<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 recursion(vector<int>arr , int i , int buy, int fee){

        if(i >= arr.size()) return 0;

        int profit = 0;
        if(buy == 0){
            int op1 = recursion(arr , i+1 , 1 , fee) - arr[i];
            int op2 = recursion(arr , i+1 , 0 , fee);
            profit = max(op1 , op2);
        }
        else{
            int op1 = (recursion(arr , i+1 , 0 , fee) + arr[i] - fee);
            int op2 = recursion(arr ,i+1 , 1 , fee);
            profit = max(op1 , op2);
        }

        return profit;
    }

    int memo(vector<int>arr , int i , int buy, int fee, vector<vector<int>>&dp){

        if(i >= arr.size()) return 0;

        if(dp[i][buy] != -1) return dp[i][buy];

        int profit = 0;
        if(buy == 0){
            int op1 = memo(arr , i+1 , 1 , fee, dp) - arr[i];
            int op2 = memo(arr , i+1 , 0 , fee, dp);
            profit = max(op1 , op2);
        }
        else{
            int op1 = memo(arr , i+1 , 0 , fee, dp) + arr[i] - fee;
            int op2 = memo(arr , i+1 , 1 , fee, dp);
            profit = max(op1 , op2);
        }

        return dp[i][buy] = profit;
    }

    int tabu(vector<int>arr , int n , int fee){

        vector<vector<int>>dp(n+1 , vector<int>(3 , 0));

        for(int i = n-1 ; i>=0 ; i--){
            for(int j = 0 ; j<=1 ; j++){
                
                int profit = 0;
                if(j == 0){
                    int op1 = dp[i+1][1] - arr[i];
                    int op2 = dp[i+1][0];
                    profit = max(op1 , op2);
                }
                else{
                    int op1 = dp[i+1][0] + arr[i] - fee;
                    int op2 = dp[i+1][1];

                    profit = max(op1 , op2);
                }

                dp[i][j] = profit;
            }
        }

         return dp[0][0];
    }

    int maxProfit(vector<int>& prices, int fee) {
        
        // recursion
        // return recursion(prices , 0 , 0, fee);

        // memoization
        // vector<vector<int>>dp(prices.size() , vector<int>(2 , -1));
        // return memo(prices , 0 , 0, fee , dp);

        // tabulation
        return tabu(prices , prices.size() , fee);

    }
};
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