Sum of elements in an array with frequencies greater than or equal to that element

Given an array arr[] of N integers. The task is to find the sum of the elements which have frequencies greater than or equal to that element in the array.

Examples:

Input: arr[] = {2, 1, 1, 2, 1, 6}
Output: 3
The elements in the array are {2, 1, 6}
Where,
 2 appear 2 times which is greater than equal to 2 itself.
 1 appear 3 times which is greater than 1 itself.
 But 6 appears 1 time which is not greater than or equals to 6.
So, sum = 2 + 1 = 3.

Input: arr[] = {1, 2, 3, 3, 2, 3, 2, 3, 3}
Output: 6

Approach:



Below is the implementation of the above approach:

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// C++ program to find sum of elements
// in an array having frequency greater
// than or equal to that element
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the sum of elements
// in an array having frequency greater
// than or equal to that element
int sumOfElements(int arr[], int n)
{
    bool prime[n + 1];
    int i, j;
  
    // Map is used to store
    // element frequencies
    unordered_map<int, int> m;
    for (i = 0; i < n; i++)
        m[arr[i]]++;
  
    int sum = 0;
  
    // Traverse the map using iterators
    for (auto it = m.begin(); it != m.end(); it++) {
  
        // Calculate the sum of elements
        // having frequencies greater than
        // or equal to the element itself
        if ((it->second) >= (it->first)) {
            sum += (it->first);
        }
    }
  
    return sum;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 3, 2, 3, 2, 3, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << sumOfElements(arr, n);
  
    return 0;
}
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// Java program to find sum of elements
// in an array having frequency greater
// than or equal to that element
import java.util.*;
class Solution
{
  
// Function to return the sum of elements
// in an array having frequency greater
// than or equal to that element
static int sumOfElements(int arr[], int n)
{
    boolean prime[] = new boolean[n + 1];
    int i, j;
  
    // Map is used to store
    // element frequencies
    HashMap<Integer, Integer> m= new HashMap<Integer,Integer>();
    for (i = 0; i < n; i++)
        {
            if(m.get(arr[i])==null)
            m.put(arr[i],1);
            else
            m.put(arr[i],m.get(arr[i])+1);
        }
  
    int sum = 0;
        // Getting an iterator 
        Iterator hmIterator = m.entrySet().iterator(); 
    
    
    // Traverse the map using iterators
        while (hmIterator.hasNext()) { 
            Map.Entry mapElement = (Map.Entry)hmIterator.next();
  
        // Calculate the sum of elements
        // having frequencies greater than
        // or equal to the element itself
        if (((int)mapElement.getValue()) >= ((int)mapElement.getKey())) {
            sum += ((int)mapElement.getKey());
        }
    }
  
    return sum;
}
  
// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 2, 3, 3, 2, 3, 2, 3, 3 };
    int n = arr.length;
  
    System.out.println(sumOfElements(arr, n));
  
 }
}
//contributed by Arnab Kundu
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# Python3 program to find sum of elements 
# in an array having frequency greater 
# than or equal to that element
  
# Function to return the sum of elements 
# in an array having frequency greater 
# than or equal to that element 
def sumOfElements(arr, n) :
  
    # dictionary is used to store 
    # element frequencies 
    m = dict.fromkeys(arr, 0)
  
    for i in range(n) :
            m[arr[i]] += 1
  
    sum = 0
  
    # traverse the dictionary
    for key,value in m.items() :
  
        # Calculate the sum of elements 
        # having frequencies greater than 
        # or equal to the element itself 
        if value >= key :
                sum += key
  
    return sum
  
# Driver code
if __name__ == "__main__" :
  
    arr = [1, 2, 3, 3, 2, 3, 2, 3, 3]
    n = len(arr)
  
    print(sumOfElements(arr, n))
  
# This code is contributed by Ryuga
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// C# program to find sum of elements
// in an array having frequency greater
// than or equal to that element
using System;
using System.Collections.Generic;
  
class GFG
{
  
    // Function to return the sum of elements
    // in an array having frequency greater
    // than or equal to that element
    static int sumOfElements(int []arr, int n)
    {
        bool []prime = new bool[n + 1];
        int i;
      
        // Map is used to store
        // element frequencies
        Dictionary<int, int> m= new Dictionary<int,int>();
        for (i = 0; i < n; i++)
            {
                if(!m.ContainsKey(arr[i]))
                    m.Add(arr[i],1);
                else
                {
                    var val = m[arr[i]];
                    m.Remove(arr[i]);
                    m.Add(arr[i], val + 1);
                }
              
            }
      
            int sum = 0;
            // Calculate the sum of elements
            // having frequencies greater than
            // or equal to the element itself
            foreach(KeyValuePair<int, int> entry in m)
            {
                if(entry.Value >= entry.Key)
                {
                    sum+=entry.Key;
                }
            }
      
        return sum;
    }
      
    // Driver code
    public static void Main(String []args)
    {
        int []arr = { 1, 2, 3, 3, 2, 3, 2, 3, 3 };
        int n = arr.Length;
      
        Console.WriteLine(sumOfElements(arr, n));
    }
}
  
// This code has been contributed by 29AjayKumar
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Output:
6

Time complexity : O(n)

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