<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)

        # Transaction type - (buy,sell): 0,0 - 1,0 - 1,1 - 2,1 - 2,2 - ...
        f = [[[0 if q <= 2 else -math.inf for q in range(1 + k * 2)] for i in range(n)] for j in range(4)]


        # [0, -prices[0], 0, -math.inf, -math.inf, ...]                 # buy
        # [0, 0, 0, -math.inf, -math.inf, ...]                          # sell
        # [0, 0, 0, -math.inf, -math.inf, ...]                          # skip buy
        # [-math.inf, -math.inf, -math.inf, -math.inf, -math.inf, ...]  # skip sell

        f[0][0][1] = -prices[0]
        f[3][0][0] = -math.inf
        f[3][0][1] = -math.inf
        f[3][0][2] = -math.inf
        

        for i in range(1, n):
            for j in range(1, 1 + k * 2, 2):
                f[0][i][j - 1] = max(f[1][i - 1][j - 1], f[2][i - 1][j - 1])
                f[0][i][j] = f[0][i][j - 1] - prices[i]

                f[2][i][j - 1] = max(f[1][i - 1][j - 1], f[2][i - 1][j - 1])
                f[2][i][j] = max(f[1][i - 1][j], f[2][i - 1][j])

            for j in range(2, 1 + k * 2, 2):
                f[1][i][j - 1] = max(f[0][i - 1][j - 1], f[3][i - 1][j - 1])
                f[1][i][j] = f[1][i][j - 1] + prices[i]

                f[3][i][j - 1] = max(f[0][i - 1][j - 1], f[3][i - 1][j - 1])
                f[3][i][j] = max(f[0][i - 1][j], f[3][i - 1][j])


        return max(val for l in f[1] for val in l)
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