<filename>leetcode/easy/Arrays and Strings/BuySellStock.py
# 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:
    def maxProfit(self, prices):
        idxOfLowest = 0
        maxDiff = 0

        for i in range(1, len(prices)):
            if prices[i] < prices[idxOfLowest]:
                idxOfLowest = i
            else:
                maxDiff = max(maxDiff, prices[i] - prices[idxOfLowest])

        return maxDiff

<filename>leetcode/easy/Arrays and Strings/BuySellStock.py
# 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:
    def maxProfit(self, prices):
        idxOfLowest = 0
        maxDiff = 0

        for i in range(1, len(prices)):
            if prices[i] < prices[idxOfLowest]:
                idxOfLowest = i
            else:
                maxDiff = max(maxDiff, prices[i] - prices[idxOfLowest])

        return maxDiff

# 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.