<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)
#         dp = [[[0, 0] for _ in range(k+1)] for _ in range(n+1)]

#         for i in reversed(range(n)):
#             # representing the index of the price
#             for j in range(1, k + 1):
#                 # representing which transaction it's on
#                 # 0 - not holding a stock, 1 - holding a stock
#                 dp[i][j][0] = max(-prices[i] + dp[i+1][j][1], dp[i+1][j][0])
#                 dp[i][j][1] = max(prices[i] + dp[i+1][j-1][0], dp[i+1][j][1])

#         print(dp)
#         return dp[0][k][0]
        
class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        n = len(prices)
        dp = [[[0, 0] for _ in range(k)] for _ in range(n)]

        # how much we can earn at last price? We can only sell at last price
        dp[-1] = [[0, prices[-1]] for _ in range(k)]

        for i in reversed(range(n-1)):
            for j in range(k):
                if j == k - 1:
                    dp[i][j][1] = max(prices[i], dp[i+1][j][1]) # last transaction, cannot sell anymore
                else:
                    dp[i][j][1] = max(prices[i] + dp[i+1][j+1][0], dp[i+1][j][1])
                
                dp[i][j][0] = max(-prices[i] + dp[i+1][j][1], dp[i+1][j][0])


        print(dp)

        return dp[0][0][0]

#[[[8, 9], [11, 12]], [[8, 9], [11, 12]], [[7, 9], [8, 12]], [[7, 9], [7, 12]], [[7, 9], [7, 9]], [[0, 9], [0, 9]], [[0, 7], [0, 7]]]
        
32701