# 122 Best Time to Buy and Sell Stock II

You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `ith` day.

On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.

Find and return the maximum profit you can achieve.

Example 1:

``````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.
``````

Example 2:

``````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.
``````

Example 3:

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

Constraints:

• `1 <= prices.length <= 3 * 104`
• `0 <= prices[i] <= 104`
 ``````1 2 3 4 5 6 7 `````` ``````class Solution: def maxProfit(self, prices: List[int]) -> int: total = 0 for i in range(1, len(prices)): if prices[i] > prices[i-1]: total += prices[i] - prices[i-1] return total``````
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 `````` ``````func maxProfit(prices []int) int { if prices == nil || len(prices) == 0 { return 0 } profit := 0 for i := 0; i < len(prices)-1; i++ { if prices[i+1] > prices[i] { profit += prices[i+1] - prices[i] } } return profit }``````