Manchester | 25-SDC-Nov | Rahwa Haile | Sprint 2 | improve_with_precomputing#128
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RahwaZeslusHaile wants to merge 3 commits intoCodeYourFuture:mainfrom
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Manchester | 25-SDC-Nov | Rahwa Haile | Sprint 2 | improve_with_precomputing#128RahwaZeslusHaile wants to merge 3 commits intoCodeYourFuture:mainfrom
RahwaZeslusHaile wants to merge 3 commits intoCodeYourFuture:mainfrom
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Code looks good.
Can you use complexity to explain how the new implementation are better than the original implementation?
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Thanks @cjyuan for that feedback! I've documented the complexity analysis in the README. |
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| - Time Complexity: **O(n² × m)** - Compares every string with every other string (n²) pairs, each taking O(m) comparisons | ||
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| **Optimized Approach (Sort + Adjacent Pairs):** | ||
| - Time Complexity: **O(n log n + n × m)** - Sort once (n log n), then compare only adjacent pairs (n comparisons) |
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If we are factoring in the length of the strings, m, then the complexity of sorting won't just be O(nlogn). nlogn measures only the number of comparisons needed.
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| **Optimized Approach (Sort + Adjacent Pairs):** | ||
| - Time Complexity: **O(n log n + n × m)** - Sort once (n log n), then compare only adjacent pairs (n comparisons) | ||
| - **Why better:** The longest common prefix will be found in the first and last strings after sorting, eliminating unnecessary comparisons. Reduces from O(n²) to O(n) pair comparisons. |
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What do you mean by first and last strings? That sounds like only two comparisons are needed to find the longest prefix after sorting.
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Thanks @cjyuan for the detailed feedback! You're absolutely right on both points. I've updated the README to correct these issues
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Good analysis. Well done! |
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Manchester | 25-SDC-Nov | Rahwa Haile | Sprint 2 | improve_with_precomputing
Learners, PR Template
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Description
Improves the performance of two algorithms using precomputing strategies without modifying existing tests.
Changelist
Common Prefix:
Pre-sort strings once, then compare only adjacent pairs
Reduces comparisons from O(n²) to O(n log n) sort + O(n) comparisons
Count Letters:
Precompute lowercase letters into a set during the first pass
Replace repeated letter.lower() in s checks with O(1) set lookups
Reduces time complexity from O(n²) to O(n)
I really appreciate the time you take to review this PR