121. Best Time to Buy and Sell Stock

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

Note :

 

1 Initialize the max profit with 0, because according to the describtion, no actions will be taken when the profit is minus. However, in other problems, if the profit can go below 0, the initiated value of max profit should be prices[1] - prices[0].

 

2 Initialize the  min price with  float('inf').    

 

float('inf') means a infinite big number.  Any other number will be smaller. Thus, after the min comparison, min_price will get a reasonable value. 

 

Can't initialize min_price with price[0], for it will trigger runtime error. 

 

Runtime Error Message:Line 9: IndexError: list index out of range

Last executed input:[]

 

 

Guess that's because I do not check edge cases? 

 

3 It' s okay to switch the order of min_price and max_profit.

 

4 One content of max_profit is the max_profit itself.....Why bother introduce a cur_profit variable?  Don't code based on memory. 

 

 

Sol 1' :

 

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

 

It's pretty much the same solution. However, if we initialize the min_price with the first element in price list. Check if it is an empty list !