<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 f(int idx, int buy, int txn, vector<int>& prices, vector<vector<vector<int>>>&dp) {
    //     //bcase

    //     if(txn == 0 || idx == prices.size()) return 0;

    //     if(dp[idx][buy][txn] != -1) return dp[idx][buy][txn];
    //     if(buy)
    //                                                  //cannot buy further             //can buy further
    //         return dp[idx][buy][txn] = max(f(idx+1,0,txn,prices,dp) - prices[idx], 0 + f(idx+1,1,txn,prices,dp));

    //     else {
    //         //not buy i.e sell
    //                                                 //sell                      //not sell
    //         return dp[idx][buy][txn] = max(prices[idx] + f(idx+1,1,txn-1,prices,dp), 0 + f(idx+1,0,txn,prices,dp));
    //     }
    // }
    // int maxProfit(vector<int>& prices) {
    //     int n = prices.size();
    //     // vector<vector<int>>dp(n,vector<int>(2,-1));
    //     // Use a 3D dp array to include txn as a dimension
    //     vector<vector<vector<int>>> dp(n, vector<vector<int>>(2, vector<int>(3, -1)));
    //     return f(0,1,2,prices,dp);
    // }

    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        // vector<vector<int>>dp(n,vector<int>(2,-1));
        // Use a 3D dp array to include txn as a dimension
        vector<vector<vector<int>>> dp(n+1, vector<vector<int>>(2, vector<int>(3, 0)));
        
        // dp[n][0][0] = 0; // from recursion // APNI GALTIYA!!
        // dp[n][1][0] = 0;
        
        for(int i = 0;i<=n;i++) { //jab txn == 0 hojaye tabh
            for(int j=0;j<=1;j++) {
                dp[i][j][0] = 0;
            }
        }

        for(int i=0;i<=1;i++) { //jab idx == prices.size() hota wo condition ka tabulative form
            for(int j=0;j<=2;j++) {
                dp[n][i][j] = 0;
            }
        }

       for(int idx = n-1;idx>=0;idx--) {
            for(int buy = 1;buy>=0;buy--) {
                for(int txn = 2;txn>0;txn--) { //yaha >= nahi , sirf > chahiye tha coz base case me 0th value ko nipta dia
                    if(buy)
                                                            //cannot buy further             //can buy further
                        dp[idx][buy][txn] = max(dp[idx+1][!buy][txn] - prices[idx], 0 + dp[idx+1][buy][txn]);

                    else {
                    //not buy i.e sell
                                                            //sell                      //not sell
                        dp[idx][buy][txn] = max(prices[idx] + dp[idx+1][!buy][txn-1], 0 + dp[idx+1][buy][txn]);
                    }
                }
            }
        }
        return dp[0][1][2]; 
    }
};
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