<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)
        if n==1:
            return 0

        @lru_cache(maxsize=None)
        def dp(i, j, bought):
            if i>=n:
                return 0
            if j==0:
                return 0
            
            if bought:
                return max(dp(i+1, j, bought), prices[i]+dp(i, j-1, 0))
            else:
                return max(dp(i+1, j, bought), -prices[i]+dp(i, j, 1))

        # dp = [[[0 for _ in range(k+1)] for _ in range(n)] for _ in range(2)]
        # dp[1][0]
        # for i in range(n):
        #     for j in range(k+1):
        #         for sold in range(2):
        #             dp[sold][i][j] = dp[sold][i-1][j] if i-1 else float('-inf')
        #             if sold:
        #                 dp[1][i][j] = max(dp[sold][i][j], -1*prices[i]+dp[0][i][j-1]) if j else float('-inf')
        #             else:
        #                 dp[0][i][j] = max(dp[0][i][j], prices[i]+dp[1][i-1][j-1]) if j else 0

        return dp(0, k, 0)

        return dp[0][n-1][k]


# dp(i, j) max profit from prices[0..i] with <=j transctations 
# dp(0, j) = 0
# dp(1, j) = 0 if j==0 or max(0, prices[1]-prices[0]) if j>=1 
# dp(i, 0) = 0
# dp(i, j) = max(dp(i-1, j), nums[i]-nums[i-1]+dp(i-2, j-1) if j>=1 
71629