<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:
        HOLD = 1
        FREE = 0
        dp = defaultdict(lambda: defaultdict(int))
        dp[0][(0, FREE)] = 0
        dp[0][(0, HOLD)] = -prices[0]
        maxprofit = 0
        for day in range(len(prices)-1):
            #print(day, 'day')
            for (action, holding), value in dp[day].items():
                #print('interpreting', action, holding, 'value', value)
                if holding:
                    dp[day+1][(action+1, FREE)] = value + prices[day+1]
                elif action < k:
                    dp[day+1][(action, HOLD)] = value - prices[day+1]
            
            for (action, holding), value in dp[day].items():
                dp[day+1][(action, holding)] = max(dp[day+1][(action, holding)], value)

        #print(dp[0])
        #print(dp[1])
        #print(dp[2])
        #print(dp[3])
        #print(dp[4])
        #print(dp[5])
        #print(dp[6])
        #print(dp[7])
        
        return max(dp[len(prices)-1].values())       




'''        @cache
        def dp(i, holding, remain):
            if i == len(prices) or remain == 0:
                return 0
            
            sell = buy = 0
            skip = dp(i + 1, holding, remain)
            if holding:
                sell = prices[i] + dp(i+1, False, remain-1)
            else:
                buy = -prices[i] + dp(i+1, True, remain)
            return max(sell, buy, skip)

        return dp(0, False, k)'''
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