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Count the elements having frequency equals to its value

  • Difficulty Level : Easy
  • Last Updated : 09 Jul, 2021

Given an array of integers arr[] of size N, the task is to count all the elements of the array which have a frequency equals to its value.

Examples: 

Input: arr[] = {3, 2, 2, 3, 4, 3} 
Output: 2 
Frequency of element 2 is 2 
Frequency of element 3 is 3 
Frequency of element 4 is 1 
2 and 3 are elements which have same frequency as it’s value

Input: arr[] = {1, 2, 3, 4, 5, 6} 
Output: 1 

Approach: Store the frequency of every element of the array using the map, and finally count all of that elements whose frequency is equal to their value.

Below is the implementation of the above approach:

C++




// C++ program to count the elements
// having frequency equals to its value
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the count
int find_maxm(int arr[], int n)
{
    // Hash map for counting frequency
    map<int, int> mpp;
 
    for (int i = 0; i < n; i++) {
 
        // Counting freq of each element
        mpp[arr[i]] += 1;
    }
 
    int ans = 0;
    for (auto x : mpp) {
        int value = x.first;
        int freq = x.second;
 
        // Check if value equals to frequency
        // and increment the count
        if (value == freq) {
            ans++;
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 2, 2, 3, 4, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << find_maxm(arr, n);
 
    return 0;
}
Java



// Java program to count the elements
// having frequency equals to its value
import java.util.*;
 
class GFG{
  
// Function to find the count
static int find_maxm(int arr[], int n)
{
    // Hash map for counting frequency
    HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();
  
    for (int i = 0; i < n; i++) {
  
        // Counting freq of each element
        if(mp.containsKey(arr[i])){
            mp.put(arr[i], mp.get(arr[i])+1);
        }else{
            mp.put(arr[i], 1);
    }
    }
  
    int ans = 0;
    for (Map.Entry<Integer,Integer> x : mp.entrySet()){
        int value = x.getKey();
        int freq = x.getValue();
  
        // Check if value equals to frequency
        // and increment the count
        if (value == freq) {
            ans++;
        }
    }
  
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 2, 3, 4, 3 };
    int n = arr.length;
  
    // Function call
    System.out.print(find_maxm(arr, n));
}
}
 
// This code is contributed by Princi Singh
Python3



# Python3 program to count the elements
# having frequency equals to its value
 
# Function to find the count
def find_maxm(arr, n):
     
    # Hash map for counting frequency
    mpp = {}
     
    for i in range (0, n):
         
        # Counting freq of each element
        if arr[i] in mpp:
            mpp[arr[i]] = mpp[arr[i]] + 1
        else:
            mpp[arr[i]] = 1
     
    ans = 0
     
    for key in mpp:
        value = key
        freq = mpp[key]
         
        # Check if value equals to frequency
        # and increment the count
        if value == freq:
            ans = ans + 1
     
    return ans
 
# Driver code
if __name__ == "__main__":
     
    arr = [ 3, 2, 2, 3, 4, 3 ]
    n = len(arr)
     
    # Function call
    print(find_maxm(arr, n))
   
# This code is contributed by akhilsaini
C#



// C# program to count the elements
// having frequency equals to its value
using System;
using System.Collections.Generic;
 
class GFG{
   
// Function to find the count
static int find_maxm(int []arr, int n)
{
    // Hash map for counting frequency
    Dictionary<int,int> mp = new Dictionary<int,int>();
   
    for (int i = 0; i < n; i++) {
   
        // Counting freq of each element
        if(mp.ContainsKey(arr[i])){
            mp[arr[i]] = mp[arr[i]] + 1;
        }else{
            mp.Add(arr[i], 1);
    }
    }
   
    int ans = 0;
    foreach (KeyValuePair<int,int> x in mp){
        int value = x.Key;
        int freq = x.Value;
   
        // Check if value equals to frequency
        // and increment the count
        if (value == freq) {
            ans++;
        }
    }
   
    return ans;
}
   
// Driver code
public static void Main(String[] args)
{
    int []arr = { 3, 2, 2, 3, 4, 3 };
    int n = arr.Length;
   
    // Function call
    Console.Write(find_maxm(arr, n));
}
}
 
// This code is contributed by PrinciRaj1992
Javascript



<script>
 
// JavaScript program to count the elements
// having frequency equals to its value
 
 
// Function to find the count
function find_maxm(arr, n)
{
    // Hash map for counting frequency
    let mpp = new Map();
 
    for (let i = 0; i < n; i++) {
 
        // Counting freq of each element
        if(mpp.has(arr[i])){
            mpp.set(arr[i], mpp.get(arr[i]) + 1)
        }else{
            mpp.set(arr[i], 1)
        }
    }
 
    let ans = 0;
    for (let x of mpp) {
        let value = x[0];
        let freq = x[1];
 
        // Check if value equals to frequency
        // and increment the count
        if (value == freq) {
            ans++;
        }
    }
 
    return ans;
}
 
// Driver code
 
let arr = [ 3, 2, 2, 3, 4, 3 ];
let n = arr.length;
 
// Function call
document.write(find_maxm(arr, n));
 
// This code is contributed by _saurabh_jaiswal
 
</script>
Output: 
2
 
Method #2:Using collections.Counter()
We can solve this problem quickly using python Counter() method. Approach is very simple.
  • First create a dictionary using Counter method having elements as keys and their frequencies as values
  • count all of that elements whose frequency is equal to their value(key)
Below is the implementation of above approach:
Python3



# Python3 program to count the elements
# having frequency equals to its value
# importing counter from collections
from collections import Counter
 
# Function to find the count
def findElements(arr, n):
   
    # Now create dictionary using counter method
    # which will have elements as key and their
    # frequencies as values
    Element_Counter = Counter(arr)
    ans = 0
 
    for key in Element_Counter:
        value = key
        freq = Element_Counter[key]
 
        # Check if value equals to frequency
        # and increment the count
        if value == freq:
            ans = ans + 1
 
    return ans
 
 
# Driver code
arr = [3, 2, 2, 3, 4, 3]
n = len(arr)
 
# Function call
print(findElements(arr, n))
 
# This code is contributed by vikkycirus
Output:
2

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mohit kumar 29
@mohit kumar 29
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