Best time to buy and sell stock iii

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        dp = [
            [0] * 5
            for _ in range(len(prices))
        ]
        dp[0][0] = 0           # No ops
        dp[0][1] = -prices[0]  # 1st Have
        dp[0][2] = 0           # 1st Not Have
        dp[0][3] = -prices[0]  # 2nd Have
        dp[0][4] = 0           # 2nd Not Have

        for i in range(1, len(prices)):
            # dp[i][0] = dp[i - 1][0]
            dp[i][1] = max(dp[i - 1][1],  - prices[i])
            dp[i][2] = max(dp[i - 1][2], dp[i - 1][1] + prices[i])
            dp[i][3] = max(dp[i - 1][3], dp[i - 1][2] - prices[i])
            dp[i][4] = max(dp[i - 1][4], dp[i - 1][3] + prices[i])

        return dp[len(prices) - 1][4]

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 i^th 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