Most frequent element in an array

Given an array, find the most frequent element in it. If there are multiple elements that appear a maximum number of times, print any one of them.

Examples:

Input : arr[] = {1, 3, 2, 1, 4, 1}
Output : 1
Explanation: 1 appears three times in array which is maximum frequency.

Input : arr[] = {10, 20, 10, 20, 30, 20, 20}
Output : 20

A simple solution is to run two loops. The outer loop picks all elements one by one. The inner loop finds the frequency of the picked element and compares it with the maximum so far. 

Implementation:

C++

// CPP program to find the most frequent element
// in an array.
#include <bits/stdc++.h>
using namespace std;

int mostFrequent(int *arr, int n) {
  // code here
  int maxcount=0;
  int element_having_max_freq;
  for(int i=0;i<n;i++)
  {
    int count=0;
    for(int j=0;j<n;j++)
    {
      if(arr[i] == arr[j])
        count++;
    }

    if(count>maxcount)
    {
      maxcount=count;
      element_having_max_freq = arr[i];
    }

  }

  return element_having_max_freq;
}

// Driver program
int main()
{
    int arr[] = { 40,50,30,40,50,30,30};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << mostFrequent(arr, n);
    return 0;
}

// This code is contributed by Arpit Jain

Java

public class GFG
{

  // Java program to find the most frequent element
  // in an array.
  public static int mostFrequent(int[] arr, int n)
  {
    int maxcount = 0;
    int element_having_max_freq = 0;
    for (int i = 0; i < n; i++) {
      int count = 0;
      for (int j = 0; j < n; j++) {
        if (arr[i] == arr[j]) {
          count++;
        }
      }

      if (count > maxcount) {
        maxcount = count;
        element_having_max_freq = arr[i];
      }
    }

    return element_having_max_freq;
  }

  // Driver program
  public static void main(String[] args)
  {
    int[] arr = { 40, 50, 30, 40, 50, 30, 30 };
    int n = arr.length;
    System.out.print(mostFrequent(arr, n));
  }
}

// This code is contributed by Aarti_Rathi

Python3

# Python3 program to find the most
# frequent element in an array.
def mostFrequent(arr, n):
  maxcount = 0;
  element_having_max_freq = 0;
  for i in range(0, n):
    count = 0
    for j in range(0, n):
      if(arr[i] == arr[j]):
        count += 1
    if(count > maxcount):
      maxcount = count
      element_having_max_freq = arr[i]

  return element_having_max_freq;

# Driver Code
arr = [40,50,30,40,50,30,30]
n = len(arr)
print(mostFrequent(arr, n))

# This code is contributed by Arpit Jain

C#

// C# program to find the most frequent element
// in an array
using System;
public class GFG
{

  // C# program to find the most frequent element
  // in an array.
  public static int mostFrequent(int[] arr, int n)
  {
    int maxcount = 0;
    int element_having_max_freq = 0;
    for (int i = 0; i < n; i++) {
      int count = 0;
      for (int j = 0; j < n; j++) {
        if (arr[i] == arr[j]) {
          count++;
        }
      }

      if (count > maxcount) {
        maxcount = count;
        element_having_max_freq = arr[i];
      }
    }

    return element_having_max_freq;
  }

  // Driver program
  public static void Main(String[] args)
  {
    int[] arr = { 40, 50, 30, 40, 50, 30, 30 };
    int n = arr.Length;
    Console.Write(mostFrequent(arr, n));
  }
}

// This code is contributed by Abhijeet Kumar(abhijeet19403)

Output

30

The time complexity of this solution is O(n²) since 2 loops are running from i=0 to i=n we can improve its time complexity by taking a visited  array and skipping numbers for which we already calculated the frequency.

A better solution is to do the sorting. We first sort the array, then linearly traverse the array. 

Implementation:

C++

// CPP program to find the most frequent element
// in an array.
#include <bits/stdc++.h>
using namespace std;

int mostFrequent(int arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);

    // Find the max frequency using linear traversal
    int max_count = 1, res = arr[0], curr_count = 1;
    for (int i = 1; i < n; i++) {
        if (arr[i] == arr[i - 1])
            curr_count++;
        else
            curr_count = 1;

        if (curr_count > max_count) {
            max_count = curr_count;
            res = arr[i - 1];
        }
    }

    return res;
}

// Driver program
int main()
{
    int arr[] = { 40,50,30,40,50,30,30};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << mostFrequent(arr, n);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// CPP program to find the most frequent element
// in an array.
#include <stdio.h>
#include <stdlib.h>

int cmpfunc(const void* a, const void* b)
{
    return (*(int*)a - *(int*)b);
}

int mostFrequent(int arr[], int n)
{
    // Sort the array
    qsort(arr, n, sizeof(int), cmpfunc);

