<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 profit(vector<int>& prices, int idx, bool haveStock, int fee, vector<vector<int>>& dp ){
        if(idx == prices.size())
            return 0; 
        if(dp[haveStock][idx] != -1)
            return dp[haveStock][idx]; 
        int buy = 0, notBuy = 0, sell = 0, notSell = 0; 
        if(haveStock == false){
            buy = profit(prices, idx + 1, true, fee, dp) - prices[idx] - fee; 
            notBuy = profit(prices, idx + 1, haveStock, fee, dp); 
        }
        else{
            sell = profit(prices, idx + 1, false, fee, dp) + prices[idx];
            notSell = profit(prices, idx + 1, haveStock, fee, dp); 
        }
        return dp[haveStock][idx] = max(max(buy, notBuy), max(notSell, sell)); 
    }
    int maxProfit(vector<int>& prices, int fee) {
        vector<vector<int>> dp(2, vector<int>(prices.size(), -1)); 
        return profit(prices, 0, false, fee, dp); 
    }
};
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