<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>

"""
[1,5,4,9,8,12,1,6]


[
    [8]
    [9]
]
 
you can store the take 
"""
class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        prices2 = []
        l, r = 0, 1

        while l < len(prices) and r < len(prices):
            while (l < len(prices) - 1) and prices[l] > prices[l + 1]:
                l += 1
            r = l + 1
            prices2.append(prices[l])
            while (r < len(prices) - 1) and prices[r] < prices[r + 1]:
                r += 1
            if r < len(prices):
                prices2.append(prices[r])
            l = r + 1
        
        cache = {}
        def search(idx, buying, t):
            if idx == len(prices2) or t == 2:
                return 0
            
            if (idx, buying, t) in cache:
                return cache[(idx, buying, t)]
            
            if buying:
                result = max(
                    search(idx + 1, not buying, t) - prices2[idx],
                    search(idx + 1, buying, t),
                )
            else:
                result = max(
                    search(idx + 1, not buying, t + 1) + prices2[idx],
                    search(idx + 1, buying, t),
                )
            
            cache[(idx, buying, t)] = result
            return result
        
        return search(0, True, 0)

        
        # dp = [[], []]
        # min_val = sys.maxsize
        # for idx in range(1, len(prices2), 2):
        #     current_diff = prices[idx] - prices[idx - 1]
        #     min_val = 
        #     dp[0] = max(dp[0], prices[idx] - min_val)
        #     dp[1] = max(dp[1], dp[0] + current_diff)

        
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