<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 stocks(int i, int limit , int buy ,vector<int>& prices , vector<vector<vector<long>>>& dp){

        if(i == prices.size() || limit==0) return 0;
        if(dp[i][buy][limit] != -1) return dp[i][buy][limit];

        long profit = 0;
        
            if(buy == false){
                
                int buy = stocks(i+1,limit,true,prices,dp)-prices[i];
                int notbuy = stocks(i+1,limit,false,prices,dp);

                profit = max(buy,notbuy);
            }
            else{
                int sell = stocks(i+1,limit-1,false,prices,dp)+prices[i];
                int notsell = stocks(i+1,limit,true,prices,dp);

                profit = max(sell,notsell);
            }
        return dp[i][buy][limit] = profit;
        }
    
    int maxProfit(vector<int>& prices) {
        
        int n = prices.size();
        vector<vector<vector<long>>> dp(n+1, vector<vector<long>>(2, vector<long>(3, 0)));

        for(int i=n-1 ; i>=0 ; i--)
        {
            for(int limit = 1 ; limit<3 ; limit++)
            
            {
                for(int buy=0 ; buy<2 ; buy++){

                    long profit = 0;
                
                    if(buy == 0){
                        
                        int buy = dp[i+1][1][limit]-prices[i];
                        int notbuy = dp[i+1][0][limit];

                        profit = max(buy,notbuy);
                    }
                    else{
                        int sell = dp[i+1][0][limit-1]+prices[i];
                        int notsell = dp[i+1][1][limit];

                        profit = max(sell,notsell);
                    }
                    dp[i][buy][limit] = profit;
                }
            }
        }
        return dp[0][0][2];
    }
};
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