<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, and an integer <code>fee</code> representing a transaction fee.</p>

<p>Find the maximum profit you can achieve. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction.</p>

<p><strong>Note:</strong></p>

<ul>
	<li>You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).</li>
	<li>The transaction fee is only charged once for each stock purchase and sale.</li>
</ul>

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

<pre>
<strong>Input:</strong> prices = [1,3,2,8,4,9], fee = 2
<strong>Output:</strong> 8
<strong>Explanation:</strong> The maximum profit can be achieved by:
- Buying at prices[0] = 1
- Selling at prices[3] = 8
- Buying at prices[4] = 4
- Selling at prices[5] = 9
The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
</pre>

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

<pre>
<strong>Input:</strong> prices = [1,3,7,5,10,3], fee = 3
<strong>Output:</strong> 6
</pre>

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

<ul>
	<li><code>1 &lt;= prices.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= prices[i] &lt; 5 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= fee &lt; 5 * 10<sup>4</sup></code></li>
</ul>

class Solution:
    def maxProfit(self, prices: List[int], fee: int) -> int:
        N = len(prices)

        buy_has_cache, sell_has_cache = [False] * N, [False] * N
        buy_cache, sell_cache = [None] * N, [None] * N

        def buy(day: int) -> int:
            if day == N:
                return 0
            if buy_has_cache[day]:
                return buy_cache[day]
            buying = sell(day+1) - (prices[day] + fee)
            not_buying = buy(day+1)
            buy_has_cache[day] = True
            buy_cache[day] = max(buying, not_buying)
            return buy_cache[day]

        def sell(day: int) -> int:
            if day == N:
                return 0
            if sell_has_cache[day]:
                return sell_cache[day]
            selling = buy(day+1) + prices[day]
            not_selling = sell(day+1)
            sell_has_cache[day] = True
            sell_cache[day] = max(selling, not_selling)
            return sell_cache[day]
        
        return buy(0)
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