<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 dp[50000][2];
    int solve(vector<int>&arr,int fee,int index,int buy) {
        if(index>=arr.size()) {
            return 0;
        }
        if(dp[index][buy]!=-1) {
            return dp[index][buy];
        }
        int b=0,s=0;
        if(buy==0) {
            b=solve(arr,fee,index+1,1)-arr[index];
        }
        else{
            s=solve(arr,fee,index+1,0)+arr[index]-fee;
        }
        int skip=solve(arr,fee,index+1,buy);
        return dp[index][buy]=max(skip,max(s,b)); 

    }

    int tab(vector<int>&arr,int fee) {
        int n=arr.size();
        vector<vector<int>>dp(n+1,vector<int>(2,0));
        for(int index=n-1;index>-1;index--) {
            for(int buy=0;buy<2;buy++) {
                int b=0,s=0;
                if(buy==0) {
                    b=dp[index+1][1]-arr[index];
                }
                else{
                    s=dp[index+1][0]+arr[index]-fee;
                }
                int skip=dp[index+1][buy];
                dp[index][buy]=max(skip,max(s,b));
            }
        }
        return dp[0][0];
    }

    int maxProfit(vector<int>& prices, int fee) {
        // memset(dp,-1,sizeof(dp));
        // int ans=solve(prices,fee,0,0);

        int ans=tab(prices,fee);
        return ans;
    }
};
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