[LeetCode Sync] Runtime - 266 ms (56.69%), Memory - 28.7 MB (84.96%)

Author: NinePiece2

Solutions/0123-best-time-to-buy-and-sell-stock-iii/README.md

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+<p>You are given an array <code>prices</code> where <code>prices[i]</code> is the price of a given stock on the <code>i<sup>th</sup></code> day.</p>
+
+<p>Find the maximum profit you can achieve. You may complete <strong>at most two transactions</strong>.</p>
+
+<p><strong>Note:</strong> You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [3,3,5,0,0,3,1,4]
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
+Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [1,2,3,4,5]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
+Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [7,6,4,3,1]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> In this case, no transaction is done, i.e. max profit = 0.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= prices.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= prices[i] &lt;= 10<sup>5</sup></code></li>
+</ul>

Solutions/0123-best-time-to-buy-and-sell-stock-iii/solution.py

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+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        b1, s1, b2, s2 = -prices[0], 0, -prices[0], 0
+        for price in prices[1:]:
+            b1 = max(b1, -price)
+            s1 = max(s1, b1 + price)
+            b2 = max(b2, s1 - price)
+            s2 = max(s2, b2 + price)
+        
+        return s2