Maximize profit from a single buy/sell of a stock given daily prices.

Interview question: Best Time to Buy and Sell Stock

You are given an array `prices` where `prices[i]` is the price of a given stock on day `i`.

You want to maximize your profit by choosing a single day to buy one stock and a different day in the future to sell it. Return the maximum profit you can achieve. If no profit is possible, return `0`.

- You need choose one day to buy and a later day to sell.
- Profit is `sell_price - buy_price`.
- If every later price is lower, the best valid answer is `0` because you can skip the trade.

Understand the Goal

- A direct solution checks every possible buy day with every later sell day.
- This finds the correct answer, but it repeats a lot of work.
- With two nested loops, the time complexity is O(n²).

Consider the Brute Force Idea

- When scanning prices from left to right, the best buy day before today is simply the lowest price seen so far.
- For each current price, calculate the profit from selling today after buying at that lowest earlier price.
- Then update the best profit if today gives a better one.

Track the Cheapest Buy Price

- Initialize `min_price` to a very large value and `best` to `0`.
- For each price, first update the minimum price seen so far.
- Then compare the current possible profit against the best answer.

Build the One-Pass Loop

- The scan preserves the rule that buying happens before selling because `min_price` only comes from days already visited.
- Time complexity is O(n) because each price is processed once.
- Space complexity is O(1) because only two variables are stored.