How do you solve Compare Version Numbers on LeetCode?

Compare Version Numbers (LeetCode #165) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Version segments are compared as integers, ignoring leading zeros.

What is the time complexity of Compare Version Numbers?

The optimal solution for Compare Version Numbers runs in O(n) time and uses O(1) space. The complexity is O(n) because we traverse each version string only once, making it efficient for longer strings.

What is the brute force solution for Compare Version Numbers?

The brute force approach for Compare Version Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves splitting both version strings into their individual revision components and comparing them one by one. This method is straightforward but can be inefficient due to nested loops. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Compare Version Numbers?

Compare Version Numbers uses the following concepts: Two Pointers, String. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Compare Version Numbers?

Always clarify the problem statement and edge cases before coding. Think about how to handle different lengths of version strings. Practice breaking down the problem into smaller parts and solving each part.

What are common mistakes when solving Compare Version Numbers?

Failing to handle leading zeros properly when converting to integers. Not considering the case where one version has more segments than the other.

#165

Compare Version Numbers

Medium

Two PointersStringTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses two pointers to traverse both version strings simultaneously. This method is efficient as it avoids unnecessary comparisons and directly handles missing revisions.

⚙️

Algorithm

3 steps

1Step 1: Initialize two pointers for each version string.
2Step 2: Use a loop to compare segments of both versions until both pointers reach the end.
3Step 3: For each segment, convert to integer and compare, treating missing segments as 0.

solution.py24 lines

1# Full working Python code
2version1 = '1.2'
3version2 = '1.10'
4
5p1, p2 = 0, 0
6while p1 < len(version1) or p2 < len(version2):
7    v1_revision = 0
8    v2_revision = 0
9    while p1 < len(version1) and version1[p1] != '.':
10        v1_revision = v1_revision * 10 + int(version1[p1])
11        p1 += 1
12    while p2 < len(version2) and version2[p2] != '.':
13        v2_revision = v2_revision * 10 + int(version2[p2])
14        p2 += 1
15    if v1_revision < v2_revision:
16        print(-1)
17        break
18    elif v1_revision > v2_revision:
19        print(1)
20        break
21    p1 += 1
22    p2 += 1
23else:
24    print(0)

ℹ

Complexity note: The complexity is O(n) because we traverse each version string only once, making it efficient for longer strings.

1Version segments are compared as integers, ignoring leading zeros.
2Missing segments are treated as zero, allowing for flexible versioning.

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