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.