# 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

prices = [7, 6, 4, 3, 1]


class Solution:
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        # initialize min price as maximum integer value
        minPrice = 2 ** 32
        # initialize max profit
        maxProfit = 0
        # the list is ordered by date so we must always check values to the right of the current price.
        # we are trying to find the difference between current min and values to the right of it
        for i in range(len(prices)):
            # if the current price is lower than the absolute min price, update the abs min price
            if prices[i] < minPrice:
                minPrice = prices[i]
            # otherwise check to see if the profit is greater than maxprofit
            elif prices[i] - minPrice > maxProfit:
                maxProfit = prices[i] - minPrice
        return maxProfit
    #  O(n) solution since we only interate through the list 1 time.


soln = Solution()
print(soln.maxProfit(prices))
