123. Best Time to Buy and Sell Stock III

Problem:

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.

Solution:

Multiple transactions may not be engaged in at the same time. That means if we view the days that involed in the same transaction as a group, there won’t be any intersection. We may complete at most two transactions, so divide the days into two groups, [0...k] and [k...n-1]. Notice k exists in both groups because technically we can sell out then immediately buy in at the same day.

Define p1(i) to be the max profit of day [0...i]. This is just like the problem of 121. Best Time to Buy and Sell Stock.

p1(0) = 0
p1(i) = max( p1(i-1), prices[i] - min(prices[0], ..., prices[i-1]) ), 0 < i <= n-1

Define p2(i) to be the max profit of day [i...n-1]. This is the mirror of p1.

p2(n-1) = 0
p2(i) = max( p2(i+1), max(prices[i], ..., prices[n-1]) - prices[i] ), n-1 > i >= 0

Define f(k) to be p1(k) + p2(k). We need to get max( f(0), ..., f(n-1) ).

/**
 * @param {number[]} prices
 * @return {number}
 */
let maxProfit = function (prices) {
  const len = prices.length;
  if (len <= 1) {
    return 0;
  }

  const dp = [0];

  let min = prices[0];
  for (let i = 1; i < len; i++) {
    dp[i] = Math.max(dp[i - 1], prices[i] - min);
    min = Math.min(prices[i], min);
  }

  let p2 = 0;
  let max = prices[len - 1];
  for (let i = len - 2; i >= 0; i--) {
    max = Math.max(prices[i], max);
    p2 = Math.max(p2, max - prices[i]);
    dp[i] += p2;
  }

  return Math.max(...dp);
};

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