<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:
    // dp[index][ith] -> max 
    // index=0, return 0, 1 based indexing 
    // this is an interesting problem
    // alright bitch

    int rec(int index,int k,vector<int>&prices,vector<vector<int>>&dp){
        if(index<0){
            return 0;
        }
        if(k==0){
            return 0;
        }
        if(dp[index][k]!=-1){
            return dp[index][k];
        }
        int ans = rec(index-1,k,prices,dp);
        int mn = 1001;
        for(int i=index-1;i>=0;i--){
             mn = min(mn,prices[i]);
            ans = max(ans,rec(i-1,k-1,prices,dp) + max(0,prices[index]-prices[i]));
        }
        return dp[index][k] = ans;
    }

    int maxProfit(int k, vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>>dp(n,vector<int>(k+3,-1));
        return rec(n-1,k,prices,dp);
    }
};
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