# [LeetCode] Top Interview 150 Problems Solving # 121 Best Time to Buy and Sell Stock

# **Understanding the Problem**

It is to find the best time to sell the stock with the max profit. A given list indicates the stock prices. Each element is a stock price in time series. For instance, if `prices = [7,1,5,3,6,4]`, the best time to buy the stock is when the price is `1` and the best time to sell it is when the price is `6`, so the profit will be `5`, which is a returned number.

# **Approach**

My idea was to loop it twice with `for` to find the best profit. But I faced the *time limit* error and could not pass with *O(n²)*. I had to make it to *O(n)*.

# **Solution**

Well, my goal was to find the min number and the max difference to find the best profit. So first initialize `min_num` with `float(“inf”)`. The code will start the loop one by one of the prices.

If the element `p` is lower than `min_num`, set `p` to `min_num` for subtraction comparison. if `p - min_num` is greater than `max_diff`, which means the profit is greater, set `max_diff` with `p - min_num`.

```python
class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        min_num = float("inf")
        max_diff = 0

        for p in prices:
            if p < min_num:
                min_num = p
            elif p - min_num > max_diff:
                max_diff = p - min_num
        return max_diff
```

# **Reflection**

The code was simple but it took some time for me to solve the problem. I was busy visiting my parents and I was looking for other jobs to move to, which means I could not focus on the problem solving algorithm. Anyways, time efficiency this time again.