<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 {
public:
    int n;
    int fee;
    vector<vector<int>> memo;
    vector<int> prices;
    int maxProfit(vector<int>& prices, int fee) 
    {
        this->prices = prices;
        n = prices.size();
        this->fee = fee;
        memo = vector<vector<int>> (n, vector<int> (2, -1));
        return dp(0, 0);
    }

    int dp(int i, int held)
    {
        if(i==n)
            return 0;

        if(memo[i][held]!=-1)
            return memo[i][held];

        int ans = 0;
        if(held)
        {
            ans = max(dp(i+1, 1),dp(i+1, 0) + prices[i] - fee);
        }
        else
        {
            ans = max(dp(i+1, 1)-prices[i], dp(i+1, 0));
        }
        memo[i][held] = ans;
        return ans;
    }
};
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