<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 getMaxProfit(self, index: int, holding_stock: bool, txn_count: int, prices: List[int], cache: Dict, k: int) -> int:
        # since we reached end of array
        if index == len(prices):
            return 0

        elif (index, holding_stock, txn_count) in cache:
            return cache[(index, holding_stock, txn_count)]
        
        max_profit = 0

        if holding_stock:
            # do not sell the stock currently we are holding
            case_1 = self.getMaxProfit(index + 1, holding_stock, txn_count, prices, cache, k)

            # else I sell that stock
            if txn_count < k - 1:
                case_2 = prices[index] + self.getMaxProfit(index + 1, False, txn_count + 1, prices, cache, k)

            # means we have performed 2 transactions now so do not call functions again
            elif txn_count == k - 1:
                case_2 = prices[index]

            max_profit = max(case_1, case_2)
            cache[(index, holding_stock, txn_count)] = max_profit

        else:
            # do not buy the current stock
            case_1 = self.getMaxProfit(index + 1, holding_stock, txn_count, prices, cache, k)
            
            # else I buy that stock
            case_2 = self.getMaxProfit(index + 1, True, txn_count, prices, cache, k) - prices[index]

            max_profit = max(case_1, case_2)
            cache[(index, holding_stock, txn_count)] = max_profit

        return max_profit

    def maxProfit(self, k: int, prices: List[int]) -> int:
        cache = dict()
        return self.getMaxProfit(0, False, 0, prices, cache, k)
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