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

Author: jeremymanning

problems/3573/gpt5-mini.md

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+# [Problem 3573: Best Time to Buy and Sell Stock V](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-v/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We can do at most k transactions. A transaction can be a normal buy-then-sell or a short-sell (sell then buy back). For any pair of days i < j the profit from a transaction between i and j is |prices[j] - prices[i]| (we would only take the positive direction). Transactions must be non-overlapping and cannot start on the same day the previous transaction ended (so if previous ended on day j, next must start at day > j).
+
+This looks like a dynamic programming problem similar to the classic "at most k transactions" stock problems (LeetCode 188). The naive DP dp[t][i] = max(dp[t][i-1], max_{s < i} dp[t-1][s] + |p[i]-p[s]|) is O(k * n^2) which is too slow for n up to 1000 and k up to ~500. For the standard problem without short-selling we optimize by tracking a running "best" value of dp[t-1][s-1] - price[s]. Here, the absolute value splits into two linear terms depending on sign: p[i]-p[s] or p[s]-p[i], so we can similarly maintain two running maxima to get O(k*n).
+
+Need to be careful about the "cannot start on same day previous ended" constraint: when adding a transaction that starts at s and ends at i, we must use dp[t-1][s-1] (profit up to day s-1) — not dp[t-1][s] — to avoid allowing immediate reuse of day s.
+
+So split the absolute into two cases and maintain:
+- best1 = max over s in [0..i-1] of dp_prev[s-1] - p[s]  (handles p[i] + best1)
+- best2 = max over s in [0..i-1] of dp_prev[s-1] + p[s]  (handles -p[i] + best2)
+
+Then dp_cur[i] = max(dp_cur[i-1], p[i] + best1, -p[i] + best2). Update best1/best2 after computing dp_cur[i] using dp_prev[i-1] (dp_prev[-1] = 0 for s = 0).
+
+## Refining the problem, round 2 thoughts
+- We must ensure indices are handled so that transactions don't start on the same day a previous transaction ended: use dp_prev[s-1] in the expressions (with dp_prev[-1] = 0).
+- Base cases: dp[0][*] = 0; dp[*][0] = 0 (no profit with zero days or zero transactions).
+- Time complexity O(k * n) and space O(n) if we keep only previous and current dp rows.
+- Edge cases: small n (2), k up to n//2 per constraints; prices large but only used in additions/subtractions so Python int fine.
+
+This yields a correct and efficient algorithm.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def maxProfit(self, k: int, prices: List[int]) -> int:
+        n = len(prices)
+        if n < 2 or k == 0:
+            return 0
+
+        # If k is large enough we could capture every adjacent absolute diff by using
+        # transactions on every adjacent pair, but constraints guarantee k <= n//2.
+        # We'll use O(k * n) DP.
+
+        # dp_prev[i] = max profit using at most (t-1) transactions up to and including day i
+        dp_prev = [0] * n
+
+        for _ in range(1, k + 1):
+            dp_cur = [0] * n
+            # best1 = max(dp_prev[s-1] - prices[s]) for s in [0..i-1]
+            # best2 = max(dp_prev[s-1] + prices[s]) for s in [0..i-1]
+            # initialize as -inf so that if no s exists (i=0) we don't use them
+            best1 = float("-inf")
+            best2 = float("-inf")
+
+            for i in range(n):
+                # carry forward the best without using a transaction ending at i
+                if i > 0:
+                    dp_cur[i] = dp_cur[i - 1]
+                else:
+                    dp_cur[i] = 0
+
+                # use a transaction that ends at i and starts at some s in [0..i-1]
+                if best1 != float("-inf"):
+                    dp_cur[i] = max(dp_cur[i], prices[i] + best1)
+                if best2 != float("-inf"):
+                    dp_cur[i] = max(dp_cur[i], -prices[i] + best2)
+
+                # now allow future i' > i to consider starting at s = i:
+                prev_val = dp_prev[i - 1] if i - 1 >= 0 else 0
+                best1 = max(best1, prev_val - prices[i])
+                best2 = max(best2, prev_val + prices[i])
+
+            dp_prev = dp_cur
+
+        return dp_prev[-1]
+```
+- Approach notes:
+  - We use dp over number of transactions. For iteration t we build dp_cur from dp_prev = dp for t-1 transactions.
+  - For a transaction that starts at s and ends at i (s < i), the gain is |prices[i] - prices[s]| plus dp_prev[s-1] (profit before s). Splitting abs into two linear forms yields two running maxima that let us compute the inner max in O(1) per i.
+- Time complexity: O(k * n) where n = len(prices).
+- Space complexity: O(n) (we keep two arrays of length n, dp_prev and dp_cur).