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maxProfit.py
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maxProfit.py
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# -*- coding: utf-8 -*-
"""
@author : zenithude
Best Time to Buy and Sell Stock III
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).
Example 1:
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.
Example 2:
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.
Example 3:
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):
"""
Parameters
----------
prices : List[int]
Returns
-------
int
"""
if not prices:
return 0
sell, buyd, n = [0], [0], len(prices)
minprices, maxprices = prices[0], prices[-1]
for i in range(1, n):
minprices = min(minprices, prices[i])
maxprices = max(maxprices, prices[n - i - 1])
sell.append(max(sell[i - 1], prices[i] - minprices))
buyd.append(max(buyd[i - 1], maxprices - prices[n - i - 1]))
return max(sell[i] + buyd[n - i - 1] for i in range(n))
obj = Solution()
print(obj.maxProfit([3, 3, 5, 0, 0, 3, 1, 4]))