Skip to content
Related Articles

Related Articles

Improve Article

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

  • Difficulty Level : Medium
  • Last Updated : 31 May, 2021

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++




// 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;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
    static int getCount(int[] v, int n)
    {
 
        // Sorting the vector
        Arrays.sort(v);
        int cnt = 0;
        for (int i = 0; i < n; i++)
        {
 
            // Count of numbers which
            // are greater than v[i]
            int tmp = n - 1 - upperBound(v, n, v[i] - 1);
            if (tmp == v[i])
                cnt++;
        }
        return cnt;
    }
 
    // Function to implement upper_bound()
    static int upperBound(int[] array, 
                          int length, int value)
    {
        int low = 0;
        int high = length;
        while (low < high)
        {
            final int mid = (low + high) / 2;
            if (value >= array[mid])
            {
                low = mid + 1;
            }
            else
            {
                high = mid;
            }
        }
        return low;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 4;
        int[] v = { 1, 2, 3, 4 };
        System.out.println(getCount(v, n));
    }
}
 
// This code is contributed by
// sanjeev2552

Python3




# 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

C#




// C# implementation of the approach
using System;
 
class GFG
{
    static int getCount(int[] v, int n)
    {
 
        // Sorting the vector
        Array.Sort(v);
        int cnt = 0;
        for (int i = 0; i < n; i++)
        {
 
            // Count of numbers which
            // are greater than v[i]
            int tmp = n - 1 - upperBound(v, n, v[i] - 1);
            if (tmp == v[i])
                cnt++;
        }
        return cnt;
    }
 
    // Function to implement upper_bound()
    static int upperBound(int[] array,
                          int length, int value)
    {
        int low = 0;
        int high = length;
        while (low < high)
        {
            int mid = (low + high) / 2;
            if (value >= array[mid])
            {
                low = mid + 1;
            }
            else
            {
                high = mid;
            }
        }
        return low;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int n = 4;
        int[] v = { 1, 2, 3, 4 };
        Console.WriteLine(getCount(v, n));
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// JavaScript implementation of the approach
 
    function getCount(v,n)
    {
        // Sorting the vector
        v.sort(function(a,b){return a-b;});
        let cnt = 0;
        for (let i = 0; i < n; i++)
        {
   
            // Count of numbers which
            // are greater than v[i]
            let tmp = n - 1 - upperBound(v, n, v[i] - 1);
            if (tmp == v[i])
                cnt++;
        }
        return cnt;
    }
     
    // Function to implement upper_bound()
    function upperBound(array,length,value)
    {
        let low = 0;
        let high = length;
        while (low < high)
        {
            let mid = Math.floor((low + high) / 2);
            if (value >= array[mid])
            {
                low = mid + 1;
            }
            else
            {
                high = mid;
            }
        }
        return low;
    }
     
    // Driver Code
    let n = 4;
    let v=[1, 2, 3, 4];
    document.write(getCount(v, n));
 
 
// This code is contributed by unknown2108
 
</script>
Output: 
1

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :