Leetcode 121. 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 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.

By using brute force, it can lead to time out

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        max_profit = 0

        for i in range(0, len(prices)):
            for j in range(i, len(prices)):
                profit = prices[j] - prices[i]

                max_profit =  profit if max_profit < profit else max_profit
        
        return max_profit

This can be solved by leveraging 2 pointer technique

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if len(prices) <= 1:
            return 0

        max_profit = 0

        buy, sell = 0, 1

        while sell < len(prices):
            profit = prices[sell] - prices[buy]
            max_profit = profit if profit > max_profit else max_profit
            
            if prices[sell] < prices[buy]:
                buy = sell

            sell += 1

        return max_profit