<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 {
    vector<vector<int>> dp;
    vector<int> prices;
    int fee;

    int helper(int i, int canBuy) {
        int n = prices.size();
        if(i >= n) {
            return 0;
        }
        if(dp[i][canBuy] != INT_MIN) {
            return dp[i][canBuy];
        }

        if(canBuy) {
            return dp[i][canBuy] = max({
                helper(i+1, 1), // didn't buy
                -1*prices[i] - fee + helper(i+1, 0) // buy
            });
        }

        return dp[i][canBuy] = max({
            prices[i] + helper(i+1, 1), // sold the stock
            helper(i+1, 0)
        });
    }

public:
    int maxProfit(vector<int>& Prices, int Fee) {
        prices = Prices;
        fee = Fee;
        int n = prices.size();
        dp = vector<vector<int>>(n+1, vector<int>(2, 0));

        for(int i=n-1; i>=0; i--) {
            dp[i][1] = max(dp[i+1][1], dp[i+1][0] - prices[i] - fee);
            dp[i][0] = max(dp[i+1][0], dp[i+1][1] + prices[i]);
        }

        return dp[0][1];
        
        // top down
        return helper(0, 1);
    }
};
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