<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>

class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        n = len(prices)
        # hold - 1, not hold - 0
        # dp[1][i] = max(dp[0][i + 2] + price, dp[1][i + 1])
        # dp[0][i] = max(dp[1][i + 1] - price, dp[0][i + 1])
        # dp_hold = [[0 for _ in range(k + 1)] for _ in range(n + 1)]
        # dp_nohold = [[0 for _ in range(k + 1)] for _ in range(n + 1)]
        # for i in range(n - 1, -1, -1):
        #     for times in range(1, k + 1):
        #         print(i, times)
        #         dp_hold[i][times] = max(dp_nohold[i + 1][times - 1] + prices[i], dp_hold[i + 1][times])
        #         dp_nohold[i][times] = max(dp_hold[i + 1][times] - prices[i], dp_nohold[i + 1][times])
        @cache
        def dfs(i, hold, k):
            if i >= n:
                return 0

            if hold:
                # can sell
                if k > 0:
                    return max(dfs(i + 1, False, k - 1) + prices[i], dfs(i + 1, True, k))
                else:
                    return dfs(i + 1, hold, k)
            else:
                # can not sell
                return max(dfs(i + 1, False, k), dfs(i + 1, True, k) - prices[i])
        return dfs(0, False, k)
82630