<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(object):
    def maxValueOfCoins(self, piles, k):
        """
        :type piles: List[List[int]]
        :type k: int
        :rtype: int
        """
        # we first skip all nodes to start from last assume i have 4 piles 
        # (try last piles , then try 3rd and last pile , etc)
        # with base case if i ==n or k ==0 return 0
        # in for loop i try all possible coins in cur pile to last pile in limit with k
        # cur_val save the value of cur try , ans res save max res of all tries
        cache ={}
        def dfs(i,k ):
            if (i,k) in cache:
                return cache[(i,k)]
            if k==0 or i == len(piles):
                return 0
            # skip piles until reach end , then try all possible from last to first 
            # dfs(i+1 , k) 
            res , cur_val = dfs(i+1 , k),0
            for j in range(min(k , len(piles[i]))):
                cur_val += piles[i][j]
                # print(cur_val ,res )
                res = max(res , cur_val + dfs(i+1 , k-j-1))
            cache[(i,k)]=res
            return cache[(i,k)]
        return dfs(0,k)
4698