"""
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.
"""


class Solution:
	def maxProfit(self, prices):
		if len(prices) <= 1: return 0
		n, k = len(prices), 2

		B = [prices[i + 1] - prices[i] for i in range(len(prices) - 1)]
		if k > len(prices) // 2: return sum(x for x in B if x > 0)

		dp = [[0] * (k + 1) for _ in range(n - 1)]
		max_profit = [[0] * (k + 1) for _ in range(n - 1)]

		dp[0][1], max_profit[0][1] = B[0], B[0]

		for i in range(1, n - 1):
			for j in range(1, k + 1):
				dp[i][j] = max(max_profit[i - 1][j - 1], dp[i - 1][j]) + B[i]
				max_profit[i][j] = max(dp[i][j], max_profit[i - 1][j])

		return max(max_profit[-1])


if __name__ == '__main__':
	S = Solution()

	print('6 -> ', S.maxProfit([3, 3, 5, 0, 0, 3, 1, 4]))
	print('4 -> ', S.maxProfit([1, 2, 3, 4, 5]))
	print('0 -> ', S.maxProfit([7, 6, 4, 3, 1]))
