<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 n = prices.size();
        vector<vector<vector<long long>>> memo(2, vector<vector<long long>>(k + 1, vector<long long>(n + 1)));
        for (int index = 0 ; index <= n ; index++)
            memo[0][0][index] = memo[1][0][index] = 0;
        for (int i = 0 ;i < 2 ; i++)
            for (int j = 0; j < k; j++)
                memo[i][j][n] = 0;
                for (int index = n - 1 ; index >= 0 ; index--) {
        for (int canBuy = 1 ; canBuy >= 0 ; canBuy--) {
            for (int transactions = k ; transactions >= 0; transactions--) {
                    long long sign = canBuy ? -1 : 1;
                    long long ans = memo[canBuy][transactions][index + 1];
                    if (transactions >= (canBuy ^ 1)) {
                        ans = max(ans, sign * prices[index] + memo[canBuy ^ 1][transactions - (canBuy ^ 1)][index + 1]);
                    }
                    memo[canBuy][transactions][index] = ans;
                }
            }
        }
        return memo[1][k][0];
    }
};
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