<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:
        '''dfs + 2 param to represent state'''

        # dp[(i, k)]
        # the max amt earned from piles[i:] given that we've picked k coins
        dp = {}

        def dfs(pile_idx, n_coin):
            # termination
            if pile_idx == len(piles) or n_coin == k:
                return 0

            # if subproblem has been solved 
            if (pile_idx, n_coin) in dp:
                return dp[(pile_idx, n_coin)]

            # not include current pile's coin
            dp[(pile_idx, n_coin)] = dfs(pile_idx + 1, n_coin)

            # include current pile's coin until
            # 1. all coins in current pile has been used up
            # 2. we've selected k coins
            total_val = 0
            for coin_idx, coin_val in enumerate(piles[pile_idx]):
                total_val += coin_val
                dp[(pile_idx, n_coin)] = max(
                    dp[(pile_idx, n_coin)],
                    total_val + dfs(pile_idx + 1, n_coin + coin_idx + 1)
                )

                # check if selected k coins -> no need to select more coins
                if n_coin + coin_idx + 1 == k:
                    break

            return dp[(pile_idx, n_coin)]
        
        return dfs(0, 0)

            

        
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