0123 Best Time to Buy and Sell Stock III

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).

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.

Constraints:

1 <= prices.length <= 105
0 <= prices[i] <= 105

def maxProfit(prices):

    # 1. Define the state
    # dp[i][j][k] represents the max profit at day i, with j transactions, and k stocks in hand
    # j = 0, 1, 2
    # k = 0, 1
    # 0 <= i < len(prices)
    dp = [[[0 for _ in range(2)] for _ in range(3)] for _ in range(len(prices))]

    # 2. Initialize the state
    # dp[0][0][0] = 0
    # dp[0][0][1] = -prices[0]
    # dp[0][1][0] = 0
    # dp[0][1][1] = -prices[0]
    # dp[0][2][0] = 0
    # dp[0][2][1] = -prices[0]
    for j in range(3):
        # profit at day 0 with j transactions and 0 stock in hand is 0
        dp[0][j][0] = 0
        # profit at day 0 with j transactions and 1 stock in hand is -prices[0]
        dp[0][j][1] = -prices[0]

    # 3. Define the loop
    for i in range(1, len(prices)):
        for j in range(1, 3):
            # no stock in hand at day i with j transactions: sell stock at day i-1 or do nothing
            dp[i][j][0] = max(dp[i-1][j][0], dp[i-1][j][1] + prices[i])
            # stock in hand at day i with j transactions: buy stock at day i-1 or do nothing
            dp[i][j][1] = max(dp[i-1][j][1], dp[i-1][j-1][0] - prices[i])
    
    # 4. Return the final state
    return dp[len(prices)-1][2][0]

# test  
prices = [3,3,5,0,0,3,1,4]
maxProfit(prices)

6