<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)
        dp = [[0 for _ in range(k+1)] for _ in range(n+1)]
        for p in range(1, n+1):
            for q in range(1, k+1):
                curr_sum = 0
                for curr_coin in range(min(len(piles[p-1]), q)+1):
                    prev_coin = q-curr_coin
                    if curr_coin > 0:
                        curr_sum += piles[p-1][curr_coin-1]
                    dp[p][q] = max(dp[p][q], curr_sum+dp[p-1][prev_coin])
        return dp[-1][k]


        # n = len(piles)
        # # dp[i][j] represents the max value when taking j coins from the leftmost i piles
        # dp = [[0]*(k+1) for _ in range(n+1)]
        # for pile in range(1, n+1):
        #     for coin in range(1, k+1):
        #         curr_sum = 0
        #         # iterate prev_coin reversely
        #         for prev_coin in range(coin, max(0, coin-len(piles[pile-1]))-1, -1):
        #             curr_coin = coin-prev_coin
        #             if curr_coin > 0:
        #                 # add the top coin of the current pile
        #                 curr_sum += piles[pile-1][curr_coin-1]
        #             dp[pile][coin] = max(dp[pile][coin], dp[pile-1][prev_coin]+curr_sum)
        # return dp[n][k]
        
5608