Find the number of elements greater than k in a sorted array
Last Updated :
05 Aug, 2021
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 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
# 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# 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 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
<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>
Time Complexity: O(log(n)) where n is the number of elements in the array.
Auxiliary Space: O(1)
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