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In this post, you will find the solution for the **Best Time to Buy and Sell Stock** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Best Time to Buy and Sell Stock– LeetCode Problem

Best Time to Buy and Sell Stock– LeetCode Problem

**Problem:**

You are given an array `prices`

where `prices[i]`

is the price of a given stock on the `i`

day.^{th}

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:5Explanation: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:0Explanation:In this case, no transactions are done and the max profit = 0.

**Constraints:**

`1 <= prices.length <= 10`

^{5}`0 <= prices[i] <= 10`

^{4}

Best Time to Buy and Sell Stock– LeetCode Solutions

Best Time to Buy and Sell Stock Solution in C++:

class Solution { public: int maxProfit(vector<int>& prices) { int sellOne = 0; int holdOne = INT_MIN; for (const int price : prices) { sellOne = max(sellOne, holdOne + price); holdOne = max(holdOne, -price); } return sellOne; } };

Best Time to Buy and Sell Stock Solution in Java:

class Solution { public int maxProfit(int[] prices) { int sellOne = 0; int holdOne = Integer.MIN_VALUE; for (final int price : prices) { sellOne = Math.max(sellOne, holdOne + price); holdOne = Math.max(holdOne, -price); } return sellOne; } }

Best Time to Buy and Sell Stock Solution in Python:

class Solution: def maxProfit(self, prices: List[int]) -> int: sellOne = 0 holdOne = -inf for price in prices: sellOne = max(sellOne, holdOne + price) holdOne = max(holdOne, -price) return sellOne