding    // window of size k    static int[] maxSlidingWindow(int[] arr, int k)    {        int n = arr.length;        int[] ans = new int[n - k + 1];        Deque maxInWindow = new ArrayDeque<>();         // Initialize maxInWindow with the first k elements        for (int i = 0; i < k; i++) {            while (!maxInWindow.isEmpty()                   && arr[i]                          >= arr[maxInWindow.peekLast()]) {                maxInWindow.removeLast();            }            maxInWindow.addLast(i);        }         // The maximum element in the first window        ans[0] = arr[maxInWindow.peekFirst()];         // Process the remaining elements        for (int i = k; i < n; i++) {            // Remove elements that are out of the current            // window            while (!maxInWindow.isEmpty()                   && maxInWindow.peekFirst() <= i - k) {                maxInWindow.removeFirst();            }             // Remove elements that are not needed in the            // current window            while (!maxInWindow.isEmpty()                   && arr[i]                          >= arr[maxInWindow.peekLast()]) {                maxInWindow.removeLast();            }             maxInWindow.addLast(i);            ans[i - k + 1] = arr[maxInWindow.peekFirst()];        }         return ans;    }     // Driver program    public static void main(String[] args)    {        int[] arr = { 2, 3, 7, 9, 5, 1, 6, 4, 3 };        int k = 3;         // Find the maximum element in each sliding window        // of size k        int[] result = maxSlidingWindow(arr, k);         // Print the results        System.out.print(            "Maximum elements in each sliding window: ");        for (int i : result)            System.out.print(i + " ");    }}

## Python

 from collections import deque # Function to find the maximum element in each sliding window of size k  def maxSlidingWindow(arr, k):    ans = []    deq = deque()     # Initialize the deque with the first k elements    for i in range(k):        while deq and arr[i] >= arr[deq[-1]]:            deq.pop()        deq.append(i)     # The maximum element in the first window    ans.append(arr[deq[0]])     # Process the remaining elements    for i in range(k, len(arr)):        # Remove elements that are out of the current window        while deq and deq[0] <= i - k:            deq.popleft()         # Remove elements that are less than the current element        while deq and arr[i] >= arr[deq[-1]]:            deq.pop()         deq.append(i)         # The maximum element in the current window        ans.append(arr[deq[0]])     return ans  # Main functionif __name__ == "__main__":    arr = [2, 3, 7, 9, 5, 1, 6, 4, 3]    k = 3     # Find the maximum element in each sliding window of size k    result = maxSlidingWindow(arr, k)     # Print the results    print(result)

## C#

 using System;using System.Collections.Generic; class Program {    // Function to find the maximum element in each sliding    // window of size k    static List MaxSlidingWindow(List arr, int k)    {        List ans = new List();        List maxInWindow = new List();         // Initialize maxInWindow with the first k elements        for (int i = 0; i < k; i++) {            while (                maxInWindow.Count > 0                && arr[i]                       >= arr[maxInWindow[maxInWindow.Count                                          - 1]]) {                maxInWindow.RemoveAt(maxInWindow.Count - 1);            }            maxInWindow.Add(i);        }         // The maximum element in the first window        ans.Add(arr[maxInWindow[0]]);         // Process the remaining elements        for (int i = k; i < arr.Count; i++) {            // Remove elements that are out of the current            // window            while (maxInWindow.Count > 0                   && maxInWindow[0] <= i - k) {                maxInWindow.RemoveAt(0);            }             // Remove elements that are not needed in the            // current window            while (                maxInWindow.Count > 0                && arr[i]                       >= arr[maxInWindow[maxInWindow.Count                                          - 1]]) {                maxInWindow.RemoveAt(maxInWindow.Count - 1);            }             maxInWindow.Add(i);            ans.Add(arr[maxInWindow[0]]);        }         return ans;    }     static void Main(string[] args)    {        List arr            = new List{ 2, 3, 7, 9, 5, 1, 6, 4, 3 };        int k = 3;         List result = MaxSlidingWindow(arr, k);         Console.WriteLine(            "Maximum elements in each sliding window:");        foreach(int i in result) { Console.Write(i + " "); }    }}

## Javascript

 function maxSlidingWindow(arr, k) {    const ans = [];    const maxInWindow = [];     // Initialize maxInWindow with the first k elements    for (let i = 0; i < k; i++) {        while (maxInWindow.length > 0 && arr[i] >= arr[maxInWindow[maxInWindow.length - 1]]) {            maxInWindow.pop();        }        maxInWindow.push(i);    }     // The maximum element in the first window    ans.push(arr[maxInWindow[0]]);     // Process the remaining elements    for (let i = k; i < arr.length; i++) {        // Remove elements that are out of the current window        while (maxInWindow.length > 0 && maxInWindow[0] <= i - k) {            maxInWindow.shift();        }         // Remove elements that are not needed in the current window        while (maxInWindow.length > 0 && arr[i] >= arr[maxInWindow[maxInWindow.length - 1]]) {            maxInWindow.pop();        }         maxInWindow.push(i);        ans.push(arr[maxInWindow[0]]);    }     return ans;} const arr = [2, 3, 7, 9, 5, 1, 6, 4, 3];const k = 3; const result = maxSlidingWindow(arr, k); console.log("Maximum elements in each sliding window: " + result.join(' '));

Output
7 9 9 9 6 6 6

Time Complexity: O(NlogN), Where N is the size of the array.
Auxiliary Space: O(K), since size of set does not never exceeds K.

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

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:Â

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 kvoid 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 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 codeint 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 codeif __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 Qi = new List();         /* 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 kfunction 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).Â 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 a total of 2n operations.
Auxiliary Space: O(K).Â Elements stored in the dequeue take O(K) space.

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

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