Most frequent element in an array
Given an array, find the most frequent element in it. If there are multiple elements that appear maximum number of times, print any one of them.
Examples:
Input : arr[] = {1, 3, 2, 1, 4, 1}
Output : 1
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 frequency of the picked element and compares with the maximum so far. Time complexity of this solution is O(n2)
A better solution is to do sorting. We first sort the array, then linearly traverse the array.
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 { if (curr_count > max_count) { max_count = curr_count; res = arr[i - 1]; } curr_count = 1; } } // If last element is most frequent if (curr_count > max_count) { max_count = curr_count; res = arr[n - 1]; } return res;}// driver programint main(){ int arr[] = { 1, 5, 2, 1, 3, 2, 1 }; int n = sizeof(arr) / sizeof(arr[0]); cout << mostFrequent(arr, n); return 0;} |
Java
//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 { if (curr_count > max_count) { max_count = curr_count; res = arr[i - 1]; } curr_count = 1; } } // If last element is most frequent if (curr_count > max_count) { max_count = curr_count; res = arr[n - 1]; } return res; } // Driver program public static void main (String[] args) { int arr[] = {1, 5, 2, 1, 3, 2, 1}; 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.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 : if (curr_count > max_count): max_count = curr_count res = arr[i - 1] curr_count = 1 # If last element is most frequent if (curr_count > max_count): max_count = curr_count res = arr[n - 1] return res# Driver Codearr = [1, 5, 2, 1, 3, 2, 1]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 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 { if (curr_count > max_count) { max_count = curr_count; res = arr[i - 1]; } curr_count = 1; } } // If last element is most frequent if (curr_count > max_count) { max_count = curr_count; res = arr[n - 1]; } return res; } // Driver code public static void Main () { int []arr = {1, 5, 2, 1, 3, 2, 1}; 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]) $curr_count++; else { if ($curr_count > $max_count) { $max_count = $curr_count; $res = $arr[$i - 1]; } $curr_count = 1; } } // If last element // is most frequent if ($curr_count > $max_count) { $max_count = $curr_count; $res = $arr[$n - 1]; } return $res;}// Driver Code{ $arr = array(1, 5, 2, 1, 3, 2, 1); $n = sizeof($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 { if (curr_count > max_count) { max_count = curr_count; res = arr[i - 1]; } curr_count = 1; } } // If last element is most frequent if (curr_count > max_count) { max_count = curr_count; res = arr[n - 1]; } return res; } // Driver Code let arr = [1, 5, 2, 1, 3, 2, 1]; let n = arr.length; document.write(mostFrequent(arr,n));// This code is contributed by code_hunt.</script> |
Output :
1
Time Complexity : O(n Log 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 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 programint main(){ int arr[] = { 1, 5, 2, 1, 3, 2, 1 }; int n = sizeof(arr) / sizeof(arr[0]); cout << mostFrequent(arr, n); return 0;} |
Java
//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[] = {1, 5, 2, 1, 3, 2, 1}; 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 mtdef 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 = [ 1, 5, 2, 1, 3, 2, 1]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 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[]{1, 5, 2, 1, 3, 2, 1}; int n = arr.Length; Console.Write(mostFrequent(arr, n)); }}// This code is contributed by// Manish Shaw(manishshaw1) |
Output:
1
Time Complexity : O(n)
Auxiliary Space : O(n)
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