Most Frequent Even Element

Most Frequent Even Element - Problem

Given an integer array nums, return the most frequent even element.

If there is a tie, return the smallest one. If there is no such element, return -1.

Input & Output

Example 1 — Basic Case with Tie

$ Input: nums = [0,1,2,2,5,4,4,6]

› Output: 2

💡 Note: Even numbers: 0(1×), 2(2×), 4(2×), 6(1×). Both 2 and 4 appear twice, but 2 < 4, so return 2.

Example 2 — Clear Winner

$ Input: nums = [4,4,4,9,2,4]

› Output: 4

💡 Note: Even numbers: 4(4×), 2(1×). 4 appears most frequently, so return 4.

Example 3 — No Even Numbers

$ Input: nums = [29,47,21,41,13,37,25,7]

› Output: -1

💡 Note: All numbers are odd, so there are no even elements to return.

Constraints

1 ≤ nums.length ≤ 2000
-10⁵ ≤ nums[i] ≤ 10⁵

Visualization

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Asked in

F Facebook 15 G Google 12

The key insight is to count frequencies of only even numbers and handle tie-breaking by choosing the smallest value. Best approach is hash map frequency counting for O(n) time and clear logic. Time: O(n), Space: O(k) where k is number of unique even elements.

Common Approaches

✓ Brute Force - Count Each Even Number

⏱️ Time: O(n²) Space: O(1)

Iterate through all numbers, and for each even number, count how many times it appears in the array. Track the most frequent even number, using the smallest value to break ties.

Hash Map Frequency Counting

⏱️ Time: O(n) Space: O(k)

Iterate through the array once, counting frequencies of even numbers using a hash map. Then find the most frequent even number, using the smallest value to break ties.

Brute Force - Count Each Even Number — Algorithm Steps

Step 1: Iterate through each number in the array
Step 2: If number is even, count its total occurrences
Step 3: Track the best candidate (most frequent, smallest for ties)

Visualization

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Step-by-Step Walkthrough

Find Even Numbers

Identify even numbers in the array

Count Frequencies

For each even number, count total occurrences

Find Winner

Select most frequent (smallest for ties)

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int bestNum = -1;
    int maxCount = 0;
    
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] % 2 == 0) {  // Only consider even numbers
            int count = 0;
            for (int j = 0; j < numsSize; j++) {
                if (nums[j] == nums[i]) {
                    count++;
                }
            }
            
            // Update best if this is better
            if (count > maxCount || (count == maxCount && (bestNum == -1 || nums[i] < bestNum))) {
                bestNum = nums[i];
                maxCount = count;
            }
        }
    }
    
    return bestNum;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[100];
    int numsSize = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, "[],\n");
    }
    
    printf("%d\n", solution(nums, numsSize));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops: outer loop O(n) × inner counting loop O(n)

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track best candidate

✓ Linear Space

12.5K Views

Medium Frequency

~15 min Avg. Time

285 Likes

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Count Each Even Number — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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