# LeetCode 121. Best Time to Buy and Sell Stock

## Description

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`

## Explanation

The problem is seeking the max profit in buying and selling stocks.

This is a simple modeling stock trading in real life. The essence of the problem is looking for the max value of prices[j] – prices[i], where j > i.

It’s easy to solve the problem by keep tracking of the minimum price and max profit when iterating the array.

## Java Solution

``````public class Solution {
public int maxProfit(int[] prices) {
int maxProfit = 0;
int minPrice = Integer.MAX_VALUE;
for (int i = 0; i < prices.length; i++) {
minPrice = Math.min(minPrice, prices[i]);
maxProfit = Math.max(maxProfit, prices[i] - minPrice);
}
return maxProfit;
}
}
﻿``````

## Python Solution

``````class Solution:
def maxProfit(self, prices: List[int]) -> int:
if not prices:
return 0

max_profit = 0
min_price = prices

for price in prices[1:]:
profit = price - min_price
max_profit = max(max_profit, profit)
min_price = min(min_price, price)

return max_profit``````
• Time complexity: O(N).
• Space complexity: O(1).