<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:

        n = len(prices)

        if n == 1:
            return 0
        
        if n == 2:
            return max(0, prices[1] - prices[0])
        
        if n == 3:
            return max(0, prices[2] - prices[0], prices[2] - prices[1], prices[1] - prices[0])

        dp = [[float('-inf') for _ in range(5)] for _ in range(n)]

        dp[0][0] = 0
        dp[0][1] = -prices[0]

        dp[2][3] = -prices[2]

        for i in range(1, n):
            dp[i][0] = dp[i - 1][0]
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] - prices[i])
            dp[i][2] = max(dp[i - 1][2], dp[i - 1][1] + prices[i])
            dp[i][3] = max(dp[i - 1][3], dp[i - 1][2] - prices[i], dp[i][3])
            dp[i][4] = max(dp[i - 1][4], dp[i - 1][3] + prices[i])
        
        return max(dp[-1])

        @lru_cache(None)
        def helper(idx, rem):

            if idx == n or rem == 0:
                return 0
            
            if rem & 1:
                return max(prices[idx] + helper(idx + 1, rem - 1), helper(idx + 1, rem))
            else:
                return max(-prices[idx] + helper(idx + 1, rem - 1), helper(idx + 1, rem))
        
        return helper(0, 4)

        
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