Best time to buy and sell stock with cooldown

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        n = len(prices)
        if n < 2:
            return 0

        # 定义三种状态的动态规划数组
        dp = [[0] * 3 for _ in range(n)]
        dp[0][0] = -prices[0]  # 持有股票的最大利润
        dp[0][1] = 0           # 不持有股票，且处于冷冻期的最大利润
        dp[0][2] = 0           # 不持有股票，不处于冷冻期的最大利润

        for i in range(1, n):
            # 当前持有股票的最大利润等于前一天持有股票的最大利润或者前一天不持有股票且不处于冷冻期的最大利润减去当前股票的价格
            dp[i][0] = max(dp[i-1][0], dp[i-1][2] - prices[i])
            # 当前不持有股票且处于冷冻期的最大利润等于前一天持有股票的最大利润加上当前股票的价格
            dp[i][1] = dp[i-1][0] + prices[i]
            # 当前不持有股票且不处于冷冻期的最大利润等于前一天不持有股票的最大利润或者前一天处于冷冻期的最大利润
            dp[i][2] = max(dp[i-1][2], dp[i-1][1])

        # 返回最后一天不持有股票的最大利润
        return max(dp[-1][1], dp[-1][2])

Best Time to Buy and Sell Stock with Cooldown

Difficulty: Medium

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 as many transactions as you like (i.e., buy one and sell one share of the stock multiple times) with the following restrictions:

  • After you sell your stock, you cannot buy stock on the next day (i.e., cooldown one day).

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

 

Example 1:

Input: prices = [1,2,3,0,2]
Output: 3
Explanation: transactions = [buy, sell, cooldown, buy, sell]

Example 2:

Input: prices = [1]
Output: 0

 

Constraints:

  • 1 <= prices.length <= 5000
  • 0 <= prices[i] <= 1000