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

Author: jeremymanning

problems/1611/gpt5-mini.md

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+# [Problem 1611: Minimum One Bit Operations to Make Integers Zero](https://leetcode.com/problems/minimum-one-bit-operations-to-make-integers-zero/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+This problem describes constrained bit flips: you can flip bit 0 anytime; you can flip bit i (>0) only when bit (i-1) == 1 and bits (i-2)..0 == 0. It smells like a structured traversal of the space of k-bit states (like a Gray-code / reflected sequence or a Tower-of-Hanoi style sequence). The question asks for minimum number of operations to go from n to 0, so we want the shortest path length in that graph.
+
+A direct BFS is possible for small bit-length, but n can be up to 1e9 (≈30 bits) and BFS over 2^30 states is impractical. The structure suggests a recursive/reflective property: top-half states (those with highest bit set) often have distances related to distances of their low-half counterpart in a mirrored way.
+
+I suspect there's a simple recurrence based on the highest set bit: split n into high bit 2^k and remainder r = n - 2^k, and express result for n in terms of result for r (or of a reflected index). Try to find that recursion.
+
+## Refining the problem, round 2 thoughts
+Work out small cases by hand to observe pattern:
+- g(0)=0
+- g(1)=1
+- g(2)=3
+- g(3)=2
+- g(4)=7
+- g(5)=6
+- g(6)=4
+- g(7)=5
+
+From these values, the top-half (numbers with highest bit 1) appear to be the mirror of the bottom-half under (2^{k+1}-1) complement. For numbers in range [2^k, 2^{k+1}-1], the formula seems to be:
+g(n) = (2^{k+1}-1) - g(n - 2^k)
+
+This recurrence fits all small examples and reduces the problem size (n - 2^k has smaller bit-length). Depth is at most number of bits (~30), so recursion is safe. We can memoize or rely on the small recursion depth. Time complexity O(bits) per call with memoization giving O(bits) overall, space O(bits) for recursion/memo.
+
+I'll implement a concise recursive solution with caching.
+
+## Attempted solution(s)
+```python
+from functools import lru_cache
+
+class Solution:
+    def minimumOneBitOperations(self, n: int) -> int:
+        @lru_cache(None)
+        def g(x: int) -> int:
+            if x == 0:
+                return 0
+            k = x.bit_length() - 1  # highest set bit position
+            high = 1 << k
+            # reflect relative to full mask of current bit-length
+            mask = (1 << (k + 1)) - 1
+            return mask - g(x - high)
+        return g(n)
+```
+- Notes about the solution approach:
+  - We use the recurrence g(0) = 0. For n > 0, let k be the highest bit index and high = 2^k. Then:
+    g(n) = (2^{k+1} - 1) - g(n - 2^k).
+    Intuition: states in the top half (with highest bit set) are a mirrored/reflected sequence of the bottom half; the minimal steps mirror accordingly.
+  - Complexity: Each recursive step reduces the bit-length, so there are at most O(B) recursive calls where B = number of bits in n (≤ 30). With memoization the cost is O(B) time and O(B) space.
+  - Implementation details: We use Python's bit_length to find k and a small LRU cache to avoid recomputation.