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

Author: jeremymanning

problems/3296/gpt5-mini.md

@@ -0,0 +1,55 @@
+# [Problem 3296: Minimum Number of Seconds to Make Mountain Height Zero](https://leetcode.com/problems/minimum-number-of-seconds-to-make-mountain-height-zero/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We need to distribute mountainHeight units of "work" among workers. If worker i does x units, their time is workerTimes[i] * (1 + 2 + ... + x) = workerTimes[i] * x*(x+1)/2. Workers operate simultaneously, so the total time is the maximum time any worker spends; we want to minimize that maximum subject to sum x_i = mountainHeight. This looks like a feasibility check: for a given time T, each worker can do at most k_i units where workerTimes[i] * k_i*(k_i+1)/2 <= T. If sum k_i >= mountainHeight, T is feasible. So binary search on T is natural. Need to compute k_i efficiently (solve quadratic or use integer math). Watch out for large numbers (T up to ~5e15), use integer sqrt (math.isqrt) to avoid floating point error.
+
+## Refining the problem, round 2 thoughts
+- For a candidate T, compute M = 2*T // workerTimes[i], solve k^2 + k <= M => k = floor((-1 + sqrt(1 + 4*M))/2). Using integer math (math.isqrt) is safe and exact.
+- Early exit the capacity sum when it reaches mountainHeight.
+- Binary search bounds: low = 0, high = max(workerTime) * mountainHeight*(mountainHeight+1)//2 (one worker does everything) — fits in 64-bit integer for given constraints.
+- Complexity: O(n * log(maxT)). With n <= 1e4 and log(maxT) ~ ~52, this is fine.
+- Edge cases: single worker, many workers faster than needed (sum capacities exceed height quickly), and very large workerTimes produce small capacities for given T.
+
+## Attempted solution(s)
+```python
+import math
+from typing import List
+
+class Solution:
+    def minimumTime(self, mountainHeight: int, workerTimes: List[int]) -> int:
+        # Helper: given time T, compute total units of height workers can reduce
+        def capacity(T: int) -> int:
+            total = 0
+            for w in workerTimes:
+                if T < w:  # can't even do 1 unit
+                    continue
+                # Solve k*(k+1)/2 * w <= T  => k^2 + k <= 2*T//w (use integer division)
+                M = (2 * T) // w
+                # discriminant for k^2 + k - M <= 0 is 1 + 4*M
+                d = 1 + 4 * M
+                sqrt_d = math.isqrt(d)
+                k = (sqrt_d - 1) // 2
+                total += k
+                if total >= mountainHeight:
+                    return total
+            return total
+
+        # Binary search on time T
+        lo = 0
+        # Upper bound: slowest single worker doing all mountainHeight
+        max_w = max(workerTimes)
+        hi = max_w * mountainHeight * (mountainHeight + 1) // 2
+
+        while lo < hi:
+            mid = (lo + hi) // 2
+            if capacity(mid) >= mountainHeight:
+                hi = mid
+            else:
+                lo = mid + 1
+        return lo
+```
+- Notes about the solution:
+  - Approach: binary search on time T; for each T compute how many units each worker can remove using integer solution of quadratic inequality.
+  - Time complexity: O(n * log U) where U is the search space upper bound (roughly max worker time * H*(H+1)/2). With constraints this is efficient (n ≤ 1e4, log U ≲ 60).
+  - Space complexity: O(1) extra space (besides input).
+  - Implementation detail: use math.isqrt on integer discriminant to avoid floating-point inaccuracies and ensure correctness for large values. Early exit when accumulated capacity reaches mountainHeight to save work.