Most frequent element in an array
Last Updated : 26 Jul, 2025
Given an integer array arr[], find the element that appears most frequently. If multiple elements have the same highest frequency, return the largest among 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 appears four times in array which is maximum frequency
Input: arr[] = [1, 2, 2, 4, 1]
Output: 2
Explanation: 1 and 2 both appear two times, so return 2 as it's value is bigger.
[Naive Approach] Nested Loops - O(n2) Time and O(1) Space
The naive approach involves using two nested loops: the outer loop picks each element, and the inner loop counts the frequency of the picked element. Update if count is higher than current max count or same but value is larger than current value.
C++ #include <iostream> #include <vector> using namespace std; int mostFreqEle(vector<int> &arr) { int n = arr.size(), maxcount = 0; int res; 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 is greater or if count // is same but value is bigger. if (count > maxcount || (count == maxcount && arr[i] > res)) { maxcount = count; res = arr[i]; } } return res; } int main() { vector<int> arr = { 40, 50, 30, 40, 50, 30, 30 }; cout << mostFreqEle(arr); return 0; } class GfG { static int mostFreqEle(int[] arr) { int n = arr.length, maxcount = 0; int res = 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 is greater or if count // is same but value is bigger. if (count > maxcount || (count == maxcount && arr[i] > res)) { maxcount = count; res = arr[i]; } } return res; } public static void main(String[] args) { int[] arr = { 40, 50, 30, 40, 50, 30, 30 }; System.out.println(mostFreqEle(arr)); } } def mostFreqEle(arr): n = len(arr) maxcount = 0 res = 0 for i in range(n): count = 0 for j in range(n): if arr[i] == arr[j]: count += 1 # If count is greater or if count # is same but value is bigger. if count > maxcount or (count == maxcount and arr[i] > res): maxcount = count res = arr[i] return res if __name__ == "__main__": arr = [40, 50, 30, 40, 50, 30, 30] print(mostFreqEle(arr))
using System; class GfG { static int mostFreqEle(int[] arr) { int n = arr.Length, maxcount = 0; int res = 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 is greater or if count // is same but value is bigger. if (count > maxcount || (count == maxcount && arr[i] > res)) { maxcount = count; res = arr[i]; } } return res; } static void Main() { int[] arr = { 40, 50, 30, 40, 50, 30, 30 }; Console.WriteLine(mostFreqEle(arr)); } } function mostFreqEle(arr) { let n = arr.length, maxcount = 0; let res = 0; for (let i = 0; i < n; i++) { let count = 0; for (let j = 0; j < n; j++) { if (arr[i] === arr[j]) count++; } // If count is greater or if count // is same but value is bigger. if (count > maxcount || (count === maxcount && arr[i] > res)) { maxcount = count; res = arr[i]; } } return res; } // Driver Code let arr = [40, 50, 30, 40, 50, 30, 30]; console.log(mostFreqEle(arr)); [Better Approach] Using Sorting - O(n × log n) Time and O(1) Space
This approach begins by sorting the array, which groups identical elements together. Once sorted, a linear traversal is performed to count the frequency of each element. Since identical elements appear consecutively after sorting, tracking their frequency becomes straightforward.
C++ #include <iostream> #include <vector> #include <algorithm> using namespace std; int mostFreqEle(vector<int>& arr) { // Sort the array sort(arr.begin(), arr.end()); int maxCnt = 1, res = arr[0], currCnt = 1; for (int i = 1; i < arr.size(); i++) { // previous element if equal to current element if (arr[i] == arr[i - 1]) currCnt++; else currCnt = 1; // Update if curr count is higher or same but value is larger if (currCnt > maxCnt || (currCnt == maxCnt && arr[i] > res)) { maxCnt = currCnt; res = arr[i]; } } return res; } int main() { vector<int> arr = { 40, 50, 30, 40, 50, 30, 30 }; cout << mostFreqEle(arr); return 0; } import java.util.Arrays; class GfG { static int mostFreqEle(int[] arr) { // Sort the array Arrays.sort(arr); int maxCnt = 1, res = arr[0], currCnt = 1; for (int i = 1; i < arr.length; i++) { // previous element if equal to current element if (arr[i] == arr[i - 1]) currCnt++; else currCnt = 1; // Update if curr count is higher or same but value is larger if (currCnt > maxCnt || (currCnt == maxCnt && arr[i] > res)) { maxCnt = currCnt; res = arr[i]; } } return res; } public static void main(String[] args) { int[] arr = { 40, 50, 30, 40, 50, 30, 30 }; System.out.println(mostFreqEle(arr)); } } def mostFreqEle(arr): # Sort the array arr.sort() maxCnt, res, currCnt = 1, arr[0], 1 for i in range(1, len(arr)): # previous element if equal to current element if arr[i] == arr[i - 1]: currCnt += 1 else: currCnt = 1 # Update if curr count is higher or same but value is larger if currCnt > maxCnt or (currCnt == maxCnt and arr[i] > res): maxCnt = currCnt res = arr[i] return res if __name__ == "__main__": arr = [40, 50, 30, 40, 50, 30, 30] print(mostFreqEle(arr))
