## Description

https://leetcode.com/problems/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 `i`

day.^{th}

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:5Explanation: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:0Explanation:In this case, no transactions are done and the max profit = 0.

**Constraints:**

`1 <= prices.length <= 10`

^{5}`0 <= prices[i] <= 10`

^{4}

## 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[0]
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).

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