LeetCode 165: Compare Version Numbers

Meta Data

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Difficulty: medium First Attempt: 2022-11-19

• Submissions: Accepted (first try)

Approach

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Split on . and convert each revision to int (dropping leading zeros), then compare with one pointer per version (Two Pointers Template 2). Missing revisions of the shorter version count as 0, so 1.0 equals 1.

My Solution

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class Solution:
    def compareVersion(self, version1: str, version2: str) -> int:
        a = list(map(int, version1.split(".")))
        b = list(map(int, version2.split(".")))
        for i in range(max(len(a), len(b))):
            x = a[i] if i < len(a) else 0
            y = b[i] if i < len(b) else 0
            if x != y:
                return 1 if x > y else -1
        return 0

Complexity

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Time O(n + m), space O(n + m).

Mistakes to Avoid

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• Compare as integers: "01" equals "1".
• Pad missing revisions with 0 instead of assuming the longer version is larger.

Interview Explanation

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I split on dots, convert each revision to an integer, and compare them in order, treating missing revisions as 0; the first difference decides. Time O(n + m).