How do you solve Most Frequent Even Element on LeetCode?

Most Frequent Even Element (LeetCode #2404) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient counting of elements.

What is the brute force solution for Most Frequent Even Element?

The brute force approach for Most Frequent Even Element is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every even number in the array and counting its occurrences. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Most Frequent Even Element?

Most Frequent Even Element uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Most Frequent Even Element?

Explain your thought process clearly while coding. Consider edge cases, such as arrays with no even numbers. Always check for ties and how to handle them.

What are common mistakes when solving Most Frequent Even Element?

Not checking if the number is even before counting. Forgetting to handle the case where no even numbers exist.

#2404

Most Frequent Even Element

Easy

ArrayHash TableCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a hash map allows us to count occurrences of each even number in a single pass through the array, making the solution much more efficient.

⚙️

Algorithm

5 steps

1Step 1: Initialize a hash map to store the frequency of each even number.
2Step 2: Loop through the array and for each even number, increment its count in the hash map.
3Step 3: After populating the hash map, initialize variables to track the maximum frequency and the smallest even number.
4Step 4: Loop through the hash map to find the most frequent even number, considering ties by choosing the smaller number.
5Step 5: Return the result or -1 if no even number was found.

solution.py18 lines

1# Full working Python code
2
3def most_frequent_even(nums):
4    freq_map = {}
5    for num in nums:
6        if num % 2 == 0:
7            if num in freq_map:
8                freq_map[num] += 1
9            else:
10                freq_map[num] = 1
11    max_freq = 0
12    result = -1
13    for num, count in freq_map.items():
14        if count > max_freq or (count == max_freq and num < result):
15            max_freq = count
16            result = num
17    return result
18

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once to build the frequency map and then once to find the most frequent even number. The space complexity is O(n) due to the hash map storing the counts.

1Using a hash map allows for efficient counting of elements.
2When dealing with ties, always consider the smallest element.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.