Best Time to Buy and Sell Stock III
Created: November 29, 2019 by [lek-tin]
Last updated: November 29, 2019
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).
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.
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.
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: List[int]) -> int: if not prices: return 0 n = len(prices) dp = [ * n for _ in range(3)] for i in range(1, 3): # money spent at first day prev = dp[i - 1] - prices for j in range(1, n): # money spent if sell stock today deal = prev + prices[j] # compare money spent if don't sell stock today with sell stock today dp[i][j] = max(dp[i][j - 1], deal) # compare i - 1 deals during j days, and don't buy stock today # with i - 1 deals during j days, and buy stock today prev = max(prev, dp[i - 1][j] - prices[j]) return dp[-1][-1]