<p>You are given an array <code>prices</code> where <code>prices[i]</code> is the price of a given stock on the <code>i<sup>th</sup></code> day.</p>

<p>Find the maximum profit you can achieve. You may complete <strong>at most two transactions</strong>.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> prices = [3,3,5,0,0,3,1,4]
<strong>Output:</strong> 6
<strong>Explanation:</strong> 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.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> prices = [1,2,3,4,5]
<strong>Output:</strong> 4
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= prices.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= prices[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:

    def __init__(self):
        self.memo = {}

    def f(self, prices, idx, held, n):
        """
        idx: the idx (day) we are currently on
        held: 0 or 1 to indicate whether we currently holding onto stock or not
        n: the number of transactions we have currently made
        """
        # if we have already made 2 transactions, we cannot buy or sell anymore
        # so return 0
        if n == 4:
            return 0
        
        # else if we can still make a transaction
        # if we are on the last day, then best option is to just sell if we are holding on
        if idx == len(prices) - 1:
            if held:
                return prices[idx]
            else:
                return 0
        
        if (idx, held, n) in self.memo:
            return self.memo[(idx, held, n)]

        # else, we have a few options
        options = []

        # if we do not currently hold any stock, we can either buy today, or just pass
        if not held:
            # buy today
            options.append(-prices[idx] + self.f(prices, idx + 1, 1, n + 1))
            # hold out
            options.append(self.f(prices, idx + 1, 0, n))
        
        # if we current hold onto stock, can either sell it now, or hold out
        else:
            # sell it now
            options.append(prices[idx] + self.f(prices, idx + 1, 0, n + 1))
            # hold out
            options.append(self.f(prices, idx + 1, 1, n))

        self.memo[(idx, held, n)] = max(options)
        return self.memo[(idx, held, n)]


    def maxProfit(self, prices: List[int]) -> int:
        return self.f(prices, 0, 0, 0)
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