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