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In this post, you will find the solution for the **Best Time to Buy and Sell Stock III** **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 III– LeetCode Problem

Best Time to Buy and Sell Stock III– LeetCode Problem

**Problem:**

You are given an array `prices`

where `prices[i]`

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

day.^{th}

Find the maximum profit you can achieve. You may complete **at most two transactions**.

**Note:** You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

**Example 1:**

Input:prices = [3,3,5,0,0,3,1,4]Output:6Explanation:Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

**Example 2:**

Input:prices = [1,2,3,4,5]Output:4Explanation:Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.

**Example 3:**

Input:prices = [7,6,4,3,1]Output:0Explanation:In this case, no transaction is done, i.e. max profit = 0.

**Constraints:**

`1 <= prices.length <= 10`

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

^{5}

Best Time to Buy and Sell Stock III– LeetCode Solutions

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

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

Best Time to Buy and Sell Stock III Solution in Java:

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

Best Time to Buy and Sell Stock III Solution in Python:

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

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