https://leetcode.com/problems/best-time-to-buy-and-sell-stock/#/solutions

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 (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 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

In this case, no transaction is done, i.e. max profit = 0.

Sol :

Keep track of two variables. The current max profit and the current min price.

Since we are going to solve this problem with O(n) time complexity, we can only iterate one pointer. max profit = max price - min price without considering that the min price should be ahead of the max price. Thus, in this case, max price should be the pointer that iterate through the list, and min price and max profit will be recorded and compared as the price pointer scans. 

max profit = each price - min price ! 

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        max_profit = 0
        # cur_profit = 0
        min_price = float(‘inf‘)
        
        for price in prices:
            
            min_price = min(min_price, price)
            max_profit = max(max_profit, price - min_price)
            
        return max_profit

121. Best Time to Buy and Sell Stock