123. Best Time to Buy and Sell Stock III

1. Description

You are given an array prices where prices[i] is the price of a given stock on the ith day.
Find the maximum profit you can achieve. You may complete at most two transactions.
Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

2. Example

Example 1

Input: prices = [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: prices = [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: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

3. Constraints

• 1 <= prices.length <= $10^5$
• 0 <= prices[i] <= $10^5$

4. Solutions

Dynamic Programming

n = prices.size()
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    int maxProfit(const vector<int> &prices) {
        const int n = prices.size();
        array<array<int, 4>, 2> profits = {};
        profits[0][0] = -prices[0];
        profits[0][2] = numeric_limits<int>::min();
        // 0: holding first stock
        // 1: after selling first stock
        // 2: holding second stock
        // 3: after selling second stock

        for (int i = 1; i < n; ++i) {
            profits[1][0] = max(profits[0][0], -prices[i]);
            profits[1][1] = max(profits[0][1], profits[0][0] + prices[i]);
            profits[1][2] = max(profits[0][2], profits[0][1] - prices[i]);
            profits[1][3] = max(profits[0][3], profits[0][2] + prices[i]);

            swap(profits[0], profits[1]);
        }

        return max(profits[0][1], profits[0][3]);
    }
};