<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:
    vector<int> vec;
    int fee;
    int n;
    int fun[50009][5];

    int dp(int indx,int bought) {

        if(indx >= n)
            return 0;

        if(fun[indx][bought] != -1)
            return fun[indx][bought];

        if(bought == 0) {
            int val1 = (-1)*vec[indx] + dp(indx+1,1);
            int val2 = dp(indx+1,bought);
            fun[indx][bought] = max(val1,val2);
        }
        else {
            int val1 = vec[indx] - fee + dp(indx+1,0);
            int val2 = dp(indx+1,bought);
            fun[indx][bought] = max(val1,val2);
        }
        return fun[indx][bought];
        
    }
    int maxProfit(vector<int>& prices, int f) {
        
        vec.clear();
        for(auto p : prices)
            vec.push_back(p);
        n = vec.size();
        fee = f;

        for(int i=0;i<n+5;i++){
            for(int j=0;j<3;j++){
                fun[i][j] = -1;
            }
        }
        return dp(0,0);
    }
};
68090