In this post, we are going to solve the 121. Best Time to Buy and Sell Stock problem of Leetcode. This problem 121. Best Time to Buy and Sell Stock is a Leetcode easy level problem. Let’s see code, 121. Best Time to Buy and Sell Stock – Leetcode Solution.
Problem
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
Now, let’s see the code of 121. Best Time to Buy and Sell Stock – Leetcode Solution.
Best Time to Buy and Sell Stock – Leetcode Solution
121. Best Time to Buy and Sell Stock – Solution in Java
class Solution { public int maxProfit(int[] prices) { int minCostToBuyAt = prices[0]; int profit = 0; for(int i=1; i< prices.length; i++){ minCostToBuyAt = Math.min(minCostToBuyAt,prices[i]); profit = Math.max(profit,prices[i] - minCostToBuyAt); } return profit; } }
121. Best Time to Buy and Sell Stock – Solution in C++
class Solution { public: int maxProfit(vector<int>& prices) { int minCostToBuyAt = prices[0]; int profit = 0; for(int i=1; i< prices.size(); i++){ minCostToBuyAt = min(minCostToBuyAt,prices[i]); profit = max(profit,prices[i] - minCostToBuyAt); } return profit; } };
121. Best Time to Buy and Sell Stock – Solution in Python
class Solution: def maxProfit(self,prices): left = 0 #Buy right = 1 #Sell max_profit = 0 while right < len(prices): currentProfit = prices[right] - prices[left] if prices[left] < prices[right]: max_profit =max(currentProfit,max_profit) else: left = right right += 1 return max_profit
Note: This problem 121. Best Time to Buy and Sell Stock is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.