<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, const vector<int>& P)
    {
        int n = P.size();
        vector<vector<int>> F(k, vector<int>(n));
        vector<vector<int>> M(k, vector<int>(n));
        M[0][n - 1] = P[n - 1];
        for (int j = n - 2; j >= 0; --j)
        {
            M[0][j] = max(P[j], M[0][j + 1]);
        }
        for (int j = n - 2; j >= 0; --j)
        {
            F[0][j] = max(F[0][j + 1], M[0][j + 1] - P[j]);
        }
        for (int i = 1; i < k; ++i)
        {
            for (int j = n - 3; j >= 0; --j)
            {
                M[i][j] = max(M[i][j + 1], P[j + 1] + F[i - 1][j + 2]);
            }
            for (int j = n - 3; j >= 0; --j)
            {
                F[i][j] = max(max(F[i - 1][j], F[i][j + 1]), M[i][j] - P[j]);
            }
        }
        return F[k - 1][0];
    }
};
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