68 lines
1.9 KiB
Markdown
68 lines
1.9 KiB
Markdown
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---
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layout: page
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title: Buy and Sell Stock 2
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last_solved: 2020-06-02
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category: array
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leetcode_url: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii
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status: Solved
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---
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Problem
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{% highlight text %}
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Say you have an array prices for which the ith element is the price of a given stock on day i.
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Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).
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Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
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Example 1:
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Input: [7,1,5,3,6,4]
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Output: 7
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Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
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Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
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Example 2:
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Input: [1,2,3,4,5]
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Output: 4
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Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
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Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
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engaging multiple transactions at the same time. You must sell before buying again.
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Example 3:
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Input: [7,6,4,3,1]
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Output: 0
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Explanation: In this case, no transaction is done, i.e. max profit = 0.
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{% endhighlight %}
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Solution
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----------
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This one is actually easier than the first one (Buy and Sell Stock 1) because you're not restricted to making only one trade.
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Since you can make multiple trades, you can buy and sell at every peak and valley. In that case, just sum up all the positive days and sum it up.
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{% highlight python %}
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def maxProfit(self, prices: List[int]) -> int:
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positive_days = []
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for i in range(len(prices)-1):
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if prices[i+1]-prices[i]>0:
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positive_days.append( prices[i+1]-prices[i] )
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return sum(positive_days)
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{% endhighlight %}
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We can make it constant space by changing ```positive_days``` to a variable that we continously sum up.
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![image1]()
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