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

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

Examples:

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++

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// 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])
            cnt++;
    }
    return cnt;
}
  
// Driver code
int main()
{
    ll int n;
    n = 4;
    vector<ll int> v;
    v.push_back(1);
    v.push_back(2);
    v.push_back(3);
    v.push_back(4);
  
    cout << getCount(v, n);
    return 0;
}

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Python3

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# Python3 implementation of the approach
from bisect import bisect as upper_bound
  
def getCount(v, n):
      
    # Sorting the vector
    v = sorted(v)
    cnt = 0
    for i in range(n):
  
        # Count of numbers which
        # are greater than v[i]
        tmp = n - 1 - upper_bound(v, v[i] - 1)
  
        if (tmp == v[i]):
            cnt += 1
    return cnt
  
# Driver codemain()
n = 4
v = []
v.append(1)
v.append(2)
v.append(3)
v.append(4)
  
print(getCount(v, n))
  
# This code is contributed by Mohit Kumar

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Output:

1


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Improved By : mohit kumar 29