    // find the max frequency using linear traversal
    int max_count = 1, res = arr[0], curr_count = 1;
     for (int i = 1; i < n; i++) {
        if (arr[i] == arr[i - 1])
            curr_count++;
        else
            curr_count = 1;

        if (curr_count > max_count) {
            max_count = curr_count;
            res = arr[i - 1];
        }
    }

    return res;
}

// driver program
int main()
{
    int arr[] = { 40,50,30,40,50,30,30};
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("%d", mostFrequent(arr, n));
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program to find the most frequent element
// in an array
import java.util.*;

class GFG {

    static int mostFrequent(int arr[], int n)
    {
        // Sort the array
        Arrays.sort(arr);

        // find the max frequency using linear traversal
        int max_count = 1, res = arr[0];
        int curr_count = 1;

        for (int i = 1; i < n; i++) {
            if (arr[i] == arr[i - 1])
                curr_count++;
            else
                curr_count = 1;

            if (curr_count > max_count) {
                max_count = curr_count;
                res = arr[i - 1];
            }
        }
        return res;
    }

    // Driver program
    public static void main(String[] args)
    {
        int arr[] = { 40,50,30,40,50,30,30};
        int n = arr.length;
        System.out.println(mostFrequent(arr, n));
    }
}

// This code is contributed by Aditya Kumar (adityakumar129)

Python3

# Python3 program to find the most
# frequent element in an array.


def mostFrequent(arr, n):

    # Sort the array
    arr.sort()

    # find the max frequency using
    # linear traversal
    max_count = 1
    res = arr[0]
    curr_count = 1

    for i in range(1, n):
        if (arr[i] == arr[i - 1]):
            curr_count += 1
        else:
            curr_count = 1

         # If last element is most frequent
        if (curr_count > max_count):
            max_count = curr_count
            res = arr[i - 1]

    return res


# Driver Code
arr = [40,50,30,40,50,30,30]
n = len(arr)
print(mostFrequent(arr, n))

# This code is contributed by Smitha Dinesh Semwal.

C#

// C# program to find the most
// frequent element in an array
using System;

class GFG {

    static int mostFrequent(int[] arr, int n)
    {

        // Sort the array
        Array.Sort(arr);

        // find the max frequency using
        // linear traversal
        int max_count = 1, res = arr[0];
        int curr_count = 1;

        for (int i = 1; i < n; i++) {
            if (arr[i] == arr[i - 1])
                curr_count++;
            else
                curr_count = 1;

            // If last element is most frequent
            if (curr_count > max_count) {
                max_count = curr_count;
                res = arr[i - 1];
            }
        }

        return res;
    }

    // Driver code
    public static void Main()
    {

        int[] arr = {40,50,30,40,50,30,30 };
        int n = arr.Length;

        Console.WriteLine(mostFrequent(arr, n));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find the
// most frequent element
// in an array.

function mostFrequent( $arr, $n)
{

    // Sort the array
    sort($arr);
    sort($arr , $n);

    // find the max frequency
    // using linear traversal
    $max_count = 1;
    $res = $arr[0];
    $curr_count = 1;
    for ($i = 1; $i < $n; $i++)
    {
        if ($arr[$i] == $arr[$i - 1])
            $curr_count++;
        else
              $curr_count = 1;

         if ($curr_count > $max_count)
         {
              $max_count = $curr_count;
              $res = $arr[$i - 1];
          }
    }

    return $res;
}

// Driver Code
{
    $arr = array(40,50,30,40,50,30,30);
    $n = sizeof($arr) / sizeof($arr[0]);
    echo mostFrequent($arr, $n);
    return 0;
}

// This code is contributed by nitin mittal
?>

Javascript

<script>
// JavaScript program to find the most frequent element
//in an array

    function mostFrequent(arr, n)
    {

        // Sort the array
        arr.sort();

        // find the max frequency using linear
        // traversal
        let max_count = 1, res = arr[0];
        let curr_count = 1;

        for (let i = 1; i < n; i++)
        {
            if (arr[i] == arr[i - 1])
                curr_count++;
            else
                curr_count = 1;

            if (curr_count > max_count)
            {
                max_count = curr_count;
                res = arr[i - 1];
            }

        }

        return res;
    }

// Driver Code

        let arr = [40,50,30,40,50,30,30];
        let n = arr.length;
        document.write(mostFrequent(arr,n));

// This code is contributed by code_hunt.
</script>

Output

30

Time Complexity: O(nlog(n)) 
Auxiliary Space: O(1)

An efficient solution is to use hashing. We create a hash table and store elements and their frequency counts as key-value pairs. Finally, we traverse the hash table and print the key with the maximum value.  

