<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:

        # @cache
        # def dfs(i, j, hold):
        #     """
        #     i: day 
        #     j: number of transactions
        #     hold: current state machine, hold or not hold
        #     """

        #     # edge case: invalid number of transaction number
        #     if j < 0:
        #         return -inf

        #     # base case
        #     if i < 0:
        #         return -inf if hold else 0

        #     if hold:
        #         # if from hold state, no transaction number increase
        #         # if from non-hold state, we will increase by one transaction number
        #         return max(dfs(i-1, j, True), dfs(i-1, j-1, False)-prices[i])

        #     # non hold state tranfer choices: i-1,j hold, i-1,j non-hold
        #     else:
        #         return max(dfs(i-1, j, True)+prices[i], dfs(i-1, j, False))

        # return dfs(len(prices)-1, k, hold=False)

        N = len(prices)
        dp = [[[-inf, -inf] for _ in range(k+2)] for _ in range(N+1)]

        # # edge case
        # dp[0][1][0] = 0 # d=-1, j=0, state=0
        # dp[0][1][1] = -inf # d=-1, j=0, state=1



        # edge case: invalid j (transaction time)
        for j in range(1, k+2): # [1, K+1]
            dp[0][j][0] = 0


        for i, price in enumerate(prices):
            for j in range(1, k+2):
                # can go from j=0 to j=1
                # but cannot go from j=-1 to j=1
                # since we are using array
                # we define j=0 as invalid
                dp[i+1][j][0] = max(dp[i][j][0], dp[i][j-1][1] + price)
                dp[i+1][j][1] = max(dp[i][j][0] - price, dp[i][j][1])

        return dp[N][k+1][0]


            
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