Most frequent element in an array (original) (raw)

Last Updated : 26 Jul, 2025

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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.

Table of Content

[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 #include

using namespace std;

int mostFreqEle(vector &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 arr = { 40, 50, 30, 40, 50, 30, 30 }; cout << mostFreqEle(arr); return 0; }

Java

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));
}

}

Python

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))

C#

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));
}

}

JavaScript

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 #include #include

using namespace std;

int mostFreqEle(vector& 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 arr = { 40, 50, 30, 40, 50, 30, 30 }; cout << mostFreqEle(arr); return 0; }

Java

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));
}

}

Python

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))

C#

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));
}

}

JavaScript

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 #include #include

using namespace std;

int mostFreqEle(vector& 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 arr = { 40, 50, 30, 40, 50, 30, 30 }; cout << mostFreqEle(arr); return 0; }

Java

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));
}

}

Python

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))

C#

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));
}

}

JavaScript

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));

`