Most frequent element in an array

15 Jul, 2022
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 : 1Explanation: 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. 
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Java
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 programint 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
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
// C# program to find the most frequent element// in an arrayusing 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)
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. 
 
C++
C
Java
Python3
C#
PHP
Javascript
// 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 programint 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)
// 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 programint 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 program to find the most frequent element// in an arrayimport 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 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 Codearr = [40,50,30,40,50,30,30]n = len(arr)print(mostFrequent(arr, n)) # This code is contributed by Smitha Dinesh Semwal.
// C# program to find the most// frequent element in an arrayusing 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 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?>
<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>
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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.  
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Java
Python3
C#
Javascript
// 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 programint main(){    int arr[] = {40,50,30,40,50,30,30 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << mostFrequent(arr, n);    return 0;}
//Java program to find the most frequent element//in an arrayimport 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 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 Codearr = [ 40,50,30,40,50,30,30]n = len(arr)print(mostFrequent(arr, n)) # This code is contributed# by Mohit kumar 29
// C# program to find the most// frequent element in an arrayusing 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)
<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 programvar arr = [40,50,30,40,50,30,30];var n = arr.length;document.write( mostFrequent(arr, n)); </script>
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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++
Java
Python3
C#
Javascript
#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}
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 codepublic 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
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 codearr = [ 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
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 codepublic 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
<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>
Element 30 occurs 3 times
Time Complexity: O(n)Auxiliary Space: O(1)
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