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Pull request overview
This pull request updates the TypeScript solution for LeetCode problem #1631 (Path With Minimum Effort). The problem involves finding the minimum effort required to travel from the top-left to bottom-right cell in a 2D grid, where effort is defined as the maximum absolute height difference between consecutive cells.
Key Changes
- Updates to TypeScript solution implementations for problem 1631
- Likely addresses inconsistencies in PriorityQueue usage patterns
- Ensures consistency between solution files and documentation
Reviewed changes
Copilot reviewed 3 out of 3 changed files in this pull request and generated no comments.
Show a summary per file
Since the actual diff is not provided in the review context, I analyzed the current state of the TypeScript solution files and identified the following:
| File | Description |
|---|---|
| Solution.ts | Union-Find based approach - appears correct and consistent with other language implementations |
| Solution2.ts | Binary Search + BFS approach - appears correct and consistent |
| Solution3.ts | Dijkstra with Priority Queue - contains PriorityQueue constructor pattern that differs from README documentation |
| README.md | Chinese documentation showing PriorityQueue with object-based constructor |
| README_EN.md | English documentation showing PriorityQueue with object-based constructor |
Note: I observed an inconsistency in the codebase where Solution3.ts uses a function-based PriorityQueue constructor new PriorityQueue<number[]>((a, b) => ...), while both README files document an object-based pattern new PriorityQueue({ compare: (a, b) => ... }). Both patterns appear to be used throughout the codebase in different problems. Without access to the actual diff showing which lines were modified, I cannot provide specific line-by-line feedback on what was changed in this PR.
The three solution approaches (Union-Find, Binary Search + BFS, and Dijkstra) are all algorithmically sound for solving this problem, with time complexities of O(m×n×log(m×n)), O(m×n×log M), and O(m×n×log(m×n)) respectively.
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