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

Author: jeremymanning

problems/2528/gpt5-mini.md

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+# [Problem 2528: Maximize the Minimum Powered City](https://leetcode.com/problems/maximize-the-minimum-powered-city/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We need to maximize the minimum "power" across all cities after adding up to k stations. Each station contributes to a contiguous window of length 2*r+1 around its position. The problem smells like a binary-search-on-answer: for a candidate minimum power X, check if we can, with ≤ k added stations, make every city's power at least X.
+
+To check feasibility for X we need to know each city's current power from existing stations (a sliding window sum). Then we must decide where to place additional stations greedily so each city meets X. A common greedy: sweep left-to-right, keep track of extra contribution from stations we already placed (use a difference array to expire contributions), and when the current city is under X, place enough stations at the rightmost position that still covers this city (i + r, capped by n-1). Placing them as far right as possible maximizes their future effect. Update the diff/ending index so recorded extra contributions expire appropriately.
+
+So plan:
+- Compute base power for each city via prefix sums (sliding window).
+- Binary search answer in [0, sum(stations) + k].
+- For each mid, simulate greedy check with diff array in O(n).
+
+Careful with indices for "expiry" of added station contributions: if we place at pos = min(n-1, i + r), a station there covers cities from pos - r to pos + r. When pos = i + r (the common case), the coverage on the sweep starting at i is from i to i + 2r, so expiry index is i + 2r + 1. Cap by n.
+
+## Refining the problem, round 2 thoughts
+Edge cases:
+- r = 0: each station affects only its own city; check reduces to raising individual values.
+- Very large k (up to 1e9) — must use 64-bit arithmetic for counts.
+- n up to 1e5, station values up to 1e5 => sum(stations) up to 1e10; sums and high bounds must be stored in Python ints (fine).
+- Binary search upper bound: sum(stations) + k is safe (no city can get more than all stations).
+- Time: O(n log(sumS + k)) which is fine (~1e5 * ~40).
+
+Implementation details:
+- Compute base array using prefix sums: base[i] = prefix[min(n-1, i+r)] - prefix[max(0, i-r)-1]
+- In feasibility check:
+  - Use diff array of length n (all zeros).
+  - running_added = 0
+  - For each i: running_added += diff[i]; curr = base[i] + running_added
+  - If curr < mid: need = mid - curr; if need > remaining_k -> fail; else remaining_k -= need; running_added += need; end = min(n-1, i + 2*r); if end + 1 < n: diff[end+1] -= need
+- Return True if loop completes.
+
+Complexities:
+- Precompute base: O(n)
+- Each check: O(n)
+- Binary search iterations: O(log(sumS + k)) up to ~40
+- Total: O(n log(sumS + k)) time, O(n) extra space.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def maxPower(self, stations: List[int], r: int, k: int) -> int:
+        n = len(stations)
+        # prefix sums
+        prefix = [0] * n
+        prefix[0] = stations[0]
+        for i in range(1, n):
+            prefix[i] = prefix[i-1] + stations[i]
+        # compute base power for each city: sum of stations in [i-r, i+r]
+        base = [0] * n
+        for i in range(n):
+            left = max(0, i - r)
+            right = min(n - 1, i + r)
+            base[i] = prefix[right] - (prefix[left - 1] if left > 0 else 0)
+
+        total_existing = prefix[-1]
+        lo, hi = 0, total_existing + k  # inclusive upper bound
+
+        def can_achieve(target: int) -> bool:
+            # difference array to manage added station contributions
+            diff = [0] * (n + 1)  # one extra for easier expiry handling
+            running_added = 0
+            remaining = k
+            for i in range(n):
+                running_added += diff[i]
+                curr = base[i] + running_added
+                if curr < target:
+                    need = target - curr
+                    if need > remaining:
+                        return False
+                    remaining -= need
+                    running_added += need
+                    # place these 'need' stations at position pos = min(n-1, i + r)
+                    # their coverage ends at end = min(n-1, pos + r)
+                    # For pos = min(n-1, i + r), this end is min(n-1, i + 2*r)
+                    end = min(n - 1, i + 2 * r)
+                    # subtract from diff at end+1 to expire their effect
+                    if end + 1 < len(diff):
+                        diff[end + 1] -= need
+                # continue
+            return True
+
+        ans = 0
+        while lo <= hi:
+            mid = (lo + hi) // 2
+            if can_achieve(mid):
+                ans = mid
+                lo = mid + 1
+            else:
+                hi = mid - 1
+        return ans
+```
+- Notes:
+  - We compute base city powers using prefix sums in O(n).
+  - The binary search tries candidate minimum powers; each feasibility check simulates greedy placements using a difference array to amortize added contributions. Each check is O(n).
+  - Time complexity: O(n log(sum(stations) + k)). Space complexity: O(n) for base and diff arrays.
+  - The greedy choice (placing needed stations at the rightmost position that still covers the current city) maximizes future benefit and is correct when sweeping left-to-right.