<p>You are given an integer 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, and an integer <code>k</code>.</p>

<p>Find the maximum profit you can achieve. You may complete at most <code>k</code> transactions: i.e. you may buy at most <code>k</code> times and sell at most <code>k</code> times.</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> k = 2, prices = [2,4,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> k = 2, prices = [3,2,6,5,0,3]
<strong>Output:</strong> 7
<strong>Explanation:</strong> Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= 100</code></li>
	<li><code>1 &lt;= prices.length &lt;= 1000</code></li>
	<li><code>0 &lt;= prices[i] &lt;= 1000</code></li>
</ul>

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        
        int n = prices.size();
        
        int dp[ n+1 ][ k +1 ][ 2 ];

        for( int i = 0 ; i<=n ;i++ )
        {
            for( int j =0;j <=k; j++ )
            {
                dp[i][j][0] = -10000000;
                dp[i][j][1] = -10000000;
            }
        }
        dp[0][0][1] = 0;
        for( int i =0; i<n; i++ )
        {
            for( int j=0; j<= k ; j++ )
            {
                   
                    dp[i+1][j][0] = dp[i][j][0];
                    dp[i+1][j][1] = dp[i][j][1];
                
         if(j>0)
                dp[i+1][j][0] = max( dp[i+1][j][0] ,dp[i][j-1][1] - prices[i] ) ;           
                dp[i+1][j][1] = max( dp[i+1][j][1], prices[i] + dp[i][j][0]);

            }
        }
       int ans =0 ;
       for( int i=0;i<=k; i++)
       {
        ans = max(ans, dp[n][i][1]);
       } 
       for( int i = 0; i<=n; i++)
       {
        for(int j =0 ;j<=k; j++)
        {
           // cout<< dp[i][j][0]<<">"<<dp[i][j][1]<<" ";
        }
        //cout<<"\n";
       }
       return ans;
    }
};
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