#165

Medium

auto_awesomeTwo-pointer scanning with invariant tracking

LeetCode Problem Workspace

Compare Version Numbers

Compare two version numbers by checking their individual revisions, considering missing revisions as zero, using a two-pointer technique.

two pointers string

Problem Statement

Given two version strings, version1 and version2, compare them. A version string consists of revisions separated by dots '.'. Each revision value is treated as an integer, ignoring any leading zeros. If one version has fewer revisions, assume missing values are 0 for the comparison.

To compare the two version strings, traverse them segment by segment using two pointers. Compare the revision values at each pointer position. If a revision is missing in one version, treat it as 0. Return -1 if version1 is smaller, 1 if version1 is greater, and 0 if they are equal.

Examples

Example 1

Input: version1 = "1.2", version2 = "1.10"

Output: -1

version1's second revision is "2" and version2's second revision is "10": 2 < 10, so version1 < version2.

Example 2

Input: version1 = "1.01", version2 = "1.001"

Output: 0

Ignoring leading zeroes, both "01" and "001" represent the same integer "1".

Example 3

Input: version1 = "1.0", version2 = "1.0.0.0"

Output: 0

version1 has less revisions, which means every missing revision are treated as "0".

Constraints

1 <= version1.length, version2.length <= 500
version1 and version2 only contain digits and '.'.
version1 and version2 are valid version numbers.
All the given revisions in version1 and version2 can be stored in a 32-bit integer.

Solution Approach

Two-pointer scanning

Use two pointers, one for each version string, and traverse them segment by segment. For each segment, compare the integer values of the corresponding revisions, treating missing revisions as zeros.

Handle missing segments

If one version has fewer segments than the other, append zeros to the shorter version. This way, all missing revisions are considered as zero for comparison.

Compare segments in order

Always compare corresponding revision segments from left to right, ensuring that no segment is skipped. Adjust the pointers as you traverse each revision in both version strings.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this approach is O(n + m), where n and m are the lengths of version1 and version2, respectively. The space complexity is O(1) as only a constant amount of space is required for the pointers and comparison logic.

What Interviewers Usually Probe

Tests the candidate's ability to handle string manipulation and parsing with an efficient solution.
Assesses understanding of two-pointer technique and how to manage different-length sequences.
Checks if the candidate can account for edge cases like missing version revisions or leading zeros.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle cases where one version has more revisions than the other.
Incorrectly handling leading zeros in revision values.
Not comparing all segments in order, which could lead to incorrect results.

Follow-up variants

Handle versions with leading zeros and unequal segment lengths.
Compare version strings that might contain a mix of long and short revision segments.
Extend the solution to compare version strings with a very large number of revisions.

FAQ

How can I compare two version numbers efficiently?

You can compare two version numbers using the two-pointer technique, comparing corresponding revision values from left to right.

What is the correct way to handle missing revisions?

When one version has fewer revisions, treat the missing revisions as zero.

How do I handle leading zeros in version numbers?

Simply ignore leading zeros by converting each revision segment to an integer before comparing them.

Why use two-pointer scanning in the Compare Version Numbers problem?

Two-pointer scanning allows you to efficiently traverse and compare the revision segments of both version strings in sync.

What is the time complexity of the solution?

The time complexity is O(n + m), where n and m are the lengths of the two version strings.

terminal

Solution

Solution 1: Two Pointers

Traverse both strings simultaneously using two pointers $i$ and $j$, which point to the current positions in each string, starting with $i = j = 0$.

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class Solution:
    def compareVersion(self, version1: str, version2: str) -> int:
        m, n = len(version1), len(version2)
        i = j = 0
        while i < m or j < n:
            a = b = 0
            while i < m and version1[i] != '.':
                a = a * 10 + int(version1[i])
                i += 1
            while j < n and version2[j] != '.':
                b = b * 10 + int(version2[j])
                j += 1
            if a != b:
                return -1 if a < b else 1
            i, j = i + 1, j + 1
        return 0

Continue Practicing

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