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 :

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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1: This is the simple method to solve the above problem.

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, k elements from starting index and print the maximum element among these k elements.
Algorithm:
1. Create a nested loop, the outer loop from starting index to n – k th elements. The inner loop will run for k iterations.
2. Create a variable to store the maximum of k elements traversed by the inner loop.
3. Find the maximum of k elements traversed by the inner loop.
4. Print the maximum element in every iteration of outer loop

Implementation:

C++

// C++ Program to find the maximum for  
// each and every contiguous subarray of size k. 
#include <bits/stdc++.h> 
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 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;  
    printKMax(arr, n, k);  
    return 0;  
} 
  
// This code is contributed by rathbhupendra 

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  

C#

// C# program to find the maximum for 
// each and every contiguous subarray of 
// size kusing System; 
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 method 
    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

<?php 
// PHP program to find the maximum  
// for each and every contiguous  
// subarray of size k 
  
function printKMax($arr, $n, $k) 
{ 
    $j; $max; 
  
    for ($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 Code 
$arr = array(1, 2, 3, 4, 5,  
             6, 7, 8, 9, 10); 
$n = count($arr); 
$k = 3; 
printKMax($arr, $n, $k); 
  
// This Code is Contributed by anuj_67. 
?> 

Output:

3 4 5 6 7 8 9 10

Complexity Analysis:
- Time Complexity: O(N * K).
  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(N * K).
- Space Complexity: O(1).
  No extra space is required.

Method 2: This method uses the uses the Self-Balancing BST to solve the given problem.

Approach:
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.
Algorithm:
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 or is greater than or equal to k then delete i-k th element from the BST
5. Print the maximum element of the BST.

Implementation:

// C++ program to delete a node from AVL Tree  
#include<bits/stdc++.h>  
using namespace std;  
  
// An AVL tree node  
class Node  
{  
    public:  
    int key;  
    Node *left;  
    Node *right;  
    int height;  
};  
  
// A utility function to get maximum  
// of two integers  
int max(int a, int b);  
  
// A utility function to get height  
// of the tree  
int height(Node *N)  
{  
    if (N == NULL)  
        return 0;  
    return N->height;  
}  
  
// A utility function to get maximum  
// of two integers  
int max(int a, int b)  
{  
    return (a > b)? a : b;  
}  
  
/* Helper function that allocates a  
new node with the given key and  
NULL left and right pointers. */
Node* newNode(int key)  
{  
    Node* node = new Node();  
    node->key = key;  
    node->left = NULL;  
    node->right = NULL;  
    node->height = 1; // new node is initially  
                    // added at leaf  
    return(node);  
}  
  
// A utility function to right  
// rotate subtree rooted with y  
// See the diagram given above.  
Node *rightRotate(Node *y)  
{  
    Node *x = y->left;  
    Node *T2 = x->right;  
  
    // Perform rotation  
    x->right = y;  
    y->left = T2;  
  
    // Update heights  
    y->height = max(height(y->left),  
                    height(y->right)) + 1;  
    x->height = max(height(x->left),  
                    height(x->right)) + 1;  
  
    // Return new root  
    return x;  
}  
  
// A utility function to left  
// rotate subtree rooted with x  
// See the diagram given above.  
Node *leftRotate(Node *x)  
{  
    Node *y = x->right;  
    Node *T2 = y->left;  
  
    // Perform rotation  
    y->left = x;  
    x->right = T2;  
  
    // Update heights  
    x->height = max(height(x->left),  
                    height(x->right)) + 1;  
    y->height = max(height(y->left),  
                    height(y->right)) + 1;  
  
    // Return new root  
    return y;  
}  
  
// Get Balance factor of node N  
int getBalance(Node *N)  
{  
    if (N == NULL)  
        return 0;  
    return height(N->left) -  
        height(N->right);  
}  
  
Node* insert(Node* node, int key)  
{  
    /* 1. Perform the normal BST rotation */
    if (node == NULL)  
        return(newNode(key));  
  
    if (key < node->key)  
        node->left = insert(node->left, key);  
    else if (key > node->key)  
        node->right = insert(node->right, key);  
    else // Equal keys not allowed  
        return node;  
  
    /* 2. Update height of this ancestor node */
    node->height = 1 + max(height(node->left),  
                        height(node->right));  
  
    /* 3. Get the balance factor of this  
        ancestor node to check whether  
        this node became unbalanced */
    int balance = getBalance(node);  
  
    // If this node becomes unbalanced,  
    // then there are 4 cases  
  
    // Left Left Case  
    if (balance > 1 && key < node->left->key)  
        return rightRotate(node);  
  
