<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>

from heapq import heappush, heappop
class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        # attempt 1: see notes and problem 123. sell/buy stock III. attempt 3
        
        n = len(prices)
        k = min(k,n//2)
        # *2 becase is transaction have 2 actions buy then sell. 
        #+1 to account for 0 trans row make algorithm smoother
        m = 2*k+1 

        # setting up row 0 => 0 transaction and column 0 => first day  based values
        t = [ [-prices[0] if i%2 else 0]*n for i in range(m) ] 

        buy_sell = -1 # -1 buy, 1 sell
        for i in range(1,m):
            for d in range(1,n):
                no_action = t[i][d-1]
                action = t[i-1][d-1] +buy_sell*prices[d]# dig = max profit at the end  of the previous half trans
                t[i][d] = max(action,no_action)
            buy_sell*=-1
        
        return t[-1][-1]
            





    


    
    
    
            

            
        

        

                





        
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