2. Best Time to Buy and Sell Stock

Problem Statement

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.

Solution

class Solution:
    def maxProfit(self, prices: list[int]) -> int:
        min_price = float('inf')
        max_profit = 0
        for price in prices:
            if price < min_price:
                min_price = price
            elif price - min_price > max_profit:
                max_profit = price - min_price
        return max_profit

Explanation

The solution iterates through the prices array, keeping track of the minimum price seen so far (min_price) and the maximum profit found (max_profit).

For each price, we first check if it's a new minimum price. If it is, we update min_price.

Otherwise, we calculate the potential profit by subtracting the min_price from the current price. If this potential profit is greater than our current max_profit, we update max_profit.

This single-pass approach has a time complexity of O(n) and a space complexity of O(1).