How do you solve First Unique Even Element on LeetCode?

First Unique Even Element (LeetCode #3866) 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: Even numbers are divisible by 2.

What is the time complexity of First Unique Even Element?

The optimal solution for First Unique Even Element runs in O(n) time and uses O(n) space. We traverse the array twice, leading to linear time complexity, while using extra space for the hash map.

What is the brute force solution for First Unique Even Element?

The brute force approach for First Unique Even Element is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each even number in the array and count its occurrences. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve First Unique Even Element?

First Unique 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 First Unique Even Element?

Clarify the problem statement. Discuss edge cases upfront. Explain your thought process clearly.

What are common mistakes when solving First Unique Even Element?

Not checking for even numbers properly. Forgetting to handle the case where no unique even exists.

#3866

First Unique 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)

Use a hash map to count occurrences of each even number, then find the first unique even number in a single pass.

⚙️

Algorithm

3 steps

1Step 1: Create a hash map to count occurrences of each even number.
2Step 2: Iterate through the array again to find the first even number with a count of 1.
3Step 3: Return that number or -1 if none exists.

solution.py9 lines

1def first_unique_even(nums):
2    count = {}
3    for num in nums:
4        if num % 2 == 0:
5            count[num] = count.get(num, 0) + 1
6    for num in nums:
7        if num % 2 == 0 and count[num] == 1:
8            return num
9    return -1

ℹ

Complexity note: We traverse the array twice, leading to linear time complexity, while using extra space for the hash map.

1Even numbers are divisible by 2.
2Counting occurrences helps identify uniqueness.

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