<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:
        memo1 = {}
        memo0 = {}
        def f_1(d, i):
            # i-times hold/buy stock at day: d 
            # return: earing
            if d == 0 and i == 1:
                return - prices[d]

            if d < i:
                return - inf

            # if i <= 0:
            #     return 0
            
            if (d, i) in memo1:
                return memo1[(d, i)]

            ans = max( f_1( d-1 , i) , f_0(d-1, i - 1) - prices[d])
            memo1[(d, i)] = ans

            return ans
        
        def f_0(d, i):
            # i-times sell/not hold stock at day: d 
            # return: earing

            if d == 0:
                return 0
            if i <= 0:
                return 0
            if (d, i) in memo0:
                return memo0[(d, i)]
            
            ans = max( f_1(d-1 , i) + prices[d], f_0(d-1, i))
            memo0[(d, i)] = ans
            return ans

        return max(f_0(len(prices) - 1 , i) for i in range(k + 1))
            
            

        
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