Auto-solve daily LeetCode problem using GPT-5-mini

Author: jeremymanning

problems/1886/gpt5-mini.md

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+# [Problem 1886: Determine Whether Matrix Can Be Obtained By Rotation](https://leetcode.com/problems/determine-whether-matrix-can-be-obtained-by-rotation/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+I need to check whether rotating mat by 0, 90, 180, or 270 degrees (clockwise) can make it equal to target. Since n is at most 10, a brute-force approach that explicitly rotates and compares is fine. Rotating a matrix 90° clockwise can be done by mapping position (i, j) -> (j, n-1-i), or by transposing and then reversing each row. I'll just build a new rotated matrix each time and compare to target; if any match, return True. Watch for the case where mat is already equal to target (0 rotations). Complexity will be small (O(n^2) per rotation).
+
+## Refining the problem, round 2 thoughts
+Refinements / alternatives:
+- We can rotate in-place (transpose + reverse rows) to avoid extra space, but building a new matrix is simpler and straightforward given the small constraints.
+- Edge case: n = 1 -> just compare single elements.
+- Stop early as soon as a rotation matches target.
+- Time complexity: up to 4 * O(n^2) comparisons/constructs. Space complexity: O(n^2) extra for a single rotated matrix (or O(1) extra if done in-place).
+- Use Python list equality for matrix comparison (works for nested lists).
+
+## Attempted solution(s)
+```python
+class Solution:
+    def findRotation(self, mat: list[list[int]], target: list[list[int]]) -> bool:
+        n = len(mat)
+        
+        def rotate90(matrix):
+            # Return new matrix rotated 90 degrees clockwise
+            return [[matrix[n - 1 - j][i] for j in range(n)] for i in range(n)]
+        
+        current = mat
+        for _ in range(4):
+            if current == target:
+                return True
+            current = rotate90(current)
+        return False
+```
+- Approach: try up to 4 rotations (including 0 rotations), building a new matrix for each 90° clockwise rotation using index mapping (i, j) <- (n-1-j, i). Compare with target using list equality; return True if any match.
+- Time complexity: O(4 * n^2) = O(n^2) since constants drop. Each rotation and comparison is O(n^2).
+- Space complexity: O(n^2) for the temporary rotated matrix each iteration (can be reduced to O(1) with in-place rotation).
+- Implementation details: function is implemented as LeetCode expects within class Solution and uses Python list comprehensions for clarity.