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

Author: jeremymanning

problems/3190/gpt5-mini.md

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+# [Problem 3190: Find Minimum Operations to Make All Elements Divisible by Three](https://leetcode.com/problems/find-minimum-operations-to-make-all-elements-divisible-by-three/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We only care about each number modulo 3. For a number x, let r = x % 3:
+- If r == 0, it's already divisible by 3 (cost 0).
+- If r == 1, we can subtract 1 (cost 1) or add 2 (cost 2) — best is 1.
+- If r == 2, we can add 1 (cost 1) or subtract 2 (cost 2) — best is 1.
+
+So each element with remainder 1 or 2 costs exactly 1 operation. That suggests the answer is simply the count of elements with remainder != 0. I briefly wondered if operations could be shared or paired to reduce cost, but operations act on single elements independently (add/subtract 1 to any element), so there's no sharing advantage.
+
+## Refining the problem, round 2 thoughts
+- Edge cases: small arrays, all already divisible by 3 -> answer 0. Works fine.
+- Alternative expression: sum(min(r, 3-r) for each remainder r) which reduces here to counting nonzero remainders since min(1,2)=1 and min(2,1)=1.
+- Time complexity: O(n) where n = len(nums).
+- Space complexity: O(1) extra (ignoring input), only a counter required.
+- Implementation detail: use generator expression to count nums with x % 3 != 0 for succinctness.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def minimumOperations(self, nums: List[int]) -> int:
+        # Each element with remainder 1 or 2 needs exactly 1 operation to become divisible by 3.
+        return sum(1 for x in nums if x % 3 != 0)
+```
+- Notes:
+  - Approach: Count how many elements have remainder 1 or 2 modulo 3. Each such element requires one +1 or -1 operation.
+  - Time complexity: O(n), single pass over the array.
+  - Space complexity: O(1) extra (generator uses constant extra memory).