<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 solve(vector<int> &prices, int ind, int buy, int n, int fee, vector<vector<int>> &dp) {
        if (ind == n) return 0;
        if (dp[ind][buy] != -1)
            return dp[ind][buy];
        int profit;
        if (buy == 0) {
            profit = max(0 + solve(prices, ind + 1, 0, n, fee, dp), -prices[ind] + solve(prices, ind + 1, 1, n, fee, dp));
        }
        if (buy == 1) { 
            profit = max(0 + solve(prices, ind + 1, 1, n, fee, dp), prices[ind] - fee + solve(prices, ind + 1, 0, n, fee, dp));
        }
        return dp[ind][buy] = profit;
    }

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

                if (buy == 1) { 
                    profit = max(0 + dp[ind + 1][1], prices[ind] - fee + dp[ind + 1][0]);
                }

                dp[ind][buy] = profit;
            }
        }

        return dp[0][0];
    }
    int maxProfit(vector<int>& prices, int fee) {
        int n = prices.size();
        // vector<vector<int>>dp(n,vector<int>(2,-1));
        // return solve(prices,0,0,n,fee,dp);

        return tab(prices,fee,n);
    }
};
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