<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:
    int help(vector<vector<int>>& piles, int idx,int k,vector<vector<int>>& cache){
        if(idx<0 or k<0)return 0;
        if(cache[idx][k]!=-1)return cache[idx][k]; 
        int ans=help(piles,idx-1,k,cache);
        for(int i=0;i<piles[idx].size();i++){
            if((k-i-1)<0){continue;}
            int take=help(piles,idx-1,k-i-1,cache)+piles[idx][i];
            ans=max(ans,take);
        }
        return cache[idx][k]=ans;
    }
    int maxValueOfCoins(vector<vector<int>>& piles, int k) {
        int n=piles.size();
        for(auto &pile: piles){
            for(int i=1;i<pile.size();i++){
                pile[i]+=pile[i-1];
            }
        }
        vector<vector<int>>cache(n,vector<int>(k+2,-1));
        vector<vector<int>>dp(n,vector<int>(2001,0));
        for(int i=0;i<piles[0].size();i++){
            if(k-i-1<0){continue;}
            dp[0][i+1]=piles[0][i];
        }
        for(int i=1;i<n;i++){
            for(int j=0;j<=2000;j++){
                for(int f=0;f<piles[i].size();f++){
                    if(j-f-1<0){continue;}
                    dp[i][j]=max(dp[i][j],dp[i-1][j-f-1]+piles[i][f]);
                }
                dp[i][j]=max(dp[i][j],dp[i-1][j]);
            }
        }
        return dp[n-1][k];
    }
};
52006