<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:
        from functools import lru_cache
        min_prices = [float("inf")] * len(prices) # min price from [0, i]
        min_so_far = float("inf")
        for i, p in enumerate(prices):
            min_so_far = min(min_so_far, p)
            min_prices[i] = min_so_far

        max_profit = [0] * len(prices) # max profit when sell at j
        for j in range(1, len(prices)):
            max_profit[j] = max(prices[j] - min_prices[j - 1], 0)
        max_profit_1 = [0] * len(prices)
        max_profit_ = 0
        for j in range(1, len(prices)):
            max_profit_ = max(max_profit_, max_profit[j])
            max_profit_1[j] = max_profit_

        @lru_cache(None)
        def max_profit_2nd_sell_at_(j):
            if j < 3:
                return max_profit_1[j]
            curr_max = prices[j] - prices[j - 1] + max_profit_1[j - 2]
            prev_max = max_profit_2nd_sell_at_(j - 1) + (prices[j] - prices[j - 1])
            return max(curr_max, prev_max)


        @lru_cache(None)
        def helper(j, k):
            if j <= 0:
                return 0
            # if k == 0:
            #     return 0
            if k == 1:
                return max_profit_1[j]
            if k == 2:
                profit_not_sell_at_j = helper(j - 1, k)
                profit_sell_at_j = max_profit_2nd_sell_at_(j)
                return max([profit_not_sell_at_j, profit_sell_at_j])
        return helper(len(prices) - 1, 2)
                            

        
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