<p>You are given an 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.</p>

<p>Find the maximum profit you can achieve. You may complete <strong>at most two transactions</strong>.</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> prices = [3,3,5,0,0,3,1,4]
<strong>Output:</strong> 6
<strong>Explanation:</strong> Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> prices = [1,2,3,4,5]
<strong>Output:</strong> 4
<strong>Explanation:</strong> Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> prices = [7,6,4,3,1]
<strong>Output:</strong> 0
<strong>Explanation:</strong> In this case, no transaction is done, i.e. max profit = 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= prices.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= prices[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        # memo = {}
        # def dp(day, hold, profit, k):
        #     res = 0
        #     if k == 0:
        #         return profit
        #     if day == len(prices):
        #         return profit
        #     print(day, profit, k)
        #     if (day, k) not in memo:
        #         for i in range(day, len(prices)):
        #             # print(i)
        #             if hold != -1:
        #                 if (prices[i] - prices[hold] > 0):
        #                     res = max(res, dp(i + 1, -1, profit + prices[i] - prices[hold], k - 1), dp(i + 1, i, profit, k))
        #             else:
        #                 res = max(res, dp(i + 1, i, profit, k), dp(i + 1, -1, profit, k))
        #         memo[(day, k)] = res
        #     return memo[(day, k)]
        # return dp(0, -1, 0, 2)

        # ^ wrong because we count the holding state multiple times eg in the for loop when we jump to ith day, we are techically "holding"

            # aggergate price
            memo = {}
            def dp(i, transactions_remaining, holding):
                if transactions_remaining == 0 or i == len(prices):
                    return 0
                if (i, transactions_remaining, holding) not in memo:
                    stalling = dp(i + 1, transactions_remaining, holding)
                    selling = 0
                    buying = 0
                    if holding:
                        selling = prices[i] + dp(i + 1, transactions_remaining - 1, 0)
                    else:
                        buying = -prices[i] + dp(i + 1, transactions_remaining, 1)
                    memo[(i, transactions_remaining, holding)] = max(stalling, selling, buying)
                return memo[(i, transactions_remaining, holding)]
            return dp(0, 2, 0)

            # @lru_cache(None)
            # def dp(i, transactions_remaining, holding):
            #     # Base case
            #     if transactions_remaining == 0 or i == len(prices): 
            #         return 0
                
            #     do_nothing = dp(i + 1, transactions_remaining, holding)
            #     do_something = 0

            #     if holding:
            #         # Sell stock
            #         do_something = prices[i] + dp(i + 1, transactions_remaining - 1, 0)
            #     else:
            #         # Buy stock
            #         do_something = -prices[i] + dp(i + 1, transactions_remaining, 1)

            #     # Recurrence relation
            #     return max(do_nothing, do_something)

            # return dp(0, 2, 0)



        
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