<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>

#include "bits/stdc++.h"
class Solution {
public:
 
 int func(int index,int k,vector<int> &prices,bool flag,vector<vector<vector<int>>> &dp){
    if(k<=0) return 0;
    if(index<0){
        return 0;
    }
    if(dp[flag][index][k]!=-1) return dp[flag][index][k];
    if(flag){
        int take=-prices[index] +func(index-1,k,prices,0,dp);
        int nottake=func(index-1,k,prices,1,dp);
        return dp[flag][index][k]=max(take,nottake);
    }
    else {
        int take= (prices[index]+func(index-1,k-1,prices,1,dp));
        int nottake=func(index-1,k,prices,0,dp);
        return dp[flag][index][k]=max(take,nottake);
    }
 }




    int maxProfit( vector<int>& prices) {
      vector<int> sex=prices;
      reverse(sex.begin(),sex.end());
      vector<vector<vector<int>>> dp(2,vector<vector<int>> (sex.size(),vector<int>(2+1,-1)));
     int ans=func(prices.size()-1,2,sex,1,dp);
     return ans;
    }
};
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