<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 helper(ind,transaction,prices,n,k):
            if ind ==n or transaction==2*k:
                return 0
            if dp[ind][transaction]!=-1:
                return dp[ind][transaction]
            profit=0
            if transaction%2==0:
                profit=max(-prices[ind]+helper(ind+1,transaction+1,prices,n,k),
                helper(ind+1,transaction,prices,n,k))
            else:
                profit=max(prices[ind]+helper(ind+1,transaction+1,prices,n,k),
                helper(ind+1,transaction,prices,n,k))
            dp[ind][transaction]=profit
            return profit
        
        n=len(prices)
        # dp=[[-1 for j in range(2*k)]for i in range(n)]
        # return helper(0,0,prices,n,k)
        dp=[[0 for j in range(2*k+1)]for i in range(n+1)]
        
        for ind in range(n-1,-1,-1):
            for transaction in range(2*k-1,-1,-1):
                profit=0
                if transaction%2==0:
                    profit=max(-prices[ind]+dp[ind+1][transaction+1],
                    dp[ind+1][transaction])
                else:
                    profit=max(prices[ind]+dp[ind+1][transaction+1],
                    dp[ind+1][transaction])
                dp[ind][transaction]=profit
        return dp[0][0]



        
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