solve 123.best-time-to-buy-and-sell-stock-iii

Author: andavid

vscode/123.best-time-to-buy-and-sell-stock-iii.java

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+/*
+ * @lc app=leetcode id=123 lang=java
+ *
+ * [123] Best Time to Buy and Sell Stock III
+ *
+ * https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/
+ *
+ * algorithms
+ * Hard (32.90%)
+ * Total Accepted:    138.3K
+ * Total Submissions: 420.4K
+ * Testcase Example:  '[3,3,5,0,0,3,1,4]'
+ *
+ * Say you have an array for which the ith element is the price of a given
+ * stock on day i.
+ *
+ * Design an algorithm to find the maximum profit. You may complete at most two
+ * transactions.
+ *
+ * Note: You may not engage in multiple transactions at the same time (i.e.,
+ * you must sell the stock before you buy again).
+ *
+ * Example 1:
+ *
+ *
+ * Input: [3,3,5,0,0,3,1,4]
+ * Output: 6
+ * Explanation: 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.
+ *
+ * Example 2:
+ *
+ *
+ * Input: [1,2,3,4,5]
+ * Output: 4
+ * Explanation: 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.
+ *
+ *
+ * Example 3:
+ *
+ *
+ * Input: [7,6,4,3,1]
+ * Output: 0
+ * Explanation: In this case, no transaction is done, i.e. max profit = 0.
+ *
+ */
+class Solution {
+  public int maxProfit(int[] prices) {
+    int buy1 = Integer.MIN_VALUE; // just buy 1nd stock
+    int buy2 = Integer.MIN_VALUE; // just buy 2nd stock
+    int sell1 = 0; // just have sold 1nd stock
+    int sell2 = 0; // just have sold 2nd stock
+
+    for (int price : prices) {
+      sell2 = Math.max(sell2, buy2 + price);
+      buy2 = Math.max(buy2, sell1 - price);
+      sell1 = Math.max(sell1, buy1 + price);
+      buy1 = Math.max(buy1, -price);
+    }
+
+    return sell2;
+  }
+}
+