C++

// CPP program to find the most frequent element
// in an array.
#include <bits/stdc++.h>
using namespace std;

int mostFrequent(int arr[], int n)
{
    // Insert all elements in hash.
    unordered_map<int, int> hash;
    for (int i = 0; i < n; i++)
        hash[arr[i]]++;

    // find the max frequency
    int max_count = 0, res = -1;
    for (auto i : hash) {
        if (max_count < i.second) {
            res = i.first;
            max_count = i.second;
        }
    }

    return res;
}

// driver program
int main()
{
    int arr[] = {40,50,30,40,50,30,30 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << mostFrequent(arr, n);
    return 0;
}

Java

//Java program to find the most frequent element
//in an array
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;

class GFG {

    static int mostFrequent(int arr[], int n)
    {

        // Insert all elements in hash
        Map<Integer, Integer> hp =
               new HashMap<Integer, Integer>();

        for(int i = 0; i < n; i++)
        {
            int key = arr[i];
            if(hp.containsKey(key))
            {
                int freq = hp.get(key);
                freq++;
                hp.put(key, freq);
            }
            else
            {
                hp.put(key, 1);
            }
        }

        // find max frequency.
        int max_count = 0, res = -1;

        for(Entry<Integer, Integer> val : hp.entrySet())
        {
            if (max_count < val.getValue())
            {
                res = val.getKey();
                max_count = val.getValue();
            }
        }

        return res;
    }

    // Driver code
    public static void main (String[] args) {

        int arr[] = {40,50,30,40,50,30,30};
        int n = arr.length;

        System.out.println(mostFrequent(arr, n));
    }
}

// This code is contributed by Akash Singh.

Python3

# Python3 program to find the most
# frequent element in an array.
import math as mt

def mostFrequent(arr, n):

    # Insert all elements in Hash.
    Hash = dict()
    for i in range(n):
        if arr[i] in Hash.keys():
            Hash[arr[i]] += 1
        else:
            Hash[arr[i]] = 1

    # find the max frequency
    max_count = 0
    res = -1
    for i in Hash:
        if (max_count < Hash[i]):
            res = i
            max_count = Hash[i]

    return res

# Driver Code
arr = [ 40,50,30,40,50,30,30]
n = len(arr)
print(mostFrequent(arr, n))

# This code is contributed
# by Mohit kumar 29

C#

// C# program to find the most
// frequent element in an array
using System;
using System.Collections.Generic;

class GFG
{
    static int mostFrequent(int []arr,
                            int n)
    {
        // Insert all elements in hash
        Dictionary<int, int> hp =
                    new Dictionary<int, int>();

        for (int i = 0; i < n; i++)
        {
            int key = arr[i];
            if(hp.ContainsKey(key))
            {
                int freq = hp[key];
                freq++;
                hp[key] = freq;
            }
            else
                hp.Add(key, 1);
        }

        // find max frequency.
        int min_count = 0, res = -1;

        foreach (KeyValuePair<int,
                    int> pair in hp)
        {
            if (min_count < pair.Value)
            {
                res = pair.Key;
                min_count = pair.Value;
            }
        }
        return res;
    }

    // Driver code
    static void Main ()
    {
        int []arr = new int[]{40,50,30,40,50,30,30};
        int n = arr.Length;

        Console.Write(mostFrequent(arr, n));
    }
}

// This code is contributed by
// Manish Shaw(manishshaw1)

Javascript

<script>

// Javascript program to find
// the most frequent element
// in an array.

function mostFrequent(arr, n)
{
    // Insert all elements in hash.
    var hash = new Map();
    for (var i = 0; i < n; i++)
    {
        if(hash.has(arr[i]))
            hash.set(arr[i], hash.get(arr[i])+1)
        else
            hash.set(arr[i], 1)
    }

    // find the max frequency
    var max_count = 0, res = -1;
    hash.forEach((value,key) => {

        if (max_count < value) {
            res = key;
            max_count = value;
        }

    });

    return res;
}

// driver program
var arr = [40,50,30,40,50,30,30];
var n = arr.length;
document.write( mostFrequent(arr, n));

</script>

Output

30

Time Complexity: O(n) 
Auxiliary Space: O(n)

An efficient solution to this problem can be to solve this problem by Moore’s voting Algorithm.

