<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 rcrsn(vector<int>& prices,vector<vector<vector<int>>>& dp,int i,int buy,int trans,int n){
        if(i==n || trans==0) return 0;
        if(dp[trans][buy][i]!=-1) return dp[trans][buy][i];
        if(!buy){
            dp[trans][buy][i]= max(-prices[i]+rcrsn(prices,dp,i+1,1,trans,n),rcrsn(prices,dp,i+1,0,trans,n));
        }
        else{
            dp[trans][buy][i]= max(prices[i]+rcrsn(prices,dp,i+1,0,trans-1,n),rcrsn(prices,dp,i+1,1,trans,n));
        }
        return dp[trans][buy][i];
    }
    int maxProfit(vector<int>& prices) {
        int n=prices.size();
        // vector<vector<vector<int>>> dp(3,vector<vector<int>>(2,vector<int>(n+1,-1)));
        // return rcrsn(prices,dp,0,false,2,n);
        //using tabulation
        vector<vector<vector<int>>> dp(3,vector<vector<int>>(2,vector<int>(n+1,0)));
        // for(int trans=0;trans<3;trans++){
        //     for(int buy=0;buy<2;buy++){
        //         dp[trans][buy][n]=0;
        //     }
        // }
        // for(int buy=0;buy<2;buy++){
        //     for(int i=0;i<n;i++){
        //         dp[0][buy][i]=0;
        //     }
        // }
                for(int i=n-1;i>=0;i--){
            for(int buy=0;buy<2;buy++){
        for(int trans=1;trans<3;trans++){
                    if(!buy){ dp[trans][buy][i]= max(-prices[i]+dp[trans][1][i+1],dp[trans][0][i+1]);
                    }
                    else{ dp[trans][buy][i]= max(prices[i]+dp[trans-1][0][i+1],dp[trans][1][i+1]);
                    }
                }
            }
        }
        return dp[2][0][0];
    }
};
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