<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(vector<int> prices,int i,int k,int buy,auto dp){
    if(k==0)return 0;
    if(i==prices.size())return 0;
    if(dp[i][k][buy]!= -1)return dp[i][k][buy];
    int profit=0;
    if(buy){
      profit=  max(f(prices,i+1,k,1,dp),-prices[i]+f(prices,i+1,k,0,dp));
    }
    else{
          profit=  max(f(prices,i+1,k,0,dp),prices[i]+f(prices,i+1,k-1,1,dp));
    }
    return dp[i][k][buy]=profit;
}
    int maxProfit(vector<int>& prices) {
        int n=prices.size();
        int p[n+1][3][2];
// memset(p,-1,(n+1)*6);
int next[3][2],curr[3][2];
for(int i=0;i<n+1;i++){
    for(int j=0;j<3;j++){
        for(int k=0;k<2;k++)p[i][j][k]=0;
    }
}
cout<<p[n-1][2][1];
                vector<vector<int>> dp(n+1,vector<int> (3,-1));
            
             for(int i=n-1;i>=0;i--){
                for(int j=1;j<3;j++){
                    for(int buy =0;buy<2;buy++){
                         int profit=0;
    if(buy){
      profit=  max(p[i+1][j][1],-prices[i]+p[i+1][j][0]);
    }
    else{
          profit=  max(p[i+1][j][0],prices[i]+p[i+1][j-1][1]);
    }
 p[i][j][buy]=profit;
                    }
                }
             }
             return p[0][2][1];
        return f(prices,0,2,1,p);
    }
};
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