<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:
        dp = [[[-1 for _ in range(k+1)] for _ in range(2)] for _ in range(len(prices))]

        def solve(i: int, buy: int, limit: int) -> int:
            if i == len(prices):
                return 0

            if limit == 0:
                return 0

            if dp[i][buy][limit] != -1:
                return dp[i][buy][limit]

            profit = 0
            if buy: # buy is allowed
                p1 = -1 * prices[i] + solve(i+1, 0, limit) # buy
                p2 = 0 + solve(i+1, 1, limit) # didnt buy
                profit = max(p1, p2)

            else: # buy is not allowed
                p1 = prices[i] + solve(i+1, 1, limit-1) # sell
                p2 = 0 + solve(i+1, 0, limit) # didnt sell
                profit = max(p1, p2)

            dp[i][buy][limit] = profit
            return profit

        def solveTab() -> int:
            dp = [[[0 for _ in range(k+1)] for _ in range(2)] for _ in range(len(prices)+1)]

            for i in range(len(prices)-1, -1, -1):
                for j in range(2):
                    for x in range(1, k+1):
                        profit = 0
                        if j: # buy is allowed
                            p1 = -1 * prices[i] + dp[i+1][0][x] # buy
                            p2 = 0 + dp[i+1][1][x] # didnt buy
                            profit = max(p1, p2)

                        else: # buy is not allowed
                            p1 = prices[i] + dp[i+1][1][x-1] # sell
                            p2 = 0 + dp[i+1][0][x] # didnt sell
                            profit = max(p1, p2)

                        dp[i][j][x] = profit

            return dp[0][1][k]


        #return solve(0, 1, k)
        return solveTab()
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