<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:
    long long fs(int n, int index, bool buy, int count, vector<int>& prices, vector<vector<vector<long long>>>& dp) {
        if (index == n || count == 0)
            return 0;
        
        long long profit = 0;
        
        if (dp[index][buy][count] != -1)
            return dp[index][buy][count];
        
        if (buy == false) {
            profit = max(fs(n, index + 1, false, count, prices, dp), 
                         fs(n, index + 1, true, count, prices, dp) - prices[index]);
        } else {
            profit = max(fs(n, index + 1, true, count, prices, dp), 
                         fs(n, index + 1, false, count - 1, prices, dp) + prices[index]);
        }
        
        return dp[index][buy][count] = profit;
    }

    int maxProfit(int k, vector<int>& prices) {
        int n = prices.size();

        cin.tie(NULL);
        ios_base::sync_with_stdio(false);

        vector<vector<vector<long long>>> dp(n + 1, vector<vector<long long>>(2, vector<long long>(k + 1, -1)));
        return fs(n, 0, false, k, prices, dp);
    }
};

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