How do you solve Finding 3-Digit Even Numbers on LeetCode?

Finding 3-Digit Even Numbers (LeetCode #2094) 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 must end with an even digit (0, 2, 4, 6, 8).

What is the brute force solution for Finding 3-Digit Even Numbers?

The brute force approach for Finding 3-Digit Even Numbers is Brute Force, with O(n³) time complexity and O(n) space complexity. The brute force approach generates all possible 3-digit combinations from the given digits and checks if they meet the criteria of being even and having no leading zeros. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Finding 3-Digit Even Numbers?

Finding 3-Digit Even Numbers uses the following concepts: Array, Hash Table, Recursion, Sorting, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Finding 3-Digit Even Numbers?

Always clarify the problem constraints and edge cases. Think about how to optimize your solution before coding. Discuss your thought process and approach with the interviewer.

What are common mistakes when solving Finding 3-Digit Even Numbers?

Not checking for leading zeros when forming numbers. Failing to handle duplicates correctly, leading to incorrect results.

#2094

Finding 3-Digit Even Numbers

Easy

ArrayHash TableRecursionSortingEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n³)	O(n)
Space	O(n)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach focuses on identifying the even digits first and then constructing valid 3-digit numbers without leading zeros. This reduces unnecessary checks and ensures we only consider valid combinations.

⚙️

Algorithm

3 steps

1Step 1: Count the occurrences of each digit using a frequency array.
2Step 2: Identify even digits and iterate through them as potential last digits of the 3-digit number.
3Step 3: For each even digit, form combinations with the remaining digits, ensuring no leading zeros.

solution.py17 lines

1from collections import Counter
2
3def findEvenNumbers(digits):
4    count = Counter(digits)
5    result = set()
6    for last_digit in [d for d in count if d % 2 == 0]:
7        count[last_digit] -= 1
8        for first_digit in count:
9            if first_digit == 0 and count[first_digit] == 0:
10                continue
11            for second_digit in count:
12                if count[first_digit] > 0 and count[second_digit] > 0:
13                    num = first_digit * 100 + second_digit * 10 + last_digit
14                    if num >= 100:
15                        result.add(num)
16        count[last_digit] += 1
17    return sorted(result)

ℹ

Complexity note: The time complexity is O(n) as we only iterate through the digits a limited number of times. The space complexity is O(n) due to the storage of unique numbers.

1Even numbers must end with an even digit (0, 2, 4, 6, 8).
2Leading zeros are not allowed, so the first digit must be non-zero.

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