<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>

import math
class Solution:
    def maxValueOfCoins(self, piles: List[List[int]], k: int) -> int:
        books=[0 for i in range(len(piles))]
        
        dp=[[-1 for i in range(k+1)] for i in range(len(piles))]

        for i in range(len(piles)):

            for j in range(1,len(piles[i])):
                piles[i][j]+=piles[i][j-1]

            if(i==0):
                books[i]=len(piles[i])
            else:
                books[i]=books[i-1]+len(piles[i])
            
        
        def dps(index,size):
            
            if(size<=0):
                return 0

            if(index==0):
                if(size<=len(piles[0])):
                    return piles[0][size-1]
                else:
                    return -math.inf
            
            if(dp[index][size]!=-1):
                return dp[index][size]
            
            maxa=0
            for i in range(len(piles[index])+1):
                if(i>size):
                    break
                if(size-i<=books[index-1]):
                    if(i==0):
                        maxa=max(maxa,dps(index-1,size))
                    else:
                        maxa=max(maxa,dps(index-1,size-i)+piles[index][i-1])
              
            
            dp[index][size]=maxa
            print(maxa,index,size)
            return maxa
        return dps(len(piles)-1,k)





        
1285