<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<int>>>dp(n+1,vector<vector<int>>(2,vector<int>(2*k,0)));
        for(int ind=n-1;ind>=0;ind--){
            for(int buy=0;buy<=1;buy++){
                for(int cap=1;cap<2*k;cap++){
                     if(buy){
                        int a=-prices[ind]+dp[ind+1][0][cap];
                        int b=dp[ind+1][1][cap];
                        dp[ind][buy][cap]=max(a,b);
                     }
                     else{
                        int a=prices[ind]+dp[ind+1][1][cap-1];
                        int b=dp[ind+1][0][cap];
                        dp[ind][buy][cap]=max(a,b);
                    }
                }
               }
            }
        
       
        return dp[0][1][k];
    }
    
};
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