123-best-time-to-buy-and-sell-stock-iii

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Description

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

Solution(Python)

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        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)] 
        mp = [[0]*(k+1) for _ in range(n-1)] 

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

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

        return max(mp[-1])