    // Right Right Case  
    if (balance < -1 && key > node->right->key)  
        return leftRotate(node);  
  
    // Left Right Case  
    if (balance > 1 && key > node->left->key)  
    {  
        node->left = leftRotate(node->left);  
        return rightRotate(node);  
    }  
  
    // Right Left Case  
    if (balance < -1 && key < node->right->key)  
    {  
        node->right = rightRotate(node->right);  
        return leftRotate(node);  
    }  
  
    /* return the (unchanged) node pointer */
    return node;  
}  
  
/* Given a non-empty binary search tree,  
return the node with minimum key value  
found in that tree. Note that the entire  
tree does not need to be searched. */
Node * minValueNode(Node* node)  
{  
    Node* current = node;  
  
    /* loop down to find the leftmost leaf */
    while (current->left != NULL)  
        current = current->left;  
  
    return current;  
}  
  
// Recursive function to delete a node  
// with given key from subtree with  
// given root. It returns root of the  
// modified subtree.  
Node* deleteNode(Node* root, int key)  
{  
      
    // STEP 1: PERFORM STANDARD BST DELETE  
    if (root == NULL)  
        return root;  
  
    // If the key to be deleted is smaller  
    // than the root's key, then it lies  
    // in left subtree  
    if ( key < root->key )  
        root->left = deleteNode(root->left, key);  
  
    // If the key to be deleted is greater  
    // than the root's key, then it lies  
    // in right subtree  
    else if( key > root->key )  
        root->right = deleteNode(root->right, key);  
  
    // if key is same as root's key, then  
    // This is the node to be deleted  
    else
    {  
        // node with only one child or no child  
        if( (root->left == NULL) ||  
            (root->right == NULL) )  
        {  
            Node *temp = root->left ?  
                        root->left :  
                        root->right;  
  
            // No child case  
            if (temp == NULL)  
            {  
                temp = root;  
                root = NULL;  
            }  
            else // One child case  
            *root = *temp; // Copy the contents of  
                        // the non-empty child  
            free(temp);  
        }  
        else
        {  
            // node with two children: Get the inorder  
            // successor (smallest in the right subtree)  
            Node* temp = minValueNode(root->right);  
  
            // Copy the inorder successor's  
            // data to this node  
            root->key = temp->key;  
  
            // Delete the inorder successor  
            root->right = deleteNode(root->right,  
                                    temp->key);  
        }  
    }  
  
    // If the tree had only one node  
    // then return  
    if (root == NULL)  
    return root;  
  
    // STEP 2: UPDATE HEIGHT OF THE CURRENT NODE  
    root->height = 1 + max(height(root->left),  
                        height(root->right));  
  
    // STEP 3: GET THE BALANCE FACTOR OF  
    // THIS NODE (to check whether this  
    // node became unbalanced)  
    int balance = getBalance(root);  
  
    // If this node becomes unbalanced,  
    // then there are 4 cases  
  
    // Left Left Case  
    if (balance > 1 &&  
        getBalance(root->left) >= 0)  
        return rightRotate(root);  
  
    // Left Right Case  
    if (balance > 1 &&  
        getBalance(root->left) < 0)  
    {  
        root->left = leftRotate(root->left);  
        return rightRotate(root);  
    }  
  
    // Right Right Case  
    if (balance < -1 &&  
        getBalance(root->right) <= 0)  
        return leftRotate(root);  
  
    // Right Left Case  
    if (balance < -1 &&  
        getBalance(root->right) > 0)  
    {  
        root->right = rightRotate(root->right);  
        return leftRotate(root);  
    }  
  
    return root;  
}  
  
// A utility function to print preorder  
// traversal of the tree.  
// The function also prints height  
// of every node  
void preOrder(Node *root)  
{  
    if(root != NULL)  
    {  
        cout << root->key << " ";  
        preOrder(root->left);  
        preOrder(root->right);  
    }  
} 
  
// Returns maximum value in a given   
// Binary Tree   
int findMax(Node* root)   
{   
    // Base case   
    if (root == NULL)   
    return INT_MIN;   
    
    // Return maximum of 3 values:   
    // 1) Root's data 2) Max in Left Subtree   
    // 3) Max in right subtree   
    int res = root->key;   
    int lres = findMax(root->left);   
    int rres = findMax(root->right);   
    if (lres > res)   
    res = lres;   
    if (rres > res)   
    res = rres;   
    return res;   
} 
  
