121. Best Time to Buy and Sell Stock

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
             Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

他的上一題是用Greegy策略，只要明天價格比較高就賣出，這題要找最大利潤，先找當前跟歷史比最大售價，再由最大售價找最大價差，有2種解法。

Normal
T: O(n), S: O(1)
class Solution:
    def maxProfit(self, prices: List[int]) -> int:
            largest_profit = 0
            min_purchase = prices[0]
            for i in range(1, len(prices)):
                if prices[i] - min_purchase > largest_profit:
                    largest_profit = prices[i] - min_purchase
                if prices[i] < min_purchase:
                    min_purchase = prices[i]
            return largest_profit


Kadane's algorithm
T: O(n), S: O(1)
class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        result = 0
        sell = 0
        for i in range(len(prices)-1,-1,-1):
            sell = max(sell, prices[i]) # get max sell price
            result = max(result, sell - prices[i]) # get max profit
        return result 

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