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

Author: jeremymanning

problems/3314/gpt5-mini.md

@@ -0,0 +1,48 @@
+# [Problem 3314: Construct the Minimum Bitwise Array I](https://leetcode.com/problems/construct-the-minimum-bitwise-array-i/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We need, for each given prime nums[i], to find the smallest nonnegative integer a such that a OR (a+1) == nums[i], otherwise output -1. A few quick observations:
+- nums[i] are small (<= 1000) and there are at most 100 numbers, so a direct search per number is feasible.
+- The operation a | (a+1) depends only on a's low bits (how many trailing 1s it has), but deriving a closed-form for every possible output value is a bit fiddly. Given the constraints, brute force search for the minimal a is simple and efficient.
+- Note: nums[i] = 2 (the only even prime) is impossible because for any a, a|(a+1) is >= 3? (in practice we will just test and get no hits -> -1).
+
+## Refining the problem, round 2 thoughts
+Refine to a straightforward algorithm:
+- For each target p in nums, iterate a from 0 up to p (or p+1) and test a | (a+1) == p.
+- Once we find the first such a, that's the minimal by construction; otherwise assign -1.
+Edge cases:
+- p = 2 should return -1 (no a satisfies condition).
+- p up to 1000 means checking up to ~1001 candidates per p is cheap.
+Time complexity: O(n * M) where M = max(nums) <= 1000, n <= 100 → at most ~100k iterations of a tiny constant-time bit op. Space: O(n) for result.
+
+This direct search is simple, robust, and efficient enough for constraints.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def minimumBitwise(self, nums: List[int]) -> List[int]:
+        res = []
+        for p in nums:
+            found = -1
+            # search minimal a such that a | (a+1) == p
+            # a cannot be much bigger than p, so iterate 0..p (inclusive)
+            for a in range(0, p + 1):
+                if (a | (a + 1)) == p:
+                    found = a
+                    break
+            res.append(found)
+        return res
+
+# Example usage:
+# s = Solution()
+# print(s.minimumBitwise([2,3,5,7]))  # [-1,1,4,3]
+# print(s.minimumBitwise([11,13,31])) # [9,12,15]
+```
+
+- Notes about the approach:
+  - We simply brute-force each possible a from 0 to p and check the bitwise condition; the first match is the minimal a.
+  - Time complexity: O(n * M) where n = len(nums) and M = max(nums) (M <= 1000). This is well within limits.
+  - Space complexity: O(n) for the output list.
+  - Implementation detail: iterating up to p is safe because a | (a+1) grows with a and will not produce values much larger than a+1; checking up to p is enough to find a solution if one exists for target p.