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tags: data_structure_python
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# Best Time to Buy and Sell Stock II <img src="https://img.shields.io/badge/-easy-brightgreen">

Say you have an array for which the $i^{th}$ element is the price of a given stock on day $i$.

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).

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

<ins>**Example 1:**</ins>

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

<ins>**Example 2:**</ins>

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

<ins>**Example 3:**</ins>

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

# Solution

### Solution 1: Leetcode solution

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if prices == []:
            return 0
        
        curr = prices[0]
        maxProfit = 0
        
        for price in prices[1:]:
            if price-curr > 0:
                maxProfit += price-curr
            curr = price
        return maxProfit

### Solution 2: Cleaner solution

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        max_profit, m = 0, len(prices)
        for i in range(1, m):
            if prices[i] > prices[i-1]:
                max_profit += prices[i] - prices[i-1]
        return max_profit