<p>There are <code>n</code> <strong>piles</strong> of coins on a table. Each pile consists of a <strong>positive number</strong> of coins of assorted denominations.</p>

<p>In one move, you can choose any coin on <strong>top</strong> of any pile, remove it, and add it to your wallet.</p>

<p>Given a list <code>piles</code>, where <code>piles[i]</code> is a list of integers denoting the composition of the <code>i<sup>th</sup></code> pile from <strong>top to bottom</strong>, and a positive integer <code>k</code>, return <em>the <strong>maximum total value</strong> of coins you can have in your wallet if you choose <strong>exactly</strong></em> <code>k</code> <em>coins optimally</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/11/09/e1.png" style="width: 600px; height: 243px;" />
<pre>
<strong>Input:</strong> piles = [[1,100,3],[7,8,9]], k = 2
<strong>Output:</strong> 101
<strong>Explanation:</strong>
The above diagram shows the different ways we can choose k coins.
The maximum total we can obtain is 101.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> piles = [[100],[100],[100],[100],[100],[100],[1,1,1,1,1,1,700]], k = 7
<strong>Output:</strong> 706
<strong>Explanation:
</strong>The maximum total can be obtained if we choose all coins from the last pile.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == piles.length</code></li>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>1 &lt;= piles[i][j] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= sum(piles[i].length) &lt;= 2000</code></li>
</ul>

class Solution:
    def maxValueOfCoins(self, piles: List[List[int]], k: int) -> int:
        n = len(piles)

    # #maximum value that you can get from ith index, i can go upto k-1
    #     def bs(pile):
    #         l = 0 
    #         r= len(pile)-1
    #         while l<=r:
    #             mid = (l+r)//2
    #             if pile[mid]>-1:
    #                 r=mid-1
    #             else:
    #                 l=mid+1
    #         return l





    #     def maxValue(i):
    #         if i==k: # you have picked k coins
    #             return 0
    #         # pick the coin from any of the piles
    #         ans = 0
    #         for pile in piles:
    #             j = bs(pile) #find first not picked coin
    #             if j!=len(pile):
    #                 coin = pile[j]
    #                 pile[j]=-1
    #                 ans = max(ans, maxValue(i+1)+coin)
    #                 pile[j]=coin
    #         return ans
    
    #     return maxValue(0)

        cache = {}

        def result(i, coins):
            if coins==0 or i ==n:
                return 0
            if (i, coins) in cache:
                return cache[(i, coins)]
            
            can_pick = min(coins, len(piles[i])) # can pick max these coins from pile i
            ans = 0
            cum_sum = 0
            for pick in range(0, can_pick+1):

                if pick>0:
                    cum_sum+=piles[i][pick-1]
                ans = max(ans, cum_sum+ result(i+1, coins-pick))
            cache[(i, coins)] = ans
            return ans

        return result(0, k)


                

        
        
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