<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:
        n = len(prices)

        '''
        memo = [ [ [-1] * (k+1) for _ in range(2) ] for _ in range(n) ]
        def memoization(idx, canBuy, transaction):
            # base cases
            if idx == n or transaction == 0:
                return 0
            if memo[idx][canBuy][transaction] != -1:
                return memo[idx][canBuy][transaction]

            # main working
            if canBuy:
                profit = max(memoization(idx+1, 1, transaction), 
                             -prices[idx] + memoization(idx+1, 0, transaction))
            else:
                profit = max( memoization(idx+1, 0, transaction), 
                              prices[idx] + memoization(idx+1, 1, transaction-1) )
            
            memo[idx][canBuy][transaction] = profit
            return memo[idx][canBuy][transaction]
        
        # trying memoization
        return memoization(0,1,k)
        '''



        def tabulation():
            dp = [ [ [0] * (k+1) for _ in range(2) ] for __ in range(n+1) ]
        
            # base case 1 - no transactions left (ie: 0) - not req as initialized with 0 alr
            # base case 2 - idx reaches len(prices) (ie: n) - not req as initialized with 0 alr

            # filling the table
            for idx in range(n-1, -1, -1):
                for canBuy in range(2):
                    for transaction in range(1, k+1):
                        if canBuy == 1:
                            dp[idx][1][transaction] = max(dp[idx+1][1][transaction], 
                                                          -prices[idx] + dp[idx+1][0][transaction])
                        else:
                            dp[idx][0][transaction] = max(dp[idx+1][0][transaction], 
                                                          prices[idx] + dp[idx+1][1][transaction-1])
            return dp[0][1][k]
        
        # trying tabulation
        return tabulation()
20008