<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:


vector<vector<vector<int>>> dp;
int rec(vector<int>& prices, int k, int idx , bool tobuy){

    if(idx==prices.size() || k==0){
        return 0;
    }

    if(dp[k][idx][tobuy] != -1){
        return dp[k][idx][tobuy];
    }

    if(tobuy){
        int a = - prices[idx] + rec(prices,k ,idx+1,!tobuy);
        int b = rec(prices,k,idx+1,tobuy);
        return dp[k][idx][tobuy] =  max(a,b);
    }else{
        int a = prices[idx] + rec(prices,k-1,idx+1,!tobuy);
        int b = rec(prices,k,idx+1,tobuy);
        return dp[k][idx][tobuy] =  max(a,b);
    }

}

    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        int k =2;

        //memo
        // dp.resize(k+1,vector<vector<int>>(n+1,vector<int>(2,-1)));
        // return rec(prices,k,0,1);

        //tabulo
        dp.resize(k+1,vector<vector<int>>(n+1,vector<int>(2)));
        
        for(int i = 0 ; i <= k ; i++){
            dp[i][n][0] = 0;
            dp[i][n][1] = 0;
        }
        for(int i = 0 ; i <= n ; i++){
            dp[0][i][0] = 0;
            dp[0][i][1] = 0;
        }

        for(int i =1 ; i<= k ; i++){
            for(int j = n-1 ; j>=0 ; j--){
                for(int tobuy = 0 ; tobuy< 2 ; tobuy++){
                    if(tobuy==1){
                        int a = - prices[j] + dp[i][j+1][!tobuy];
                        int b = dp[i][j+1][tobuy];
                        dp[i][j][tobuy] =  max(a,b);
                    }else{
                        int a = prices[j] + dp[i-1][j+1][!tobuy];
                        int b = dp[i][j+1][tobuy];
                        dp[i][j][tobuy] =  max(a,b);
                    }
                }
            }
        }

        return dp[k][0][1];
            

    }
};
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