LeetCode 2094: Finding 3-Digit Even Numbers ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2025-05-12

Submissions: Wrong Answer → Accepted

Approach
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There are only 450 three-digit even numbers (100 to 998), so enumerating answers is simpler (Hash Table Template 1):

Count the digits available: have = Counter(digits).
For each candidate x, count the digits it needs: need = Counter(str(x)).
need - have keeps only “needed more than available”; empty means enough.

Enumerating in increasing order gives a sorted result with no duplicates.

My Attempts
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I originally nested three loops over the counter keys to build the digits, excluding cases with too few copies. Attempt 1’s condition (j == k and counter[i] < 2) used the wrong variable; it should check counter[j]. For [3, 2, 2] the answer is [232, 322]; 322 ends in two 2s, and checking the count of 3 dropped it. Changing to counter[j] passed.

Solution
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class Solution:
    def findEvenNumbers(self, digits: List[int]) -> List[int]:
        have = Counter(map(str, digits))
        return [x for x in range(100, 1000, 2) if not Counter(str(x)) - have]

Complexity
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Time O(450 × 3 + n), i.e. O(n), space O(1) (at most 10 digits).

Mistakes to Avoid
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With three nested loops, check the repeat count with the matching position’s variable.
Keep Counter key types consistent: strings on both sides.

Interview Explanation
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There are only 450 three-digit even numbers, so I enumerate 100 to 998 by twos, count the digits each one needs, and compare with the digits on hand; an empty Counter difference means enough. Enumerating in order yields a sorted, duplicate-free result.