# 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.

class Solution(object):
    def maxProfit(self, prices):
        maxProfit = 0
        minPrice = float('inf')

        for p in prices:
            minPrice = min(p, minPrice)
            profit = p - minPrice
            if profit > maxProfit:
                maxProfit = profit
        return maxProfit

def main():
    s = Solution()
    print(s.maxProfit([7,1,5,3,6,4]))
    print(s.maxProfit([7,6,4,3,1]))

if __name__ == "__main__":
    main()
# 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.

class Solution(object):
    def maxProfit(self, prices):
        maxProfit = 0
        minPrice = float('inf')

        for p in prices:
            minPrice = min(p, minPrice)
            profit = p - minPrice
            if profit > maxProfit:
                maxProfit = profit
        return maxProfit

def main():
    s = Solution()
    print(s.maxProfit([7,1,5,3,6,4]))
    print(s.maxProfit([7,6,4,3,1]))

if __name__ == "__main__":
    main()
# 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.