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

/*
 * @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;
  }
}