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# 0121. Best Time to Buy and Sell Stock

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

## ac

```java
class Solution {
    public int maxProfit(int[] prices) {
        int min = Integer.MAX_VALUE, max = 0;
        for (int i = 0; i < prices.length; i++) {
            min = Math.min(min, prices[i]);
            max = Math.max(max, prices[i] - min);
        }
        return max;
    }
}
/*
divergent thinking! dynamically calculate the temp result, if oper complexity is O(n), just do it.
*/
```

```java
class Solution {
    public int maxProfit(int[] prices) {
        // edge case
        if (prices == null || prices.length <= 1) return 0;

        int min = prices[0];
        int res = 0;
        for (int i = 0; i < prices.length; i++) {
            if (prices[i] < min) min = prices[i];
            int diff = prices[i] - min; // profit if sell
            res = Math.max(res, diff);  // compare not sell vs. sell
        }

        return res;
    }
}
```