Count the elements having frequency equals to its value

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++

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// 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 frquency
    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 equls 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;
}

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Java

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// 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 frquency
    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 equls 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

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C#

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// 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 frquency
    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 equls 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

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Output:

2

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