<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 size = prices.size();
        int* c =  new int[size * size];
        int* b =  new int[size * (k+1)];
        int* d =  new int[size * (k+1)];
        int diff;
        if (size == 1) return 0;

        for(int i=0; i<size; ++i)
        {
            int curmax = 0;            
            for(int j=i+1; j<size; ++j)
            {
                diff = prices[j] - prices[i];
                
                if (diff > curmax)
                {
                    c[(i*size) + j] = diff;
                    curmax = diff;
                } 
                else c[(i*size) + j] = -10000;
            }
            b[(i*(k+1)) + 1] = curmax;
        }
        for(int i=size-2; i>0; --i)
        {
            if (b[(i+1)*(k+1) + 1] > b[i*(k+1) + 1]) b[i*(k+1) + 1] = b[(i+1)*(k+1) + 1];
        }
        
        
        for(int r=2; r<=k; ++r)
        {
            b[((size-1)*(k+1)) + r] = b[((size-1)*(k+1)) + r-1];
            for(int i=size-2; i>=0; --i)
            {
                if (b[i*(k+1) + r-1] > b[(i+1)*(k+1) + r]) b[i*(k+1) + r] = b[i*(k+1) + r-1];
                else b[(i*(k+1)) + r] = b[(i+1)*(k+1) + r];
                for(int j=i+1; j<size; ++j)
                {
                    int rightside = 0;
                    if (j+1 < size) rightside =b[((j+1)*(k+1)) + r-1];
                    int curprofit = c[(i*size) + j] + rightside;
                    //cout << i << " " << curprofit << "\n";
                    if (curprofit >  b[(i*(k+1)) + r])
                    {
                        b[(i*(k+1)) + r] = curprofit;
                    }
                }
            }
        }
        int absolutmax = 0;
        for(int i=0; i<size; ++i)
        {
            if (b[(i*(k+1)) + k] > absolutmax) absolutmax = b[(i*(k+1)) + k];
        }
        return absolutmax;
    }
};
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