Problem description:

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

this is a relatively easier way to understand this question.

  1. create a 2d array, row is total number of transaction, column is total number of days.

  2. the dp formula is
    T[i][j]= max(T[i][j-1], price[j]-price[m]+ T[i-1][m])
    it means, the maximum profit can get on transaction i in day j should be one of the following one.

    1. T[i][j-1], the same as previous day, means make no transaction at all.
    2. price[j]-price[m]+ T[i-1][m], sale the stock bought on day m with the price on day j, and add the previous transaction profit together.
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class Solution {
public:
    //transaction: k, days: n
    //O(k*n^2)
    int maxProfit(vector<int>& prices) {
        if(prices.empty()) return 0;
        int transactions= 2;
        vector<vector<int>> dp(transactions+1, vector<int>(prices.size(), 0));
        for(int r= 1; r< dp.size(); r++){
            for(int c= 1; c< dp[0].size(); c++){
                int maxval= 0;
                for(int m= 0; m< c; m++){
                    maxval= max(maxval, prices[c]-prices[m]+ dp[r-1][m]);
                }
                dp[r][c]= max(dp[r][c-1], maxval);
            }
        }
        return dp[transactions][prices.size()-1];
    }
};

However, the above solution would need many time to calculate. We can use a variable maxdiff to check whether if it’s a good time to sell the stock.

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class Solution {
public:
    //transaction: k, days: n
    //O(k*n)
    int maxProfit(vector<int>& prices) {
        if(prices.empty()) return 0;
        int transactions= 2;
        vector<vector<int>> dp(transactions+1, vector<int>(prices.size(), 0));
        for(int r= 1; r< dp.size(); r++){
            int maxdiff= -prices[0];
            for(int c= 1; c< dp[0].size(); c++){
                dp[r][c]= max(dp[r][c-1], prices[c]+maxdiff);
                maxdiff= max(maxdiff, dp[r-1][c]-prices[c]);
            }
        }
        return dp[transactions][prices.size()-1];
    }
};
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class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        sell1, sell2 = 0, 0
        hold1, hold2 = float("-inf"), float("-inf")
        for i in prices:
            sell2 = max(sell2, hold2 + i) # The maximum if we've just sold 2nd stock so far.
            hold2 = max(hold2, sell1 - i) # The maximum if we've just buy  2nd stock so far.
            sell1 = max(sell1, hold1 + i) # The maximum if we've just sold 1st stock so far.
            hold1 = max(hold1, -i)        # The maximum if we've just buy  1st stock so far.
        return sell2

reference:
https://www.youtube.com/watch?v=oDhu5uGq_ic&feature=youtu.be