<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>

'''
# Discern the problem
- maximum total value
- chose a coint from a pile, deprive the chane to chose other coins in other piles

# Base Case
- i >= len(piels): 0 value
- k <= 0: 0 value

# Rrcurrence relation
## Actions
- skip the pile
    - skip = dp(i+1, remain)
- pick from the pile 
    - action = max(dp(i+1, remain - (j+1)) + acc(pile[i][j])) , where j = (0, min(len(pile[i]), remain)]
## Conclusion
- dp(i, j) = max(skip, action)
'''

class Solution:
    def maxValueOfCoins(self, piles: List[List[int]], k: int) -> int:
        @cache
        def top_down(i, remain):
            if i >= len(piles) or remain <= 0:
                return 0
            
            skip = top_down(i + 1, remain)
            acc = 0
            action = 0
            boundary = min(len(piles[i]), remain)
            for j in range(boundary):
                acc += piles[i][j]
                action = max(action, top_down(i + 1, remain - (j + 1)) + acc)
            return max(action, skip)

        def bottom_up():
            n = len(piles)
            dp = [[0] * (k + 1) for _ in range(n + 1)]
            for i in range(n - 1, -1, -1):
                for r in range(1, k + 1):
                    skip = dp[i+1][r]
                    acc = 0
                    boundary = min(len(piles[i]), r)
                    action = 0
                    for j in range(boundary):
                        acc += piles[i][j]
                        action = max(action, dp[i + 1][r - (j + 1)] + acc)
                    dp[i][r] = max(skip, action)
            return dp[0][k]                

        # return top_down(0, k)
        return bottom_up()
4357