# 121-best-time-to-buy-and-sell-stock Try it on leetcode ## 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:

## Solution(Python) ```Python class Solution: def maxProfit(self, prices: List[int]) -> int: return self.optimized(prices) """ Time Complexity: O(n^2) Space Complexity: O(1) """ def bruteforce(self, prices: List[int]) -> int: n = len(prices) maxProfit = 0 for i in range(n): for j in range(i + 1, n): if prices[j] - prices[i] > maxProfit: maxProfit = prices[j] - prices[i] return maxProfit """ Time Complexity: O(n) Space Complexity: O(1) """ def optimized(self, prices: List[int]) -> int: buyPrice = float("inf") profit = 0 for sellprice in prices: if sellprice < buyPrice: buyPrice = sellprice elif sellprice - buyPrice > profit: profit = sellprice - buyPrice return profit ```