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

Author: jeremymanning

problems/1582/gpt5-mini.md

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+# [Problem 1582: Special Positions in a Binary Matrix](https://leetcode.com/problems/special-positions-in-a-binary-matrix/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We need to count positions (i, j) where mat[i][j] == 1 and every other element in row i and column j is 0. The brute force way would be: for every 1, scan its entire row and column to verify there are no other 1s — that would be O(m + n) work per 1 and in worst case O(mn(m+n)), which is unnecessary for constraints up to 100x100 but still suboptimal.
+
+A common pattern: precompute the count of 1s in each row and each column. If a cell is 1 and its row-count == 1 and its col-count == 1, it's special. That gives a single pass to compute counts and another pass to count special positions — overall O(mn) time.
+
+## Refining the problem, round 2 thoughts
+- Edge cases: all zeros -> answer 0. All ones -> usually 0 unless matrix is 1x1.
+- Precomputing row and column sums is straightforward and stable.
+- Space: need O(m + n) extra space for the counts, which is fine for m,n <= 100.
+- Alternative in-place trick: you could encode counts into existing matrix to save auxiliary space, but it's unnecessary here and less clear.
+- Complexity: two passes over the matrix -> O(mn) time, O(m + n) extra space.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def numSpecial(self, mat: List[List[int]]) -> int:
+        if not mat or not mat[0]:
+            return 0
+        m, n = len(mat), len(mat[0])
+        # Compute row counts
+        row_count = [sum(row) for row in mat]
+        # Compute column counts
+        col_count = [0] * n
+        for j in range(n):
+            s = 0
+            for i in range(m):
+                s += mat[i][j]
+            col_count[j] = s
+
+        # Count special positions
+        ans = 0
+        for i in range(m):
+            for j in range(n):
+                if mat[i][j] == 1 and row_count[i] == 1 and col_count[j] == 1:
+                    ans += 1
+        return ans
+```
+- Notes:
+  - Approach: precompute number of 1s in each row and column, then count cells where mat[i][j] == 1 and both corresponding counts are exactly 1.
+  - Time complexity: O(m * n) — one pass to compute row counts (O(mn) inside sum across rows) and one pass for column counts (O(mn)), plus final check pass (O(mn)).
+  - Space complexity: O(m + n) for row_count and col_count arrays.
+  - Implementation detail: using Python's sum(row) for rows is concise; columns computed by iterating rows for each column is clear and efficient for given constraints.