<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:
        #best value when considering the first n piles, and taking k coins
        #involves taking no coins from the nth pile, 1 coin from the nth pile, ... k coins from the nth pile (or however many coins possible)
        #f(n, k) = max(piles[n][0] + f(n - 1, k - 1), sum(piles[n][0:1]) + f(n -1, k - 2), etc)

        memo = [[0] * (k + 1) for _ in range(len(piles))]

        for pileIndex, pile in enumerate(piles):
            for kVal in range(k+1):
                lastPileCoins = 0
                bestValue = 0
                if pileIndex > 0:
                    bestValue = memo[pileIndex - 1][kVal]
                for takeCoins in range(1, min(kVal + 1, len(pile) + 1)):
                    coinTaken = 0
                    if takeCoins <= len(pile):
                        coinTaken = pile[takeCoins - 1]
                    lastPileCoins += coinTaken
                    if pileIndex > 0:
                        bestValue = max(bestValue, lastPileCoins + memo[pileIndex - 1][kVal - takeCoins])
                    else:
                        bestValue = max(bestValue, lastPileCoins)
                memo[pileIndex][kVal] = bestValue
        return memo[-1][-1]
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