<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 f1(int ind, int buy, vector<int>& prices, int n, vector<vector<int>>& dp, int fee){
        if(ind==n) return 0;
        if(dp[ind][buy]!=-1) return dp[ind][buy];
        long profit = 0;
        if(buy){
            profit = max(-prices[ind] + f1(ind+1, 0, prices, n, dp, fee),  0 + f1(ind+1, 1, prices, n, dp, fee));
        }else{
            profit = max(prices[ind] - fee + f1(ind+1, 1, prices, n, dp, fee),  0 + f1(ind+1, 0, prices, n, dp, fee));
        }
        return dp[ind][buy] = profit;
    }
    int maxProfit1(vector<int>& prices, int fee) {
        int n = prices.size();
        vector<vector<int>> dp(n, vector<int>(2, -1)); // Corrected DP table size
        return f1(0, 1, prices, n, dp, fee);
    }
    int maxProfit(vector<int>& prices,  int fee) {
        int n = prices.size();
        vector<vector<int>> dp(n + 1, vector<int>(2, 0)); // Corrected DP table size
        for (int i = n - 1; i >= 0; i--) {
            for (int j = 0; j <= 1; j++) {
                long profit = 0;
                if (j == 1) {
                    profit = max(-prices[i] + dp[i + 1][0], 0 + dp[i + 1][1]); 
                } else {
                    profit = max(prices[i] - fee + dp[i + 1][1], 0 + dp[i + 1][0]);
                }
                dp[i][j] = profit; // Assign the calculated profit to dp[i][j]
            }
        }
        return dp[0][1]; // Return the maximum profit when starting from day 0 with the option to buy
    }    
};
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