# K Stack Pops - Google Top Interview Questions

### Problem Statement :

```You are given two-dimensional list of integers stacks and an integer k.

Assuming each list in stacks represents a stack, return the maximum possible sum that can be achieved from popping off exactly k elements from any combination of the stacks.

Constraints

n ≤ 500 where n is the number of rows in stacks.

m ≤ 200 where m is the maximum number of elements in a stack.

k ≤ 100

Example 1

Input

stacks = [

[100, -3, -10],

,

[4, 5, 6]

]

k = 4

Output

93

Explanation

We can pop off all 3 elements from the first stack and pop the last element of the last stack to get -10
+ -3 + 100 + 6.

Example 2
Input

stacks = [

[1, 2, 4],

[2, 3, 5],

[3, 0, 6]

]

k = 3

Output

15

Explanation

Pop off the 4, 5 and 6 from each of the stacks.```

### Solution :

```                        ```Solution in C++ :

int solve(vector<vector<int>>& rows, int k) {
vector<int> dp(k + 1, -1e9);
dp = 0;

for (const auto& nums : rows) {
int n = nums.size();
auto dpc = dp;

for (int i = 1; i <= k; ++i) {
int sum = 0;
for (int j = 1; j <= min(i, n); ++j)
dpc[i] = max(dpc[i], (sum += nums[n - j]) + dp[i - j]);
}

dp = dpc;
}

return dp[k];
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
Integer[][] dp;
List<List<Integer>> total;
public int solve(int[][] stacks, int k) {
dp = new Integer[stacks.length][k + 1];
total = new ArrayList();
for (int i = 0; i < stacks.length; i++) {
List<Integer> list = new ArrayList();
for (int j = stacks[i].length - 1; j > -1 && j >= stacks[i].length - k; j--) {
}
}
return recur(0, k);
}
public int recur(int ind, int k) {
if (k == 0)
return 0;
if (ind == dp.length || k < 0)
return -1 * (int) (1e7);
if (dp[ind][k] != null)
return dp[ind][k];
int a = -1 * (int) (1e7);
for (int i = 0; i < total.get(ind).size(); i++) {
a = Math.max(a, total.get(ind).get(i) + recur(ind + 1, k - i));
}
dp[ind][k] = a;
return a;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, stacks, k):
NINF = float("-inf")
dp = [NINF] * (k + 1)
dp = 0

for stack in stacks:
P = 
for x in reversed(stack):
P.append(P[-1] + x)

for j in range(k, 0, -1):
for i in range(1, min(j + 1, len(P))):
dp[j] = max(dp[j], dp[j - i] + P[i])

return dp[k]```
```

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