<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 t[100001][2][3];
    int n;
    int solve(vector<int>& A, int in , int n , bool flag ,  int k){
        if(in>=n){
            return 0;
        }
        if(k==0){
            return 0;
        }
        if(t[in][flag][k]!=-1){
            return t[in][flag][k];
        }
        int profit = 0;
        if(flag){
            int take = solve(A,in+1,n,false,k)-A[in];
            int not_take = solve(A,in+1,n,true,k);
            profit = max(profit,max(take,not_take));
        }
        else{
            int take = A[in] + solve(A,in+1,n,true,k-1);
            int not_take = solve(A,in+1,n,false,k);
            profit = max(profit,max(take,not_take));
        }
        return t[in][flag][k] = profit;
    }
    int maxProfit(vector<int>& A) {
        n = A.size();
        int dp[(int)(1e5+1)][2][3];
        memset(dp,0,sizeof(dp));
        for(int in = n-1;in>=0;in--){
            for(int flag = 0;flag<=1;flag++){
                for(int k=1;k<=2;k++){
                    int profit = 0;
                    if(flag){
                        int take = dp[in+1][0][k] - A[in];
                        int not_take = dp[in+1][1][k];
                        profit = max(profit,max(take,not_take));
                    }
                    else{
                        int take = A[in] + dp[in+1][1][k-1];
                        int not_take = dp[in+1][0][k];
                        profit = max(profit,max(take,not_take));
                    }
                    dp[in][flag][k] = profit;
                }
            }
        }
        return dp[0][1][2];
    }
};
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