<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,f;
    int memo(vector<vector<int>>& dp,vector<int>& prices,int i,int buy){
        if(i>=n) return 0;
        if(dp[i][buy]!=-1) return dp[i][buy];
        if(buy) return dp[i][buy]=max(memo(dp,prices,i+1,buy),-prices[i]+memo(dp,prices,i+1,0));
        else return dp[i][buy]=max(memo(dp,prices,i+1,buy),-f+prices[i]+memo(dp,prices,i+1,1));
    }
    int maxProfit(vector<int>& prices, int fee) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        n=prices.size();
        // f=fee;
        // vector<vector<int>> dp(n+1,vector<int>(2,-1));
        // return memo(dp,prices,0,1);

        vector<vector<int>> dp(n+1,vector<int>(2,0));
        // dp[n-1][0]=prices[n-1];
        // dp[n-1][1]=max(dp[n][1],-prices[n-1]+dp[n][0]);

        for(int i=n-1;i>=0;i--){
            for(int buy=0;buy<2;buy++){
                if(buy) dp[i][buy]=max(dp[i+1][buy],-prices[i]+dp[i+1][0]);
                else dp[i][buy]=max(dp[i+1][buy],-fee+prices[i]+dp[i+1][1]);
            }
        }
        return dp[0][1];
    }
};
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