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
func maxProfit(prices []int, fee int) int {
	n := len(prices)
	f := make([][2]int, n)
	for i := range f {
		f[i] = [2]int{-1, -1}
	}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i >= n {
			return 0
		}
		if f[i][j] != -1 {
			return f[i][j]
		}
		ans := dfs(i+1, j)
		if j > 0 {
			ans = max(ans, prices[i]+dfs(i+1, 0)-fee)
		} else {
			ans = max(ans, -prices[i]+dfs(i+1, 1))
		}
		f[i][j] = ans
		return ans
	}
	return dfs(0, 0)
}