Best Time to Buy and Sell Stock IV
<p>You are given an integer 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>k</code>.</p>

<p>Find the maximum profit you can achieve. You may complete at most <code>k</code> transactions: i.e. you may buy at most <code>k</code> times and sell at most <code>k</code> times.</p>

<p><strong>Note:</strong> You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> k = 2, prices = [2,4,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> k = 2, prices = [3,2,6,5,0,3]
<strong>Output:</strong> 7
<strong>Explanation:</strong> Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= 100</code></li>
	<li><code>1 &lt;= prices.length &lt;= 1000</code></li>
	<li><code>0 &lt;= prices[i] &lt;= 1000</code></li>
</ul>
Array; Dynamic Programming
function maxProfit(k: number, prices: number[]): number {
    const n = prices.length;
    const f = Array.from({ length: n }, () =>
        Array.from({ length: k + 1 }, () => Array.from({ length: 2 }, () => -1)),
    );
    const dfs = (i: number, j: number, k: number): number => {
        if (i >= n) {
            return 0;
        }
        if (f[i][j][k] !== -1) {
            return f[i][j][k];
        }
        let ans = dfs(i + 1, j, k);
        if (k) {
            ans = Math.max(ans, prices[i] + dfs(i + 1, j, 0));
        } else if (j) {
            ans = Math.max(ans, -prices[i] + dfs(i + 1, j - 1, 1));
        }
        return (f[i][j][k] = ans);
    };
    return dfs(0, k, 0);
}
