123. Best Time to Buy and Sell Stock III 188. Best Time to Buy and Sell Stock IV

123. 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).

 

DP best solution:

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        n = len(prices)
        # buy1 表示买入一次
        # buy2 表示买入一次，卖出一次，然后再买入一次
        buy1 = buy2 = -prices[0]
        # sell1 表示进行过一次交易
        # sell2 表示完成了两次交易
        sell1 = sell2 = 0
        for i in range(1, n):
            buy1 = max(buy1, -prices[i])
            sell1 = max(sell1, buy1 + prices[i])
            buy2 = max(buy2, sell1 - prices[i])
            sell2 = max(sell2, buy2 + prices[i])
        return sell2

1-dimensional DP:

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if len(prices) == 0:
            return 0
        dp = [0] * 5 
        dp[1] = -prices[0]
        dp[3] = -prices[0]
        for i in range(1, len(prices)):
            dp[1] = max(dp[1], dp[0] - prices[i])
            dp[2] = max(dp[2], dp[1] + prices[i])
            dp[3] = max(dp[3], dp[2] - prices[i])
            dp[4] = max(dp[4], dp[3] + prices[i])
        return dp[4]

2-dimensional DP:

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if len(prices) == 0:
            return 0
        dp = [[0] * 5 for _ in range(len(prices))]
        dp[0][1] = -prices[0]
        dp[0][3] = -prices[0]
        for i in range(1, len(prices)):
            dp[i][0] = dp[i-1][0]
            dp[i][1] = max(dp[i-1][1], dp[i-1][0] - 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[-1][4]

188. Best Time to Buy and Sell Stock IV

You are given an integer array prices where prices[i] is the price of a given stock on the ith day, and an integer k.

Find the maximum profit you can achieve. You may complete at most k transactions: i.e. you may buy at most k times and sell at most k times.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

 attention:

1. dp = [0] * (2 * k + 1)   Buy and sell times plus one

2. dp[1,3,5,7....] = -prices[0]

1-dimensional DP:

Time complexity: O(n * k),     n is the length of prices
Space complexity: O(k)

class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        if len(prices) == 0: return 0
        dp = [0] * (2*k + 1)
        for i in range(1,2*k,2):
            dp[i] = -prices[0]
        for i in range(1,len(prices)):
            for j in range(1,2*k + 1):
                if j % 2:
                    dp[j] = max(dp[j],dp[j-1]-prices[i])
                else:
                    dp[j] = max(dp[j],dp[j-1]+prices[i])
        return dp[2*k]

 2-dimensional DP:

Time complexity: O(n * k),     n is the length of prices
Space complexity: O(n * k),

class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        if len(prices) == 0:
            return 0
        dp = [[0] * (2*k+1) for _ in range(len(prices))]
        for j in range(1, 2*k, 2):
            dp[0][j] = -prices[0]
        for i in range(1, len(prices)):
            for j in range(0, 2*k-1, 2):
                dp[i][j+1] = max(dp[i-1][j+1], dp[i-1][j] - prices[i])
                dp[i][j+2] = max(dp[i-1][j+2], dp[i-1][j+1] + prices[i])
        return dp[-1][2*k]