# LeetCode: [121] Best Time to Buy and Sell Stock

You are given an array `prices`

where `prices[i]`

is the price of a given stock on the ith day.

You want to maximize your profit by choosing a **single** day to buy one stock and choosing a **different day in the future** to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

### Example 1:

Input: prices = [7,1,5,3,6,4]

Output: 5

**Explanation:** Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.

Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

### Example 2:

Input: prices = [7,6,4,3,1]

Output: 0

**Explanation:** In this case, no transactions are done and the max profit = 0.

**Constraints:**

`1 <= prices.length <= 105`

`0 <= prices[i] <= 104`

### Solution

This problem is an ideal candidate for **dynamic programming**, which is keeping the best result until we see a better one.

In this case, we start from the end to the beginning, keeping track of the highest return and the highest stock price.

```
/**
* @param {number[]} prices
* @return {number}
*/
var maxProfit = function (prices) {
// dp from end to start
if (!prices || prices.length < 1) return 0
let maxProfit = 0
let high = 0
for (let i = prices.length; i >= 0; i--) {
let profit = high - prices[i]
maxProfit = Math.max(maxProfit, profit)
high = Math.max(prices[i], high)
}
return maxProfit
}
```

Source: https://leetcode.com/problems/best-time-to-buy-and-sell-stock/