<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 maxProfit(vector<int>& prices) {
    //     int n = prices.size();
    //     if(n<=1)
    //     {
    //       return 0;
    //     }
    //     if(n==2)
    //     {
    //         if(prices[0]< prices[1])
    //           return prices[1]- prices[0];
    //         return 0;  
    //     }
    //     vector<int> one_transection(n+1,0);
    //     int ans = 0;
    //     int mn = INT_MAX;
    //     one_transection[0] =0;
    //     one_transection[1] =0;
    //     mn = prices[0];
    //     for(int i=2;i<=n;i++)
    //     {
    //         one_transection[i] = max(one_transection[i-1],prices[i-1]-mn);
    //         mn = min(mn,prices[i-1]);
    //         ans = max(ans,one_transection[i]);
    //     }

    //     int dp[n+1][2];
    //     dp[0][0] = 0;
    //     dp[0][1] = INT_MIN;
       
        
      
    //     for(int i=1;i<n+1;i++)
    //     {
    //         dp[i][1]= max(dp[i-1][1],one_transection[i-1]- prices[i-1]);
    //         dp[i][0] = max(dp[i-1][0],dp[i-1][1]+prices[i-1]);
    //         ans = max(ans,dp[i][0]);
    //     }
                
    //      return ans;

    // }


  int maxProfit(vector<int>& prices) 
  {
      int n = prices.size();
      vector<int> one(n+1);
      one[0] = 0;
      one[1] =0;
      int temp = prices[0];
      for(int i=2;i<n+1;i++)
      {
        one[i] = max(prices[i-1]-temp,one[i-1]);
        temp = min(temp,prices[i-1]);
      }

      vector<vector<int>> dp(n+1,vector<int>(2));
      dp[0][0] = 0;
      dp[0][1] = INT_MIN;
      for(int i=1;i<n+1;i++)
      {
         dp[i][1] = max(one[i-1]-prices[i-1],dp[i-1][1]);
         dp[i][0] = max(dp[i-1][1]+prices[i-1],dp[i-1][0]);
      }
      return dp[n][0];
      
  }



















     
};
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