<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 solve(int ind,int trans,vector<int>&prices,vector<vector<int>>&dp){
        if(ind==prices.size() || trans==4) return 0;
        if(dp[ind][trans]!=-1) return dp[ind][trans];
        int profit=0;
        if(trans%2==0) profit=max(-prices[ind]+solve(ind+1,trans+1,prices,dp),solve(ind+1,trans,prices,dp));
        else profit=max(prices[ind]+solve(ind+1,trans+1,prices,dp),solve(ind+1,trans,prices,dp));
        return dp[ind][trans]=profit;
    }
    int maxProfit(vector<int>& prices) {
        int n=prices.size();
        if(n==0) return 0;
        vector<vector<int>>dp(n+1,vector<int>(5,-1));
        return solve(0,0,prices,dp);
        
    }

//Space Optimization

// vector<vector<int>>after(2,vector<int>(3,0));
//         vector<vector<int>>curr(2,vector<int>(3,0));

//         // for(int i=0;i<n;i++) for(int j=0;j<2;j++) dp[i][j][0]=0;
//         // for(int i=0;i<2;i++) for(int j=0;j<3;j++) dp[n-1][i][j]=0;
//         //No need of base casees as already initialised as 0
//         for(int i=n-1;i>=0;i--){
//             for(int b=0;b<2;b++){
//                 for(int c=1;c<3;c++){
//                     if(b==1) curr[b][c]=max(-prices[i]+after[0][c],after[1][c]);
//                     else curr[b][c]=max(prices[i]+after[1][c-1],after[0][c]);
//                 }
//             }
//             after=curr;
//         }
//         return after[1][2];



//Tabulation

// int n=prices.size();
//         if(n==0) return 0;
//         vector<vector<vector<int>>>dp(n+1,vector<vector<int>>(2,vector<int>(3,0)));
//         // for(int i=0;i<n;i++) for(int j=0;j<2;j++) dp[i][j][0]=0;
//         // for(int i=0;i<2;i++) for(int j=0;j<3;j++) dp[n-1][i][j]=0;
//         //No need of base casees as already initialised as 0
//         for(int i=n-1;i>=0;i--) for(int b=0;b<2;b++) for(int c=1;c<3;c++){
//             if(b==1) dp[i][b][c]=max(-prices[i]+dp[i+1][0][c],dp[i+1][1][c]);
//             else dp[i][b][c]=max(prices[i]+dp[i+1][1][c-1],dp[i+1][0][c]);
//         }
//         return dp[0][1][2];


//Memoization

// int solve(int ind,int buy,int cap,vector<int>&prices,vector<vector<vector<int>>>&dp){
//         if(ind==prices.size()) return 0;
//         if(cap==0) return 0;
//         if(dp[ind][buy][cap]!=-1) return dp[ind][buy][cap];
//         int profit=0;
//         if(buy){
//             profit=max(-prices[ind]+solve(ind+1,0,cap,prices,dp),solve(ind+1,1,cap,prices,dp));
//         }
//         else{
//             profit=max(prices[ind]+solve(ind+1,1,cap-1,prices,dp),solve(ind+1,0,cap,prices,dp));
//         }
//         return dp[ind][buy][cap]=profit;
//     }
//     int maxProfit(vector<int>& prices) {
//         int n=prices.size();
//         vector<vector<vector<int>>>dp(n,vector<vector<int>>(2,vector<int>(3,-1)));
//         return solve(0,1,2,prices,dp);        
//     }






};
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