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Compare Version Numbers

Difficulty: Medium

Problem Description

Given two version strings, version1 and version2, compare them. A version string consists of revisions separated by dots. The value of the revision is its integer conversion ignoring leading zeros. To compare version strings, compare their revision values in left-to-right order. If one of the version strings has fewer revisions, treat the missing revision values as 0. Return -1 if version1 < version2, return 1 if version1 > version2, otherwise return 0.

Key Insights

  • Version strings are compared by breaking them into integer revisions.
  • Leading zeros in revisions are ignored when making comparisons.
  • Shorter version strings with fewer revisions are treated as having additional zero revisions.

Space and Time Complexity

Time Complexity: O(n + m) where n and m are the lengths of the version strings. Space Complexity: O(1) since we are using a constant amount of space for comparison.

Solution

To solve this problem, we will split both version strings by the dot ('.') delimiter to get individual revision numbers. We will then iterate through the revisions of both versions, converting each revision to an integer to compare their values. If one version runs out of revisions, we treat its missing revisions as zero. We will use a two-pointer approach to iterate through the revisions of both versions until we find a discrepancy or exhaust all revisions.

Code Solutions

def compareVersion(version1: str, version2: str) -> int:
    # Split the version strings by the dot character
    v1 = version1.split('.')
    v2 = version2.split('.')
    
    # Determine the maximum length of the two version lists
    max_length = max(len(v1), len(v2))
    
    for i in range(max_length):
        # Get the current revision value or 0 if out of bounds
        rev1 = int(v1[i]) if i < len(v1) else 0
        rev2 = int(v2[i]) if i < len(v2) else 0
        
        # Compare the current revisions
        if rev1 < rev2:
            return -1
        elif rev1 > rev2:
            return 1
            
    return 0
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