human
Solve in Python:
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.
assistant
#Here is Python program to find the maximum profit.

class Solution:
     def maxProfit(self, prices):
         """
         :type prices: List[int]
         :rtype: int
         """
         if not prices:
             return 0
         
         n = len(prices)
         
         sell1 = 0
         sell2 = 0
         hold1 = -prices[0]
         hold2 = -prices[0]
         
         for i in range(1, n):
             sell2 = max(prices[i] + hold2, sell2)
             hold2 = max(sell1 - prices[i], hold2)
             sell1 = max(prices[i] + hold1, sell1)
             hold1 = max(-prices[i], hold1)
             
             
             
             #sell1[i] = max(sell1[i-1], prices[i] + hold1[i-1])
             #hold1[i] = max(hold1[i-1],  -prices[i])
             #sell2[i] = max(prices[i] + hold2[i-1], sell2[i-1] )
             #hold2[i] = max(sell1[i-1] - prices[i], hold2[i-1])
             
         return sell2