<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 arr[105][2][1005];
    int help(vector<int>& prices,int k,bool can, int idx){
        //if((! can) && idx== prices.size()) return INT_MIN;// if at end and still holding on a stock
        if (k==0 || (idx==prices.size())) return 0;// if can is false we need to sell first 
        if (arr[k][can][idx] != -1) return arr[k][can][idx] ;
        int buy = INT_MIN;
        int sell= INT_MIN;
        int pass =INT_MIN;
        int res=INT_MIN;
        if (can){
            //cout << " bought " << prices[idx]<< "  ";
            buy = (-prices[idx]) +help(prices,k,false,idx+1);//buy this 
            //no k-1 here as k==0 may happen without selling it so do k-1 after sellig it
        }
        if (!can){//sell
            //cout << " sold " << prices[idx] << "  ";
            sell= ( prices[idx])+ help(prices,k-1,true,idx+1);
        }
        //simple pass 
        pass =help(prices,k,can,idx+1);
        res = max({pass ,buy ,sell});
        return arr[k][can][idx] =res;
    }
    int maxProfit(int k, vector<int>& prices) {
        memset ( arr, -1, sizeof(arr));
        return help(prices,k,true,0);//idx changes k changes and bool changes 
    }
};
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