<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:
        
        cache = [] #day, transcation,  own
        for d in range(len(prices)): #days
            tem = []
            for i in range(k*2): #transcation
                tem.append([None, None]) #own or not
            cache.append(tem)
            
        def dfs(d, t, own):
            if d == len(prices) or t == 2*k:
                return 0
            
            if cache[d][t][own] != None:
                return cache[d][t][own] 
            
            if own == 1:
                #sell it or not today
                cache[d][t][own] = max( dfs(d + 1, t + 1, 0) + prices[d], dfs(d + 1, t, 1)) 
                
            else:
                #buy it or not today
                cache[d][t][own] = max( dfs(d + 1, t + 1, 1) - prices[d], dfs(d + 1, t, 0))

            return cache[d][t][own]
        
        return dfs(0, 0, 0)

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