<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 {
     int findMaxProfit(vector<int>&prices,bool bought,int index,int k)
    {
        if(index==prices.size() || k==0)return 0;
        //we can either buy or sell or both
        if(bought==false)//we haven't bought anything
        {
           return max( findMaxProfit(prices,false,index+1,k),findMaxProfit(prices,true,index+1,k)-prices[index]);
            
        }
    
        //we already bought
        return max(findMaxProfit(prices,false,index+1,k-1)+prices[index],findMaxProfit(prices,true,index+1,k));
        
        
    }
public:
    int maxProfit(int k, vector<int>& prices) {
      //  return findMaxProfit(prices,false,0,k);
      int n=prices.size();
      //lets first make a 3d dp then optimize it to a simple 2d one
      //vector<vector<vector<int>>>dp(n+1,vector<vector<int>>(k+1,vector<int>(2,0)));
      vector<vector<int>>next(k+1,vector<int>(2,0));
      vector<vector<int>>cur(k+1,vector<int>(2,0));
      
      
      for(int i=n;i>=0;i--)
      {
        for(int j=0;j<=k;j++)
        {
            if(i==n || j==0)
            {
                next[j][0]=0;
                next[j][1]=0;
                continue;
            }
            cur[j][0]=max( next[j][0],next[j][1]-prices[i]);
            cur[j][1]=max(next[j-1][0]+prices[i],next[j][1]);

            next=cur;
            
        }
      }
      //we are using 0 and not one because we are sending 0 in the beginnig ie bough=false;
      return next[k][0];
    }
};
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