Sliding Window Maximum (Maximum of all subarrays of size k)

• Difficulty Level : Hard
• Last Updated : 17 Jan, 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

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

Method 1: This is a 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 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);     int k = 3;     printKMax(arr, n, k);     return 0; }   // This code is contributed by rathbhupendra

C

 #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 Code int main() {     int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };     int n = sizeof(arr) / sizeof(arr);     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 code     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

 \$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. ?>

Javascript



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 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 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++14

 // C++ program to delete a node from AVL Tree #include 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 k)         {             root = deleteNode(root, arr[l++]);             c--;         }                   //size of subarray equal to k         if(c == k)         {             cout<

Java

 // JAVA program to delete a node from AVL Tree import java.io.*; import java.util.*;   class GFG {       static ArrayList findKMaxElement(int[] arr,                                               int k, int n)     {         // creating the max heap ,to get max element always         PriorityQueue queue = new PriorityQueue<>(             Collections.reverseOrder());           ArrayList res = new ArrayList<>();         int i = 0;           for (; i < k; i++)             queue.add(arr[i]);           // adding the maximum element among first k elements         res.add(queue.peek());           // removing the first element of the array         queue.remove(arr);           // iterarting for the next elements         for (; i < n; i++) {               // adding the new element in the window             queue.add(arr[i]);               // finding & adding the max element in the             // current sliding window             res.add(queue.peek());               // finally removing the first element from front             // end of queue             queue.remove(arr[i - k + 1]);         }           return res;           // this code is Contributed by Pradeep Mondal P     }       // Driver Code     public static void main(String[] args)     {         int arr[] = { 8, 5, 10, 7, 9, 4, 15, 12, 90, 13 };         int k = 4, n = arr.length;           List res = findKMaxElement(arr, k, n);         for (int x : res)             System.out.print(x + " ");     } }



C#

 // C# program to delete a node from AVL Tree using System; using System.Collections.Generic;   public class GFG {     static List findKMaxElement(int[] arr, int k,                                    int n)   {       // creating the max heap ,to get max element always     List queue = new List();       List res = new List();     int i = 0;       for (; i < k; i++)       queue.Add(arr[i]);     queue.Sort();     queue.Reverse();     // adding the maximum element among first k elements     res.Add(queue);       // removing the first element of the array     queue.Remove(arr);       // iterarting for the next elements     for (; i < n; i++) {         // adding the new element in the window       queue.Add(arr[i]);       queue.Sort();       queue.Reverse();       // finding & adding the max element in the       // current sliding window       res.Add(queue);         // finally removing the first element from front       // end of queue       queue.Remove(arr[i - k + 1]);     }       return res;       // this code is Contributed by Pradeep Mondal P   }     // Driver Code   public static void Main(String[] args)   {     int[] arr = { 8, 5, 10, 7, 9, 4, 15, 12, 90, 13 };     int k = 4, n = arr.Length;       List res = findKMaxElement(arr, k, n);     foreach(int x in res) Console.Write(x + " ");   } }   // This code is contributed by umadevi9616

Output

10 10 10 15 15 90 90

Complexity 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 Complexity: 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 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. 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. 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.
3. Now, run a loop from k to end of the array.
4. Print the front element of the deque.
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. 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.
7. Print the maximum element of the last window.

Implementation:

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 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 code int main() {     int arr[] = { 12, 1, 78, 90, 57, 89, 56 };     int n = sizeof(arr) / sizeof(arr);     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 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 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 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]) + " ", end = "")                   # Remove the elements which are         # out of this window         while Qi and Qi <= 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]))       # Driver code 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 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] + " ");               // Remove the elements which are             // out of this window             while ((Qi.Count != 0) && Qi <= 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]);     }       // 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

Complexity 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 is modification in queue implementation of two stack:

Algorithm :

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

Implementation

C++

 #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 slidingMaximum(int a[], int b,int n) {     //s2 for push     //s1 for pop     vectorans;     stacks1,s2;           //shifting all value except the last one if first window     for(int i=0;i=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; }   int main() {     int arr[] = { 8, 5, 10, 7, 9, 4, 15, 12, 90, 13 };     int n = sizeof(arr) / sizeof(arr);     int k = 4;     vector ans=slidingMaximum(arr,k,n);       for(auto x:ans) cout<

Output

78 90 90 90 89

Complexity Analysis:

• Time Complexity: O(N): This is because every element will just two types push and pop; hence time complexity is linear.
• Space Complexity: 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.

Method 5: 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.

Method 6: This method works on finding three max elements and updating them rigorously.

Approach: This method is based on finding three max elements and printing the first max from every subarray for every new element that enters and leaves the subarray.

Algorithm:

1. Run a loop from index 0 to k-1 to find the first, second, and third max (say maxA, maxB, and maxC)

2. Print maxA.

3. Run a loop from k to n-1

: Update maxA, maxB, and maxC by every element that enters the window of length of k.

: if maxA matches the element that is being removed from the window replace maxA to maxB

: print maxA

C++14

 #include using namespace std;   void printKMax(int *arr,int n,int k) {     int i,maxA=INT_MIN,maxB=INT_MIN,maxC=INT_MIN;     //base condition     //if no subarrays needs to be formed     if(k==1||k==n)     {         for(i=0;i

Java

 import java.util.*; class GFG{     static void printKMax(int []arr,int n,int k)   {     int i, maxA = Integer.MIN_VALUE,     maxB = Integer.MIN_VALUE,     maxC = Integer.MIN_VALUE;       // base condition     // if no subarrays needs to be formed     if(k == 1 || k == n)     {       for(i = 0; i < n; i++)         System.out.print(arr[i]+" ");       return;     }       // traverse k elements to find     // find second and third largest elements     for(i = 0; i < k; i++){       if(maxA

C#

 using System;   public class GFG{     static void printKMax(int []arr,int n,int k)   {     int i, maxA = int.MinValue,     maxB = int.MinValue,     maxC = int.MinValue;       // base condition     // if no subarrays needs to be formed     if(k == 1 || k == n)     {       for(i = 0; i < n; i++)         Console.Write(arr[i] + " ");       return;     }       // traverse k elements to find     // find second and third largest elements     for(i = 0; i < k; i++){       if(maxA

Time Complexity: O(n)

Auxiliary Space: O(1)

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