<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>

const int N = 1e3+5;
int dp[N][N][2];

class Solution {
public:
    int maxProfit(int L, vector<int>& a) {
        int n = a.size();
        for (int i = 0; i <= n+1; i++) {
            for (int j = 0; j <= n+1; j++) {
                for (int k = 0; k < 2; k++) {
                    dp[i][j][k] = 0;
                }
            }
        }

        for (int limit = 1; limit <= L; limit++) {
            for (int start = n; start >= 1; start--) {
                int minBuyPrice = a[start-1];
                for (int i = start; i <= n; i++) {
                    dp[start][n][limit%2] = max(
                        dp[start][n][limit%2],
                        a[i-1] - minBuyPrice + dp[i+1][n][(limit-1+2)%2]
                    );
 
                    minBuyPrice = min(
                        minBuyPrice,
                        a[i-1]
                    );
                }
            }
        }

        return dp[1][n][L%2];
    }
};
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