# Count of Array elements greater than all elements on its left and at least K elements on its right

Given an array A[ ] consisting of N distinct integers, the task is to find the number of elements which are strictly greater than all the elements preceding it and strictly greater than at least K elements on its right.

Examples:

Input: A[] = {2, 5, 1, 7, 3, 4, 0}, K = 3
Output: 2
Explanation:
The only array elements satisfying the given conditions are:

• 5: Greater than all elements on its left {2} and at least K(= 3) elements on its right {1, 3, 4, 0}
• 7: Greater than all elements on its left {2, 5, 1} and at least K(= 3) elements on its right {3, 4, 0}

Therefore, the count is 2.

Input: A[] = {11, 2, 4, 7, 5, 9, 6, 3}, K = 2
Output: 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach:
The simplest approach to solve the problem is to traverse the array and for each element, traverse all the elements on its left and check if all of them are smaller than it or not and traverse all elements on its right to check if at least K elements are smaller than it or not. For every element satisfying the conditions, increase count. Finally, print the value of count.
Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach:
The above approach can be further optimized by using Self-Balancing BST. Follow the steps below:

• Traverse the array from right to left and insert all elements one by one in an AVL Tree
• Using the AVL Tree generate an array countSmaller[] which contains the count of smaller elements on the right of every array element.
• Traverse the array and for every ith element, check if it is the maximum obtained so far and countSmaller[i] is greater than or equal to K.
• If so, increase count.
• Print the final value of count as the answer.

Below is the implementation of the above approach:

 `// C++ Program to implement ` `// the above appraoch ` `#include ` `using` `namespace` `std; ` ` `  `// Structure of an AVL Tree Node ` `struct` `node { ` `    ``int` `key; ` `    ``struct` `node* left; ` `    ``struct` `node* right; ` `    ``int` `height; ` `    ``// Size of the tree rooted ` `    ``// with this node ` `    ``int` `size; ` `}; ` ` `  `// Utility function to get maximum ` `// of two integers ` `int` `max(``int` `a, ``int` `b); ` ` `  `// Utility function to get height ` `// of the tree rooted with N ` `int` `height(``struct` `node* N) ` `{ ` `    ``if` `(N == NULL) ` `        ``return` `0; ` `    ``return` `N->height; ` `} ` ` `  `// Utility function to find size of ` `// the tree rooted with N ` `int` `size(``struct` `node* N) ` `{ ` `    ``if` `(N == NULL) ` `        ``return` `0; ` `    ``return` `N->size; ` `} ` ` `  `// Utility function to get maximum ` `// of two integers ` `int` `max(``int` `a, ``int` `b) ` `{ ` `    ``return` `(a > b) ? a : b; ` `} ` ` `  `// Helper function to allocates a ` `// new node with the given key ` `struct` `node* newNode(``int` `key) ` `{ ` `    ``struct` `node* node ` `        ``= (``struct` `node*) ` `            ``malloc``(``sizeof``(``struct` `node)); ` `    ``node->key = key; ` `    ``node->left = NULL; ` `    ``node->right = NULL; ` `    ``node->height = 1; ` `    ``node->size = 1; ` `    ``return` `(node); ` `} ` ` `  `// Utility function to right rotate ` `// subtree rooted with y ` `struct` `node* rightRotate(``struct` `node* y) ` `{ ` `    ``struct` `node* x = y->left; ` `    ``struct` `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; ` ` `  `    ``// Update sizes ` `    ``y->size = size(y->left) ` `              ``+ size(y->right) + 1; ` `    ``x->size = size(x->left) ` `              ``+ size(x->right) + 1; ` ` `  `    ``// Return new root ` `    ``return` `x; ` `} ` ` `  `// Utility function to left rotate ` `// subtree rooted with x ` `struct` `node* leftRotate(``struct` `node* x) ` `{ ` `    ``struct` `node* y = x->right; ` `    ``struct` `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; ` ` `  `    ``// Update sizes ` `    ``x->size = size(x->left) ` `              ``+ size(x->right) + 1; ` `    ``y->size = size(y->left) ` `              ``+ size(y->right) + 1; ` ` `  `    ``// Return new root ` `    ``return` `y; ` `} ` ` `  `// Function to obtain Balance factor ` `// of node N ` `int` `getBalance(``struct` `node* N) ` `{ ` `    ``if` `(N == NULL) ` `        ``return` `0; ` ` `  `    ``return` `height(N->left) ` `           ``- height(N->right); ` `} ` ` `  `// Function to insert a new key to the ` `// tree rooted with node ` `struct` `node* insert(``struct` `node* node, ``int` `key, ` `                    ``int``* count) ` `{ ` `    ``// Perform the normal BST rotation ` `    ``if` `(node == NULL) ` `        ``return` `(newNode(key)); ` ` `  `    ``if` `(key < node->key) ` `        ``node->left ` `            ``= insert(node->left, key, count); ` `    ``else` `{ ` `        ``node->right ` `            ``= insert(node->right, key, count); ` ` `  `        ``// Update count of smaller elements ` `        ``*count = *count + size(node->left) + 1; ` `    ``} ` ` `  `    ``// Update height and size of the ancestor ` `    ``node->height = max(height(node->left), ` `                       ``height(node->right)) ` `                   ``+ 1; ` `    ``node->size = size(node->left) ` `                 ``+ size(node->right) + 1; ` ` `  `    ``// Get the balance factor of the ancestor ` `    ``int` `balance = getBalance(node); ` ` `  `    ``// 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` `node; ` `} ` ` `  `// Function to generate an array which contains ` `// count of smaller elements on the right ` `void` `constructLowerArray(``int` `arr[], ` `                         ``int` `countSmaller[], ` `                         ``int` `n) ` `{ ` `    ``int` `i, j; ` `    ``struct` `node* root = NULL; ` ` `  `    ``for` `(i = 0; i < n; i++) ` `        ``countSmaller[i] = 0; ` ` `  `    ``// Insert all elements in the AVL Tree ` `    ``// and get the count of smaller elements ` `    ``for` `(i = n - 1; i >= 0; i--) { ` `        ``root = insert(root, arr[i], ` `                      ``&countSmaller[i]); ` `    ``} ` `} ` ` `  `// Function to find the number ` `// of elements which are greater ` `// than all elements on its left ` `// and K elements on its right ` `int` `countElements(``int` `A[], ``int` `n, ``int` `K) ` `{ ` ` `  `    ``int` `count = 0; ` ` `  `    ``// Stores the count of smaller ` `    ``// elements on its right ` `    ``int``* countSmaller ` `        ``= (``int``*)``malloc``(``sizeof``(``int``) * n); ` `    ``constructLowerArray(A, countSmaller, n); ` ` `  `    ``int` `maxi = INT_MIN; ` `    ``for` `(``int` `i = 0; i <= (n - K - 1); i++) { ` `        ``if` `(A[i] > maxi && countSmaller[i] >= K) { ` `            ``count++; ` `            ``maxi = A[i]; ` `        ``} ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``int` `A[] = { 2, 5, 1, 7, 3, 4, 0 }; ` `    ``int` `n = ``sizeof``(A) / ``sizeof``(``int``); ` `    ``int` `K = 3; ` ` `  `    ``cout << countElements(A, n, K); ` ` `  `    ``return` `0; ` `} `

Output:
```2
```

Time Complexity: O(NlogN)
Auxiliary Space: O(N)

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