123. Best Time to Buy and Sell Stock III

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.

// Constant space DP
int maxProfit(vector<int>& prices) { // time: O(n); space: O(1)
    int sell1 = 0, sell2 = 0, buy1 = INT_MIN, buy2 = INT_MIN;
    for (int price : prices) {
        buy1 = max(buy1, -price);
        sell1 = max(sell1, price + buy1);
        buy2 = max(buy2, sell1 - price);
        sell2 = max(sell2, price + buy2);
    }
    return sell2;
}

原本是需要2維的dp array去紀錄buy[i][k]和sell[i][k]，i代表日期，k代表交易次數。 buy[i][k] = max(buy[i - 1][k], sell[i - 1][k - 1] - prices[i]) sell[i][k] = max(sell[i - 1][k], buy[i - 1][k] + prices[i]) 但可以簡化只用buy[k]和sell[k]兩個一維arrays去紀錄DP過程的結果。 buy[k] = max(buy[k], sell[k - 1] - price) sell[k] = max(sell[k], buy[k] + price) 先loop交易日期，內層loop交易次數，要注意的是交易次數需要從大的值到小的值，因為他們需要用到前一天同樣k次的交易次數結果。注意到的是sell[k]會用到前一天的buy[k]所以在程式碼裡必須先計算sell[k]才能計算更新當天的buy[k]。

// Constant space DP
int maxProfit(vector<int>& prices) { // time: O(n); space: O(1)
    vector<int> buy(3, INT_MIN), sell(3, 0);
    for (int i = 0; i < prices.size(); ++i) {
        for (int j = 2; j >= 1; --j) {
            sell[j] = max(sell[j], buy[j] + prices[i]);
            buy[j] = max(buy[j], sell[j - 1] - prices[i]);
        }
    }
    return sell[2];
}

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