<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 {
public:
    vector<vector<int>> dp; 
    
    int solve(int ind, int k, vector<vector<int>> &p) {
        if (ind==p.size())
            return 0; 
        
        if (dp[ind][k] != -1)
            return dp[ind][k];  
        
        dp[ind][k] = solve(ind+1, k, p); 
        int till = min(k, int(p[ind].size())); 
        
        for (int i=1; i<=till; i++) {
            dp[ind][k] = max(dp[ind][k], p[ind][i-1]+solve(ind+1, k-i, p));     
        }
        
        return dp[ind][k]; 
    }
    
    int maxValueOfCoins(vector<vector<int>>& piles, int k) {
        dp.resize(piles.size(),vector<int>(k+1, -1)); 
        for (auto &i: piles) {
            for (int j=1; j<i.size(); j++)
                i[j] += i[j-1]; 
        }
        
        return solve(0, k, piles); 
        
    }
};
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