<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) {
        vector<vector<int>> dp(prices.size(), vector<int>(k + 1, 0));
        
        for(int i = 0; i < prices.size(); i++) {
            int b = i;

            for(int j = i - 1; j >= 0; j--) {
                for(int p = 0; p < k; p++)
                    dp[i][p + 1] = max(dp[i][p + 1], dp[j][p] + prices[i] - prices[b]);

                if(prices[j] < prices[b])
                    b = j;
            }

            for(int p = 0; p < k; p++)
                dp[i][p + 1] = max(dp[i][p + 1], prices[i] - prices[b]);
        }

        int ans = 0;
        for(int i = 0; i < prices.size(); i++)
            for(int j = 0; j <= k; j++)
                ans = max(ans, dp[i][j]);

        return ans;
    }
};
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