# Solving Leetcode Interviews in Seconds with AI: Best Time to Buy and Sell Stock III


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "123" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given an array prices where prices[i] is the price of a given stock on the ith day. 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: 6 Explanation: 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: 4 Explanation: 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: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.    Constraints:  1 <= prices.length <= 105 0 <= prices[i] <= 105  

	# Explanation
	Here's the breakdown of the solution:

*   **Dynamic Programming:** The problem is solved using dynamic programming to store and reuse intermediate results. We track the maximum profit after 0, 1, and 2 transactions.
*   **State Transitions:** The DP states represent the maximum profit achievable at each day given the number of transactions completed. We consider buying, selling, or doing nothing at each step.
*   **Optimization:** The code avoids unnecessary iterations and calculations by keeping track of the best buying and selling points seen so far.

*   **Runtime Complexity:** O(n), where n is the number of prices. **Storage Complexity:** O(1)

	
	# Code
	```python
	def maxProfit(prices):
    """
    Calculates the maximum profit that can be achieved with at most two transactions.

    Args:
        prices: A list of stock prices.

    Returns:
        The maximum profit achievable.
    """
    if not prices:
        return 0

    n = len(prices)
    buy1 = float('inf')
    sell1 = 0
    buy2 = float('inf')
    sell2 = 0

    for price in prices:
        buy1 = min(buy1, price)
        sell1 = max(sell1, price - buy1)
        buy2 = min(buy2, price - sell1)
        sell2 = max(sell2, price - buy2)

    return sell2
	```