**Stacks - Google Top Interview Questions**

### Problem Statement :

Given a list of list of positive integers stacks, you can take any stack(s) in stacks and pop any number of elements. Return the maximum sum that can be achieved such that all stacks have the same sum. Constraints n * m ≤ 500,000 where n and m are the number of rows and columns in stacks Example 1 Input stacks = [ [2, 3, 4, 5], [4, 5, 2, 3, 3], [9, 1, 1, 1] ] Output 9 Explanation Here are the operations we can take Pop [5] from the first stack to get [2, 3, 4] for a sum of 9. Pop [2, 3, 3] from the second stack to get [4, 5] for a sum of 9. Pop [1, 1, 1] from the first stack to get [9] for a sum of 9. Example 2 Input stacks = [ [5, 13], [7, 2], [50] ] Output 0 Explanation We have to pop all elements from every stack since there's no way to make the three stacks have equal sum otherwise.

### Solution :

` ````
Solution in C++ :
int solve(vector<vector<int>>& stacks) {
int res = 0;
unordered_map<int, int> f;
for (vector<int>& A : stacks) {
int sum = 0;
for (int i : A) {
sum += i;
if (i == 0) continue;
f[sum]++;
if (f[sum] == stacks.size()) {
res = max(res, sum);
}
}
}
return res;
}
```

` ````
Solution in Java :
import java.util.*;
import java.util.stream.Collectors;
class Solution {
public class Node {
int sum;
int row;
int idx;
public Node(int s, int r, int i) {
this.sum = s;
this.row = r;
this.idx = i;
}
}
public int solve(int[][] stacks) {
if (stacks == null || stacks.length == 0)
return 0;
PriorityQueue<Node> pq = new PriorityQueue<>((n1, n2) -> n2.sum - n1.sum);
Map<Integer, Integer> map = new HashMap();
int sum = 0;
for (int i = 0; i < stacks.length; i++) {
int[] arr = stacks[i];
sum = 0;
for (int j = 0; j < arr.length; j++) sum += arr[j];
pq.offer(new Node(sum, i, arr.length - 1));
map.put(sum, map.getOrDefault(sum, 0) + 1);
}
if (map.size() == 1)
return sum;
while (map.size() > 1) {
Node node = pq.poll();
sum = node.sum;
int row = node.row;
int idx = node.idx;
if (map.containsKey(sum)) {
if (map.get(sum) == 1)
map.remove(sum);
else
map.put(sum, map.get(sum) - 1);
}
sum -= stacks[row][idx];
map.put(sum, map.getOrDefault(sum, 0) + 1);
if (idx == 0)
return 0;
pq.offer(new Node(sum, row, idx - 1));
}
for (int key : map.keySet()) return key;
return 0;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, stacks):
if not stacks:
return 0
seen = set(accumulate(stacks[0]))
for s in stacks[1:]:
seen2 = set(accumulate(s))
seen &= seen2
return max(seen, default=0)
```

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