using System; class GfG { static int mostFreqEle(int[] arr) { // Sort the array Array.Sort(arr); int maxCnt = 1, res = arr[0], currCnt = 1; for (int i = 1; i < arr.Length; i++) { // previous element if equal to current element if (arr[i] == arr[i - 1]) currCnt++; else currCnt = 1; // Update if curr count is higher or same but value is larger if (currCnt > maxCnt || (currCnt == maxCnt && arr[i] > res)) { maxCnt = currCnt; res = arr[i]; } } return res; } static void Main() { int[] arr = { 40, 50, 30, 40, 50, 30, 30 }; Console.WriteLine(mostFreqEle(arr)); } } function mostFreqEle(arr) { // Sort the array arr.sort((a, b) => a - b); let maxCnt = 1, res = arr[0], currCnt = 1; for (let i = 1; i < arr.length; i++) { // previous element if equal to current element if (arr[i] === arr[i - 1]) currCnt++; else currCnt = 1; // Update if curr count is higher or same but value is larger if (currCnt > maxCnt || (currCnt === maxCnt && arr[i] > res)) { maxCnt = currCnt; res = arr[i]; } } return res; } // Driver Code let arr = [40, 50, 30, 40, 50, 30, 30]; console.log(mostFreqEle(arr)); [Expected Approach] Using Hashing - O(n) Time and O(n) Space
This approach uses a hash table to count the frequency of each element efficiently. After building the frequency map, the element with the highest frequency is identified and choosing the largest one in case of a tie.
C++ #include <iostream> #include <vector> #include <unordered_map> using namespace std; int mostFreqEle(vector<int>& arr) { int n = arr.size(); // Insert all elements in hash map. unordered_map<int, int> freq; for (int i = 0; i < n; i++) freq[arr[i]]++; // find the max frequency int maxCnt = 0, res = -1; for (auto i : freq) { int val = i.first, cnt = i.second; // Update if frequency is higher or same but value is larger if (maxCnt < cnt || (cnt == maxCnt && val > res)) { res = val; maxCnt = cnt; } } return res; } int main() { vector<int> arr = { 40, 50, 30, 40, 50, 30, 30 }; cout << mostFreqEle(arr); return 0; } import java.util.HashMap; class GfG { static int mostFreqEle(int[] arr) { int n = arr.length; // Insert all elements in hash map. HashMap<Integer, Integer> freq = new HashMap<>(); for (int i = 0; i < n; i++) freq.put(arr[i], freq.getOrDefault(arr[i], 0) + 1); // find the max frequency int maxCnt = 0, res = -1; for (var entry : freq.entrySet()) { int val = entry.getKey(), cnt = entry.getValue(); // Update if frequency is higher or same but value is larger if (maxCnt < cnt || (cnt == maxCnt && val > res)) { res = val; maxCnt = cnt; } } return res; } public static void main(String[] args) { int[] arr = { 40, 50, 30, 40, 50, 30, 30 }; System.out.println(mostFreqEle(arr)); } } def mostFreqEle(arr): n = len(arr) # Insert all elements in hash map. freq = {} for i in range(n): freq[arr[i]] = freq.get(arr[i], 0) + 1 # find the max frequency maxCnt, res = 0, -1 for val, cnt in freq.items(): # Update if frequency is higher or same but value is larger if maxCnt < cnt or (cnt == maxCnt and val > res): res = val maxCnt = cnt return res if __name__ == "__main__": arr = [40, 50, 30, 40, 50, 30, 30] print(mostFreqEle(arr)) using System; using System.Collections.Generic; class GfG { static int mostFreqEle(int[] arr) { int n = arr.Length; // Insert all elements in hash map. Dictionary<int, int> freq = new Dictionary<int, int>(); for (int i = 0; i < n; i++) freq[arr[i]] = freq.ContainsKey(arr[i]) ? freq[arr[i]] + 1 : 1; // find the max frequency int maxCnt = 0, res = -1; foreach (var entry in freq) { int val = entry.Key, cnt = entry.Value; // Update if frequency is higher or same but value is larger if (maxCnt < cnt || (cnt == maxCnt && val > res)) { res = val; maxCnt = cnt; } } return res; } static void Main() { int[] arr = { 40, 50, 30, 40, 50, 30, 30 }; Console.WriteLine(mostFreqEle(arr)); } } function mostFreqEle(arr) { let n = arr.length; // Insert all elements in hash map. let freq = new Map(); for (let i = 0; i < n; i++) freq.set(arr[i], (freq.get(arr[i]) || 0) + 1); // find the max frequency let maxCnt = 0, res = -1; for (let [val, cnt] of freq) { // Update if frequency is higher or same but value is larger if (maxCnt < cnt || (cnt === maxCnt && val > res)) { res = val; maxCnt = cnt; } } return res; } // Driver Code let arr = [40, 50, 30, 40, 50, 30, 30]; console.log(mostFreqEle(arr));
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