# 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:** 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:** prices = [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:** prices = [7,6,4,3,1]

**Output:** 0

**Explanation:** 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]))