<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 kk;
vector<int>v;
int dp[1000+1][205];
    int rec(int id,int rem){
        if(id==v.size())return 0;
        if(dp[id][rem]!=-1)return dp[id][rem];
        int ans=0;
        ans=max(ans,rec(id+1,rem));
        if(rem<kk*2)ans=max(ans,v[id]*(rem%2? 1:-1)+rec(id+1,rem+1));
        return dp[id][rem]=ans;
    }
    int maxProfit(int k, vector<int>& prices) {
        kk=k;
        for(int i=0;i<=prices.size();i++){
            for(int j=0;j<=k*2;j++)dp[i][j]=-1;
        }
        for(int it:prices)v.push_back(it);
        return rec(0,0);
    }
};
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