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

Author: jeremymanning

problems/1260/gpt5-mini.md

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+# [Problem 1260: Shift 2D Grid](https://leetcode.com/problems/shift-2d-grid/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+When I first read this, I think of performing the described shift operation k times. Doing that naively (simulate one shift, repeat k times) would be straightforward but could be inefficient if k were large. However, constraints show m,n ≤ 50 and k ≤ 100, so naive might actually pass here — but it's still good practice to do it in O(m*n) time regardless of k.
+
+This is basically a rotation of all elements in row-major order by k positions to the right. So flatten the 2D grid into a 1D list, rotate that list by k (with k reduced mod total elements), and then reshape back into m x n. Alternatively, we can compute index mapping directly without an intermediate list.
+
+I should be careful to take k % (m*n) and handle k == 0 specially (just return the grid as-is or a copy).
+
+## Refining the problem, round 2 thoughts
+Refinements:
+- Compute total = m * n and do k = k % total to avoid unnecessary work.
+- Flattening to a 1D list makes rotation easier: rotated = flat[-k:] + flat[:-k] (if k > 0).
+- Reconstruct the 2D result by slicing the rotated list into rows of length n.
+- Edge cases:
+  - k == 0 (or k % total == 0) -> return the original grid (or a shallow copy).
+  - m or n equals 1 -> still works with flatten+rotate approach.
+- Complexity: flattening is O(total) time and space; rotating using slicing is O(total); rebuilding is O(total). So overall O(m*n) time and O(m*n) extra space. We could do an in-place cycle-rotate with O(1) extra space, but it's unnecessary here and adds complexity.
+
+Now implement the clear, concise Python solution using flatten / slice / reshape.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
+        m = len(grid)
+        n = len(grid[0]) if m > 0 else 0
+        total = m * n
+        if total == 0:
+            return grid
+        
+        k %= total
+        if k == 0:
+            # No change needed; return a copy to avoid mutating input if that matters
+            return [row[:] for row in grid]
+        
+        # Flatten the grid in row-major order
+        flat = [grid[i][j] for i in range(m) for j in range(n)]
+        # Rotate the flattened list to the right by k
+        rotated = flat[-k:] + flat[:-k]
+        # Reconstruct the 2D grid
+        res = []
+        for i in range(m):
+            row = rotated[i * n : (i + 1) * n]
+            res.append(row)
+        return res
+```
+- Notes about the approach:
+  - We flatten the 2D grid to a 1D list in row-major order, perform a right rotation by k using slicing, and then reshape back to m x n.
+  - Time complexity: O(m * n) — flattening, slicing and rebuilding each touch each element a constant number of times.
+  - Space complexity: O(m * n) extra space for the flattened and rotated lists (can be reduced with an in-place cycle rotation, but not necessary given constraints).
+  - Important detail: reduce k with k %= m*n to handle cases where k ≥ total elements.