<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();
        if (n == 0) return 0;  // If no prices are available, no profit can be made
        
        vector<vector<long>> dp(n, vector<long>(2 * k + 1, -1));  // Initialize dp table with -1
        return maximumProfit(prices, n, k);  // Calculate the maximum profit
    }
    
private:
    // Helper function for calculating maximum profit
    long f(int ind, int k, vector<int> &prices, vector<vector<long>> &dp) {
        if (ind == prices.size() || k == 0) return 0;  // Base case: no more days or transactions left
        if (dp[ind][k] != -1) return dp[ind][k];  // Return memoized result if available
        
        long profit = 0;
        
        // Buy case
        if (k % 2 == 0) {
            // We can either buy at the current index or skip it
            profit = max(-prices[ind] + f(ind + 1, k - 1, prices, dp), f(ind + 1, k, prices, dp));
        }
        // Sell case
        else {
            // We can either sell at the current index or skip it
            profit = max(prices[ind] + f(ind + 1, k - 1, prices, dp), f(ind + 1, k, prices, dp));
        }
        
        return dp[ind][k] = profit;  // Store the result in dp table
    }
    
    // Function to calculate the maximum profit
    long maximumProfit(vector<int> &prices, int n, int k) {
        vector<vector<long>> dp(n, vector<long>(2 * k + 1, -1));  // Initialize dp table with -1
        return f(0, 2 * k, prices, dp);  // Start from day 0 with 2*k (buy and sell pairs) transactions
    }
};

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