# Buy-Sell Stocks II

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

Find the maximum profit you can achieve. 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 simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

`Input: prices = [7,1,5,3,6,4]Output: 7Explanation: 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: prices = [1,2,3,4,5]Output: 4Explanation: 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: prices = [7,6,4,3,1]Output: 0Explanation: In this case, no transaction is done, i.e., max profit = 0.`

For us to solve this problem I came up with a strategy, first we set a variable to hold our final profit, then we iterate through the prices array starting at index 1. After we subtract our index -1 from our index meaning since we start at index 1 we want to see the value of the element at index 0 minus the element at index 1, and if the value is greater than 0 meaning we gain a profit then we add that profit to our maximum, if not we continue, and the. we return our maximum.

Here is the working code to solve this problem.

`const maxProfit = (prices) => {   let max = 0;        for (let i = 1; i < prices.length; i++) {         let profit = prices[i] - prices[i - 1];        if (profit > 0) {             max += profit        } else {            continue;         }    }    return max;};console.log(maxProfit([7,6,4,3,1]))`