<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:
    """
    First solution I came up with, it's accepted but very slow: beats 5.05%.
    The DP bottom up solution of the editorial beats 40%.
    """
    def maxProfitHelper(self, prices, i, minPrice, attemptsLeft, memo):
        # It is tempting to not want to use attemptsLeft in the key of the memo table, but
        # the callers need to know how many attempts were used to calculate the solution
        # of a given prefix since they will have so many attempts left.
        if i >= len(prices) or attemptsLeft == 0:
           return 0
        elif (i, minPrice, attemptsLeft) in memo:
           return memo[(i, minPrice, attemptsLeft)]

        currentPrice = prices[i]
        if currentPrice <= minPrice:
            memo[(i, minPrice, attemptsLeft)] = self.maxProfitHelper(prices, i+1, currentPrice, attemptsLeft, memo)
        elif i+1 < len(prices)-1 and prices[i+1] > currentPrice:
            # This is an optimization based on the fact that we don't need to sell if we aren't at a peak and we won't
            # get a better solution than if we use the solution that sells at the peak, which is going to be a subproblem
            # of the current problem.
            # But it doesn't change the runtime when I submit it unfortunately.
            memo[(i, minPrice, attemptsLeft)] = self.maxProfitHelper(prices, i+1, minPrice, attemptsLeft, memo)
        else:
            memo[(i, minPrice, attemptsLeft)] = max(
                self.maxProfitHelper(prices, i+1, minPrice, attemptsLeft, memo),
                self.maxProfitHelper(prices, i+1, currentPrice, attemptsLeft-1, memo) + currentPrice - minPrice)

        return memo[(i, minPrice, attemptsLeft)]

    def maxProfit(self, k: int, prices: List[int]) -> int:
        memo = {}
        return self.maxProfitHelper(prices, 1, prices[0], k, memo)


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