Most Frequent Even Element - Practice Coding Problems

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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Understanding the Visualization

Input Array

Array with mix of odd and even numbers

Filter & Count

Count frequencies of even numbers only

Find Winner

Most frequent even number (smallest for ties)

Key Takeaway

🎯 Key Insight: Count only even numbers and use smallest value to break frequency ties

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

Approach	Time	Space	Notes
✓ Brute Force - Count Each Even Number	O(n²)	O(1)	For each even number, count its frequency by scanning the entire array
Hash Map Frequency Counting	O(n)	O(k)	Use a hash map to count frequencies of even numbers in a single pass

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

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Count Each Even Number — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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