<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>

#include <bits/stdc++.h>
class Solution {
public:
    int solve(int i, vector<vector<int>>& piles, int k,
              vector<vector<int>>& dp) {
        if (i == piles.size()) {
            return 0;
        }
        if (k == 0) {
            return 0;
        }

        if (dp[i][k] != -1) {
            return dp[i][k];
        }
        int not_take = solve(i + 1, piles, k, dp);

        int take = 0;

        int sum = 0;
        int mini = min((int)piles[i].size(), k);
        for (int j = 0; j < mini; j++) {
            sum += piles[i][j];

            take = max(take, sum + solve(i + 1, piles, k - (j + 1), dp));
        }

        return dp[i][k] = max(take, not_take);
    }

    int solvetab(vector<vector<int>>& piles, int k) {
        int n = piles.size();
        vector<vector<int>> dp(n + 1,
                               vector<int>(k + 1, 0)); // Initialize dp with 0

        for (int i = n - 1; i >= 0; i--) { // Loop from n-1 to 0
            for (int K = 0; K <= k; K++) {
                int not_take = dp[i + 1][K]; // Skipping the current pile

                int take = 0;
                int sum = 0;
                int mini =
                    min((int)piles[i].size(),
                        K); // Number of items we can take from current pile
                for (int j = 0; j < mini; j++) {
                    sum += piles[i][j]; // Accumulate the value of items taken
                    if (K >= j + 1)
                        take = max(
                            take,
                            sum + dp[i + 1]
                                    [K - (j + 1)]); // Max value after taking j
                                                    // items from pile i
                }

                dp[i][K] = max(take, not_take); // Best option: take or not take
            }
        }
        return dp[0][k]; // Result stored in dp[0][k]
    }
    int maxValueOfCoins(vector<vector<int>>& piles, int k) {
        int n = piles.size();
        vector<vector<int>> dp(n, vector<int>(k + 1, -1));
        return solvetab(piles, k);
    }
};
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