Maximum element in an array which is equal to its frequency

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

Examples:

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

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

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 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 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 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 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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 frquency 
    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 equls 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 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 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

chevron_right


Output:

3

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.




My Personal Notes arrow_drop_up

Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.