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

Hello Everyone,

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

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++ 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;`

`}`

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;`

`}`