<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>

def neg(b):
    if b != 0:
        return 0
    return 1


class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        dp = [
            [
                [[-float("inf") for _ in range(3)] for _ in range(2)]
                for _ in range(k + 1)
            ]
            for _ in range(len(prices) + 1)
        ]
        for i in range(len(prices) + 1):
            prices_i = i - 1
            for num_transactions in range(k + 1):
                for holding in range(2):
                    for action in range(3):
                        if i == 0:
                            if num_transactions == 0 and action == 0 and holding == 0:
                                dp[i][num_transactions][holding][action] = 0
                            continue
                        if num_transactions == 0 and action == 2:
                            continue
                        if holding == 1 and action == 2:
                            continue
                        if holding == 0 and action == 1:
                            continue
                        match action:
                            case 0:  # Sit -> Last action was sit, buy or sell
                                dp[i][num_transactions][holding][action] = max(
                                    dp[i - 1][num_transactions][holding][action],
                                    dp[i - 1][num_transactions][holding][1],
                                    dp[i - 1][num_transactions][holding][2],
                                )
                            case 1:  # Buy -> Last action was sit or sell
                                dp[i][num_transactions][holding][action] = -prices[
                                    prices_i
                                ] + max(
                                    dp[i - 1][num_transactions][neg(holding)][0],
                                    dp[i - 1][num_transactions][neg(holding)][2],
                                )
                            case 2:  # Sell -> Last action was sit or buy
                                dp[i][num_transactions][holding][action] = prices[
                                    prices_i
                                ] + max(
                                    dp[i - 1][num_transactions - 1][neg(holding)][0],
                                    dp[i - 1][num_transactions - 1][neg(holding)][1],
                                )
        return max(
            max(max(max(sssdp) for sssdp in sub_sub_dp) for sub_sub_dp in sub_dp)
            for sub_dp in dp
        )

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