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Maximum element in an array which is equal to its frequency

  • Last Updated : 31 Aug, 2021

Given an array of integers arr[] of size N, the task is to find the maximum element in the array whose frequency equals to it’s value 
Examples: 
 

Input: arr[] = {3, 2, 2, 3, 4, 3} 
Output:
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 and 3 is the maximum.
Input: arr[] = {1, 2, 3, 4, 5, 6} 
Output:
 

 

Approach: Store the frequency of every element of the array using the map, and finally find out the maximum of those element whose frequency is equal to their value.
Below is the implementation of the above approach:
 

CPP




// C++ program to find the maximum element
// whose frequency equals to it’s value
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum element
// whose frequency equals to it’s value
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 it is the maximum element or not
        if (value == freq) {
            ans = max(ans, value);
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 2, 2, 3, 4, 3 };
     
    // Size of array
    int n = sizeof(arr) / sizeof(arr[0]);
     
    // Function call
    cout << find_maxm(arr, n);
 
    return 0;
}

Java




// Java program to find the maximum element
// whose frequency equals to it’s value
import java.util.*;
 
class GFG{
  
// Function to find the maximum element
// whose frequency equals to it’s value
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 it is the maximum element or not
        if (value == freq) {
            ans = Math.max(ans, value);
        }
    }
  
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 2, 3, 4, 3 };
      
    // Size of array
    int n = arr.length;
      
    // Function call
    System.out.print(find_maxm(arr, n));
  
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program to find the maximum element
# whose frequency equals to it’s value
 
# Function to find the maximum element
# whose frequency equals to it’s value
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.update( {arr[i] : mpp[arr[i]] + 1} )
        else:
            mpp[arr[i]] = 1
 
    ans = 0
    for value,freq in mpp.items():
        # Check if value equals to frequency
        # and it is the maximum element or not
        if (value == freq):
            ans = max(ans, value)
 
    return ans
 
# Driver code
arr = [ 3, 2, 2, 3, 4, 3 ]
     
# Size of array
n = len(arr)
 
# Function call
print(find_maxm(arr, n))
 
# This code is contributed by Sanjit_Prasad

C#




// C# program to find the maximum element
// whose frequency equals to it’s value
using System;
using System.Collections.Generic;
 
class GFG{
   
// Function to find the maximum element
// whose frequency equals to it’s value
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 it is the maximum element or not
        if (value == freq) {
            ans = Math.Max(ans, value);
        }
    }
   
    return ans;
}
   
// Driver code
public static void Main(String[] args)
{
    int []arr = { 3, 2, 2, 3, 4, 3 };
       
    // Size of array
    int n = arr.Length;
       
    // Function call
    Console.Write(find_maxm(arr, n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript program to find the maximum element
// whose frequency equals to it’s value
 
// Function to find the maximum element
// whose frequency equals to it’s value
function find_maxm(arr, n)
{
    // Hash map for counting frequency
    var mpp = new Map();
 
    for (var 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)
    }
 
    var ans = 0;
    mpp.forEach((value, key) => {
         
        var value = value;
        var freq = key;
         
        // Check if value equals to frequency
        // and it is the maximum element or not
        if (value == freq) {
            ans = Math.max(ans, value);
        }
    });
 
    return ans;
}
 
// Driver code
var arr = [3, 2, 2, 3, 4, 3 ];
 
// Size of array
var n = arr.length;
 
// Function call
document.write( find_maxm(arr, n));
 
// This code is contributed by famously.
</script>
Output: 
3

 


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