<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 {
public:
int f(int index,int buy,int two,vector<int>& prices,vector<vector<vector<int>>>& dp)
{
    if(index==prices.size() || two==0) return 0;
   
    if(dp[index][buy][two]!=-1)
    {
        return dp[index][buy][two];
    }
    if(buy)
    {
        int profit1=-prices[index]+f(index+1,0,two,prices,dp);
        int profit2=0+f(index+1,1,two,prices,dp);
        return dp[index][buy][two]=max(profit1,profit2);
    }
    else{
        int profit1=prices[index]+f(index+1,1,two-1,prices,dp);
        int profit2=0+f(index+1,0,two,prices,dp);
        return dp[index][buy][two]=max(profit1,profit2);
    }
}
int f2(int index,int transaction,vector<int>& prices,vector<vector<int>>& dp2)
{
    if(index==prices.size()||transaction==4)
    {
        return 0;
    }
    if(dp2[index][transaction]!=-1)
    {
        return dp2[index][transaction];
    }
    if(transaction%2==0)
    {
        int profit1=-prices[index]+f2(index+1,transaction+1,prices,dp2);
        int profit2=0+f2(index+1,transaction,prices,dp2);
        return dp2[index][transaction]=max(profit1,profit2);
    }
    else{
        int profit1=prices[index]+f2(index+1,transaction+1,prices,dp2);
        int profit2=0+f2(index+1,transaction,prices,dp2);
        return dp2[index][transaction]=max(profit1,profit2);
    }
}
    int maxProfit(vector<int>& prices) {
        vector<vector<vector<int>>> dp(prices.size()+1,vector<vector<int>>(2,vector<int>(3,0)));
        vector<vector<int>> dp2(prices.size(),vector<int>(4,-1));
        // return f(0,1,2,prices,dp);
        return f2(0,0,prices,dp2);
        vector<vector<int>> front(2,vector<int>(3,0));
        vector<vector<int>> curr(2,vector<int>(3,0));
        for(int i=prices.size()-1;i>=0;i--)
        {
            for(int buy=0;buy<2;buy++)
            {
                for(int two=1;two<=2;two++)
                {
                    if(buy)
                    {
                        int profit1=-prices[i]+front[0][two];
                        int profit2=0+front[1][two];
                        curr[buy][two]=max(profit1,profit2);
                    }
                    else{
                        int profit1=prices[i]+front[1][two-1];
                        int profit2=0+front[0][two];
                        curr[buy][two]=max(profit1,profit2);
                    }
                }
            }
            front=curr;
        }
        return front[1][2];

    }
};
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