// Method to find the maximum for each  
// and every contiguous subarray of size k. 
void printKMax(int arr[], int n, int k)  
{  
    int c = 0,l=0; 
    Node *root = NULL;  
  
      
    //traverse the array ; 
    for(int i=0; i<n; i++) 
    { 
        c++; 
        //insert the element in BST 
        root = insert(root, arr[i]);  
          
        //size of subarray greater than k  
        if(c > k) 
        { 
            root = deleteNode(root, arr[l++]);  
            c--; 
        } 
          
        //size of subarray equal to k 
        if(c == k) 
        { 
            cout<<findMax(root)<<" "; 
        } 
    } 
} 
// Driver code 
int main()  
{  
    int  arr[] = {8, 5, 10, 7, 9, 4, 15, 12, 90, 13}, k = 4;  
    int n = sizeof(arr) / sizeof(arr[0]);  
    printKMax(arr, n, k);  
    return 0;  
} 

Output:

10 10 10 15 15 90 90

Compelxity Analysis:
- Time Complexity: O(N * Log k).
  Insertion, deletion and search takes log k time in a AVL tree. So the overall time Complexity is O(N * log k)
- Space Compelxity: O(k).
  The space required to store k elements in a BST is O(k).

Method 3: This method uses Deque to solve the above problem.

Approach:
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 left 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 of Qi is the smallest of current window. Thanks to Aashish for suggesting this method.
Dry run of the above approach:
Algorithm :
1. Create a deque to store k elements.
2. Run a loop and insert first k elements in the deque. While inserting the element if the element at the back of the queue is smaller than the current element remove all those elements and then insert the element.
3. Now, run a loop from k to end of the array.
4. Print the front element of the array
5. Remove the element from the front of the queue if they are out of the current window.
6. Insert the next element in the deque. While inserting the element if the element at the back of the queue is smaller than the current element remove all those elements and then insert the element.
7. Print the maximum element of the last window.

Implementation:

C++

#include <bits/stdc++.h> 
  
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()]) 
            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 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<Integer> Qi = new LinkedList<Integer>(); 
  
        /* 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()]) 
                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  

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.IndexOf(0)]) 
                Qi.RemoveAt(Qi.Count-1); // Remove from rear 
  
            // 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 code 
    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 has been contributed by 29AjayKumar 

Output:

78 90 90 90 89

Compelxity Analysis:
- Time Complexity: O(n).
  It seems more than O(n) at first look. It can be observed that every element of array is added and removed at most once. So there are total 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.

Method 4: This method uses Max-Heap to solve the above problem.

Approach:
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.
Max-heap is a suitable data structure as it provides constant-time retrieval and logarithmic time removal of both the minimum and maximum elements in it, i.e. it takes constant time to find the maximum element and insertion and deletion takes log n time.
Algorithm:
1. Pick first k elements and create a max heap of size k.
2. Perform heapify and print the root element.
3. Store the next and last element from the array
4. Run a loop from k – 1 to n
  - Replace the value of element which is got out of the window with new element which came inside the window.
  - Perform heapify.
  - Print the root of the Heap.

Implementation:

Python3

# Python program to find the maximum for  
# each and every contiguous subarray of 
# size k  
import heapq 
  
# Method to find the maximum for each 
# and every contiguous subarray of s  
# of size k 
def max_of_all_in_k(arr, n): 
    i = 0
    j = k-1
      
    # Create the heap and heapify 
    heap = arr[i:j + 1] 
    heapq._heapify_max(heap) 
      
    # Print the maximum element from  
    # the first window of size k 
    print(heap[0], end =" ") 
    last = arr[i] 
    i+= 1
    j+= 1
    nexts = arr[j] 
      
    # For every remaining element 
    while j < n: 
          
        # Add the next element of the window 
        heap[heap.index(last)] = nexts 
          
        # Heapify to get the maximum  
        # of the current window 
        heapq._heapify_max(heap) 
          
        # Print the current maximum 
        print(heap[0], end =" ") 
        last = arr[i] 
        i+= 1
        j+= 1
        if j < n: 
            nexts = arr[j] 
              
# Driver Function 
n, k = 10, 3
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 
max_of_all_in_k(arr, n) 

Output:

3 4 5 6 7 8 9 10

Complexity Analysis:
- Time Complexity: O(n * k).
  The time complexity of steps 4(a) is O(k), 4(b) is O(Log(k)) and it is in a loop that runs (n – k + 1) times. Hence, the time complexity of the complete algorithm is O((k + Log(k)) * n) i.e. O(n * k).
- Space Complexity: O(k).
  To store the elements in Heap O(k) space is used.

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

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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