<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 {
    // int dp[50000][2];
public:
    // int help(int curi,int buy,int fee,vector<int>& prices){
    //     if(curi==prices.size()) return 0;
    //     if(dp[curi][buy]!=-1) return dp[curi][buy];
    //     if(buy){
    //         // if(cd) return dp[curi][buy][cd]=help(curi+1,buy,cd-1,prices);
    //         return dp[curi][buy]=max({0,help(curi+1,0,fee,prices)-prices[curi],help(curi+1,buy,fee,prices)});
    //     }else{
    //         return dp[curi][buy]=max({0,help(curi+1,1,fee,prices)+prices[curi]-fee,help(curi+1,0,fee,prices)});
    //     }
    // }
    int maxProfit(vector<int>& prices, int fee) {
        // memset(dp,-1,sizeof(dp));
        // return help(0,1,fee,prices);
        vector<vector<int>> dp(prices.size(),vector<int>(2,0));
        dp[0][0]=0;dp[0][1]=-prices[0]-fee;
        for(int i=1;i<prices.size();i++){
            dp[i][0]=max(dp[i-1][0],dp[i-1][1]+prices[i]);
            dp[i][1]=max(dp[i-1][1],dp[i-1][0]-prices[i]-fee);
        }
        return max(dp.back()[0],dp.back()[1]);
    }
};
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