NOTE: THE ABOVE VOTING ALGORITHM ONLY WORKS WHEN THE MAXIMUM OCCURRING ELEMENT IS MORE THAN (SIZEOFARRAY/2) TIMES;

In this method, we will find the maximum occurred integer by counting the votes a number has.

C++

#include <iostream>
using namespace std;

int maxFreq(int *arr, int n) {
    //using moore's voting algorithm
    int res = 0;
    int count = 1;
    for(int i = 1; i < n; i++) {
        if(arr[i] == arr[res]) {
            count++;
        } else {
            count--;
        }

        if(count == 0) {
            res = i;
            count = 1;
        }

    }

    return arr[res];
}

int main()
{
    int arr[] = {40,50,30,40,50,30,30};
    int n = sizeof(arr) / sizeof(arr[0]);
    int freq =  maxFreq(arr , n);
    int count = 0;
    for(int i = 0; i < n; i++) {
        if(arr[i] == freq) {
            count++;
        }
    }
    cout <<"Element " << maxFreq(arr , n) << " occurs " << count << " times" << endl;;
    return 0;
    //This code is contributed by Ashish Kumar Shakya
}

Java

import java.io.*;
class GFG
{

static int maxFreq(int []arr, int n)
{

    // using moore's voting algorithm
    int res = 0;
    int count = 1;
    for(int i = 1; i < n; i++) {
        if(arr[i] == arr[res]) {
            count++;
        } else {
            count--;
        }

        if(count == 0) {
            res = i;
            count = 1;
        }

    }

    return arr[res];
}

  // Driver code
public static void main (String[] args) {
    int arr[] = {40,50,30,40,50,30,30};
    int n = arr.length;
    int freq =  maxFreq(arr , n);
    int count = 0;
    for(int i = 0; i < n; i++) {
        if(arr[i] == freq) {
            count++;
        }
    }
    System.out.println("Element " +maxFreq(arr , n) +" occurs "  +count +" times" );
}

}

// This code is contributed by shivanisinghss2110

Python3

def maxFreq(arr, n):

    # Using moore's voting algorithm
    res = 0
    count = 1

    for i in range(1, n):
        if (arr[i] == arr[res]):
            count += 1
        else:
            count -= 1

        if (count == 0):
            res = i
            count = 1

    return arr[res]

# Driver code
arr = [ 40, 50, 30, 40, 50, 30, 30 ]
n = len(arr)
freq =  maxFreq(arr, n)
count = 0

for i in range (n):
        if(arr[i] == freq):
            count += 1

print("Element ", maxFreq(arr , n),
      " occurs ", count, " times")

# This code is contributed by shivanisinghss2110

C#

using System;
class GFG
{

static int maxFreq(int []arr, int n)
{

    // using moore's voting algorithm
    int res = 0;
    int count = 1;
    for(int i = 1; i < n; i++) {
        if(arr[i] == arr[res]) {
            count++;
        } else {
            count--;
        }

        if(count == 0) {
            res = i;
            count = 1;
        }

    }

    return arr[res];
}

  // Driver code
public static void Main (String[] args) {
    int []arr = {40,50,30,40,50,30,30};
    int n = arr.Length;
    int freq =  maxFreq(arr , n);
    int count = 0;
    for(int i = 0; i < n; i++) {
        if(arr[i] == freq) {
            count++;
        }
    }
    Console.Write("Element " +maxFreq(arr , n) +" occurs "  +count +" times" );
}

}

// This code is contributed by shivanisinghss2110

Javascript

<script>

      function maxFreq(arr, n) {
        //using moore's voting algorithm
        var res = 0;
        var count = 1;
        for (var i = 1; i < n; i++) {
          if (arr[i] === arr[res]) {
            count++;
          } else {
            count--;
          }

          if (count === 0) {
            res = i;
            count = 1;
          }
        }

        return arr[res];
      }

      var arr = [40, 50, 30, 40, 50, 30, 30];
      var n = arr.length;
      var freq = maxFreq(arr, n);
      var count = 0;
      for (var i = 0; i < n; i++) {
        if (arr[i] === freq) {
          count++;
        }
      }
      document.write(
        "Element " + maxFreq(arr, n) + " occurs " +
        count + " times" + "<br>"
      );

</script>

Output

Element 30 occurs 3 times

Time Complexity: O(n)
Auxiliary Space: O(1)