Hello Everyone,

Given a sorted array arr[] of integers and an integer k, the task is to find the count of elements in the array which are greater than k. Note that k may or may not be present in the array.

Examples:

Input: arr[] = {2, 3, 5, 6, 6, 9}, k = 6

Output: 1

Input: arr[] = {1, 1, 2, 5, 5, 7}, k = 8

Output: 0

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

Approach: The idea is to perform binary search and find the number of elements greater than k.

Below is the implementation of the above approach:

// C++ implementation of the approach

#include <bits/stdc++.h>

using namespace std;

// Function to return the count of elements

// from the array which are greater than k

int countGreater(int arr[], int n, int k)

{

int l = 0;

int r = n - 1;

```
// Stores the index of the left most element
// from the array which is greater than k
int leftGreater = n;
// Finds number of elements greater than k
while (l <= r) {
int m = l + (r - l) / 2;
// If mid element is greater than
// k update leftGreater and r
if (arr[m] > k) {
leftGreater = m;
r = m - 1;
}
// If mid element is less than
// or equal to k update l
else
l = m + 1;
}
// Return the count of elements greater than k
return (n - leftGreater);
```

}

// Driver code

int main()

{

int arr[] = { 3, 3, 4, 7, 7, 7, 11, 13, 13 };

int n = sizeof(arr) / sizeof(arr[0]);

```
int k = 7;
cout << countGreater(arr, n, k);
return 0;
```

}

Output:

3

Time Complexity: O(log(n)) where n is the number of elements in the array.