# Find the number of elements greater than k in a sorted array

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 = 6Output:1Input:arr[] = {1, 1, 2, 5, 5, 7}, k = 8Output:0

**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++

`// 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;` `}` |

## Java

`// Java implementation of the approach` `class` `GFG` `{` ` ` `// Function to return the count of elements` `// from the array which are greater than k` `static` `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` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `arr[] = { ` `3` `, ` `3` `, ` `4` `, ` `7` `, ` `7` `, ` `7` `, ` `11` `, ` `13` `, ` `13` `};` ` ` `int` `n = arr.length;` ` ` `int` `k = ` `7` `;` ` ` `System.out.println(countGreater(arr, n, k));` `}` `}` `// This code is contributed by Code_Mech` |

## Python3

`# Python 3 implementation of the approach` `# Function to return the count of elements` `# from the array which are greater than k` `def` `countGreater(arr, n, k):` ` ` `l ` `=` `0` ` ` `r ` `=` `n ` `-` `1` ` ` `# Stores the index of the left most element` ` ` `# from the array which is greater than k` ` ` `leftGreater ` `=` `n` ` ` `# Finds number of elements greater than k` ` ` `while` `(l <` `=` `r):` ` ` `m ` `=` `int` `(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` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `arr ` `=` `[` `3` `, ` `3` `, ` `4` `, ` `7` `, ` `7` `, ` `7` `, ` `11` `, ` `13` `, ` `13` `]` ` ` `n ` `=` `len` `(arr)` ` ` `k ` `=` `7` ` ` `print` `(countGreater(arr, n, k))` `# This code is contributed by` `# Surendra_Gangwar` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` ` ` `// Function to return the count of elements` `// from the array which are greater than k` `static` `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` `public` `static` `void` `Main()` `{` ` ` `int` `[] arr = { 3, 3, 4, 7, 7, 7, 11, 13, 13 };` ` ` `int` `n = arr.Length;` ` ` `int` `k = 7;` ` ` `Console.WriteLine(countGreater(arr, n, k));` `}` `}` `// This code is contributed by Code_Mech` |

## PHP

`<?php` `// PHP implementation of the approach` `// Function to return the count of elements` `// from the array which are greater than k` `function` `countGreater(` `$arr` `, ` `$n` `, ` `$k` `)` `{` ` ` `$l` `= 0;` ` ` `$r` `= ` `$n` `- 1;` ` ` `// Stores the index of the left most element` ` ` `// from the array which is greater than k` ` ` `$leftGreater` `= ` `$n` `;` ` ` `// Finds number of elements greater than k` ` ` `while` `(` `$l` `<= ` `$r` `)` ` ` `{` ` ` `$m` `= ` `$l` `+ (int)((` `$r` `- ` `$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` `- ` `$leftGreater` `);` `}` `// Driver code` `$arr` `= ` `array` `(3, 3, 4, 7, 7, 7, 11, 13, 13);` `$n` `= sizeof(` `$arr` `);` `$k` `= 7;` `echo` `countGreater(` `$arr` `, ` `$n` `, ` `$k` `);` `// This code is contributed` `// by Akanksha Rai` |

## Javascript

`<script>` `// Javascript implementation of the approach` `// Function to return the count of elements` `// from the array which are greater than k` `function` `countGreater(arr, n, k)` `{` ` ` `var` `l = 0;` ` ` `var` `r = n - 1;` ` ` `// Stores the index of the left most element` ` ` `// from the array which is greater than k` ` ` `var` `leftGreater = n;` ` ` `// Finds number of elements greater than k` ` ` `while` `(l <= r) {` ` ` `var` `m = l + parseInt((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` `var` `arr = [3, 3, 4, 7, 7, 7, 11, 13, 13];` `var` `n = arr.length;` `var` `k = 7;` `document.write( countGreater(arr, n, k));` `</script>` |

**Output:**

3

**Time Complexity:** O(log(n)) where n is the number of elements in the array.**Auxiliary Space: **O(1)

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