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++
#include <bits/stdc++.h>
using namespace std;
int countGreater( int arr[], int n, int k)
{
int l = 0;
int r = n - 1;
int leftGreater = n;
while (l <= r) {
int m = l + (r - l) / 2;
if (arr[m] > k) {
leftGreater = m;
r = m - 1;
}
else
l = m + 1;
}
return (n - leftGreater);
}
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
class GFG
{
static int countGreater( int arr[], int n, int k)
{
int l = 0 ;
int r = n - 1 ;
int leftGreater = n;
while (l <= r) {
int m = l + (r - l) / 2 ;
if (arr[m] > k) {
leftGreater = m;
r = m - 1 ;
}
else
l = m + 1 ;
}
return (n - leftGreater);
}
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));
}
}
|
Python3
def countGreater(arr, n, k):
l = 0
r = n - 1
leftGreater = n
while (l < = r):
m = int (l + (r - l) / 2 )
if (arr[m] > k):
leftGreater = m
r = m - 1
else :
l = m + 1
return (n - leftGreater)
if __name__ = = '__main__' :
arr = [ 3 , 3 , 4 , 7 , 7 , 7 , 11 , 13 , 13 ]
n = len (arr)
k = 7
print (countGreater(arr, n, k))
|
C#
using System;
class GFG
{
static int countGreater( int []arr, int n, int k)
{
int l = 0;
int r = n - 1;
int leftGreater = n;
while (l <= r)
{
int m = l + (r - l) / 2;
if (arr[m] > k)
{
leftGreater = m;
r = m - 1;
}
else
l = m + 1;
}
return (n - leftGreater);
}
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));
}
}
|
PHP
<?php
function countGreater( $arr , $n , $k )
{
$l = 0;
$r = $n - 1;
$leftGreater = $n ;
while ( $l <= $r )
{
$m = $l + (int)(( $r - $l ) / 2);
if ( $arr [ $m ] > $k )
{
$leftGreater = $m ;
$r = $m - 1;
}
else
$l = $m + 1;
}
return ( $n - $leftGreater );
}
$arr = array (3, 3, 4, 7, 7, 7, 11, 13, 13);
$n = sizeof( $arr );
$k = 7;
echo countGreater( $arr , $n , $k );
|
Javascript
<script>
function countGreater(arr, n, k)
{
var l = 0;
var r = n - 1;
var leftGreater = n;
while (l <= r) {
var m = l + parseInt((r - l) / 2);
if (arr[m] > k) {
leftGreater = m;
r = m - 1;
}
else
l = m + 1;
}
return (n - leftGreater);
}
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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