<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:
        # Recursive code
        # ans = self.solve(prices, 0, 0, fee)
        # Memoization code
        # n = len(prices)
        # dp = [[0 for i in range(2)] for i in range(n+1)]
        # ans = self.solve(prices, 0, 0, dp, fee)
        # return ans
    
    # Recursive code
    # def solve(self, prices, ind, buy, fee):
    #     if ind >= len(prices):
    #         return 0
    #     if buy == 0:
    #         op1 = 0 + self.solve(prices, ind+1, 0, fee)  # Not Buy
    #         op2 = -prices[ind] + self.solve(prices, ind+1, 1, fee) # Buy
    #     elif buy == 1:
    #         op1 = 0 + self.solve(prices, ind+1, 1, fee)  # Not Sell
    #         op2 = (prices[ind] - fee) + self.solve(prices, ind+1, 0, fee)  # Sell

    #     return max(op1, op2)
    # Memoization
    # def solve(self, prices, ind, buy, dp, fee):
    #     if ind >= len(prices):
    #         return 0
    #     if dp[ind][buy]!=0:
    #         return dp[ind][buy]
    #     if buy == 0:
    #         op1 = 0 + self.solve(prices, ind+1, 0, dp, fee)  # Not Buy
    #         op2 = -prices[ind] + self.solve(prices, ind+1, 1, dp, fee) # Buy
    #     elif buy == 1:
    #         op1 = 0 + self.solve(prices, ind+1, 1, dp, fee)  # Not Sell
    #         op2 = (prices[ind] - fee) + self.solve(prices, ind+1, 0, dp, fee)  # Sell
    #     dp[ind][buy] = max(op1, op2)
    #     return dp[ind][buy]

        # Tabulation without helper function
        n = len(prices)
        dp = [[0 for i in range(2)] for i in range(n+1)]
        for ind in range(n-1, -1, -1):
            for buy in range(2):
                if ind <= 0:
                    dp[ind][buy] = 0
                if buy == 0:
                    op1 = 0 + dp[ind+1][0]
                    op2 = -prices[ind]+dp[ind+1][1]
                elif buy == 1:
                    op1 = 0 + dp[ind+1][1]
                    op2 = (prices[ind] - fee) +dp[ind+1][0]
                dp[ind][buy] = max(op1, op2)
        return dp[0][0]
        
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