309. Best Time to Buy and Sell Stock with Cooldown 714. Best Time to Buy and Sell Stock with Transa
309. Best Time to Buy and Sell Stock with Cooldown
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).
The following four states can be distinguished:
State 1: Holding the stock state (buy the stock today, or buy the stock before and then did not operate, has been holding)
Not holding the stock status, here there are two kinds of sell the stock status
State 2: Hold Sell Stock status (sold the stock two days ago to get through a one-day freeze. Or the day before is the sell stock status, has not been operated)
State 3: Sell the stock today
Status 4: Today is a frozen status, but the frozen status is not sustainable, only one day!
2-dimensional DP:
from typing import List
class Solution:
def maxProfit(self, prices: List[int]) -> int:
n = len(prices)
if n == 0:
return 0
dp = [[0] * 4 for _ in range(n)] # 创建动态规划数组，4个状态分别表示持有股票、不持有股票且处于冷冻期、不持有股票且不处于冷冻期、不持有股票且当天卖出后处于冷冻期
dp[0][0] = -prices[0] # 初始状态：第一天持有股票的最大利润为买入股票的价格
for i in range(1, n):
dp[i][0] = max(dp[i-1][0], max(dp[i-1][3], dp[i-1][1]) - prices[i]) # 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][1] = max(dp[i-1][1], dp[i-1][3]) # 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][2] = 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][3] = dp[i-1][2] # 当前不持有股票且当天卖出后处于冷冻期的最大利润等于前一天不持有股票且不处于冷冻期的最大利润
return max(dp[n-1][3], dp[n-1][1], dp[n-1][2]) # 返回最后一天不持有股票的最大利润optimal:
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])714. Best Time to Buy and Sell Stock with Transaction Fee
You are given an array prices where prices[i] is the price of a given stock on the ith day, and an integer fee representing a transaction fee.
Find the maximum profit you can achieve. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction.
Note:
- You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
- The transaction fee is only charged once for each stock purchase and sale.
AC ez:
Time complexity: O(n)
Space complexity: O(m)
class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
dp = [[0]*2 for _ in range(len(prices))]
dp[0][0] = -prices[0]
#dp[0][1] = 0 不是-free
for i in range(1, len(prices)):
dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] - prices[i])
dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i] - fee)
return dp[-1][1]