<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, kk: int, prices: List[int]) -> int:
        # daylength = len(prices)

        # everyday, I can buy, sell, or hold
        # @cache
        # def dp(startday, remainK, hasStock):
        #     if startday >= daylength - 1:
        #         if not hasStock:
        #             return 0
        #         else:
        #             return prices[startday]
        #     res = []
        #     best1 = dp(startday + 1, remainK, hasStock)
        #     res.append(best1)
        #     if not hasStock and remainK > 0:
        #         best2 = dp(startday + 1, remainK, True) - prices[startday]
        #         res.append(best2)
        #     if hasStock and remainK > 0:
        #         best3 = dp(startday + 1, remainK - 1, False) + prices[startday]
        #         res.append(best3)
        #     # print(startday, remainK, purchaseDay, res)
        #     return max(res)

        # return dp(0, k, False)

        dp = [[[0, 0] for k in range(kk + 1)] for i in range(len(prices))]
        for i in range(len(prices)):
            dp[i][0][0] = 0
            dp[i][0][1] = -prices[i]
        for k in range(kk + 1):
            if k > 0:
                dp[0][k][0] = 0
                dp[0][k][1] = -prices[0]  # -math.inf
        for i in range(1, len(prices)):
            for k in range(0, kk + 1):
                dp[i][k][1] = max(dp[i - 1][k][1], dp[i - 1][k][0] - prices[i])
                if k > 0:
                    dp[i][k][0] = max(dp[i - 1][k][0], dp[i - 1][k - 1][1] + prices[i])
        ans = 0
        print(dp)
        for k in range(kk + 1):
            print(ans, dp[-1][k][0])
            ans = max(dp[-1][k][0], dp[-1][k][1], ans)
        return ans

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