Easy
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.
Constraints
- 1 <= prices.length <= 105 - 0 <= prices[i] <= 104
If your really up to the task look into Kadane's Algorithm. For a simpler solution, try iterating two points, one starting at the front, and one starting at the back of the array.
*hover to un-blur*
# two pointer solution # (credit to Leetcode user mageshyt) class Solution: def maxProfit(self,prices): left = 0 #Buy right = 1 #Sell max_profit = 0 while right < len(prices): currentProfit = prices[right] - prices[left] #our current Profit if prices[left] < prices[right]: max_profit =max(currentProfit,max_profit) else: left = right right += 1 return max_profit # Very fast dynamic programming method # (credit to Leetcode user farhanzamanarnob) class Solution: def maxProfit(self, prices): buy = prices[0] profit = 0 for i in range(1, len(prices)): if prices[i] < buy: buy = prices[i] elif prices[i] - buy > profit: profit = prices[i] - buy return profit