<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:
        def destroy_directions(arr: List[int]):
            to_remove = set()
            for i in range(1, len(arr) - 1):
                if arr[i] == arr[i + 1]:
                    to_remove.add(i)
            arr = [x for i, x in enumerate(arr) if i not in to_remove]
            to_remove = set()
            for i in range(1, len(arr) - 1):
                if arr[i - 1] <= arr[i] <= arr[i + 1] or arr[i + 1] <= arr[i] <= arr[i - 1]:
                    to_remove.add(i)

            return [x for i, x in enumerate(arr) if i not in to_remove]

        # print(len(prices))
        prices = destroy_directions(prices)
        
        @lru_cache(None)
        def selling(start, rem):
            M, ans = prices[start], 0
            for sell in range(start + 1, len(prices)):
                if prices[sell] > M:
                    ans = max(ans, prices[sell] - prices[start] + buying(sell + 1, rem - 1))
                    M = prices[sell]
            return ans

        @lru_cache(None)
        def buying(start, rem):
            if rem == 0 or start >= len(prices) - 1: return 0
            return max(selling(start, rem), buying(start + 1, rem))

        return buying(0, k)
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