Count elements such that there are exactly X elements with values greater than or equal to X

Hello Everyone,

Given an array arr of N integers, the task is to find the number of elements that satisfy the following condition:
If the element is X then there has to be exactly X number of elements in the array (excluding the number X ) which are greater than or equal to X


Input: arr[] = {1, 2, 3, 4} Output: 1 Only element 2 satisfies the condition as there are exactly 2 elements which are greater than or equal to 2 (3, 4) except 2 itself. Input: arr[] = {5, 5, 5, 5, 5} Output: 0

Approach: The problem involves efficient searching for each arr[i] element the number of arr[j]’s (i != j) which are greater than or equal to arr[i].

  • Sort the array in ascending order.
  • For every element arr[i], using binary search get the count of all the elements that are greater than or equal to arr[i] except arr[i] itself.
  • If the count is equal to arr[i] then increment the result.
  • Print the value of the result in the end.

Below is the implementation of the above approach:

  • C++

// C++ implementation of the approach

#include <bits/stdc++.h>

using namespace std;

#define ll long long

ll int getCount(vector<ll int > v, int n)


// Sorting the vector

sort((v).begin(), (v).end());

ll int cnt = 0;

for (ll int i = 0; i < n; i++) {

// Count of numbers which

// are greater than v[i]

ll int tmp = v.end() - 1

- upper_bound((v).begin(), (v).end(), v[i] - 1);

if (tmp == v[i])



return cnt;


// Driver code

int main()


ll int n;

n = 4;

vector<ll int > v;





cout << getCount(v, n);

return 0;