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]))