Best Time to Buy and Sell Stock with Transaction Fee
<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>
Greedy; Array; Dynamic Programming
class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int n = prices.size();
        int f[n][2];
        memset(f, -1, sizeof(f));
        function<int(int, int)> dfs = [&](int i, int j) {
            if (i >= prices.size()) {
                return 0;
            }
            if (f[i][j] != -1) {
                return f[i][j];
            }
            int ans = dfs(i + 1, j);
            if (j) {
                ans = max(ans, prices[i] + dfs(i + 1, 0) - fee);
            } else {
                ans = max(ans, -prices[i] + dfs(i + 1, 1));
            }
            return f[i][j] = ans;
        };
        return dfs(0, 0);
    }
};