**Largest Elements in Their Row and Column - Google Top Interview Questions**

### Problem Statement :

You are given a two-dimensional list of integers matrix containing 1s and 0s. Return the number of elements in matrix such that: matrix[r][c] = 1 matrix[r][j] = 0 for every j ≠ c and matrix[i][c] = 0 for every i ≠ r Constraints 0 ≤ n, m ≤ 250 where n and m are the number of rows and columns in matrix Example 1 Input matrix = [ [0, 0, 1], [1, 0, 0], [0, 1, 0] ] Output 3 Explanation We have matrix[0][2], matrix[1][0] and matrix[2][1] meet the criteria. Example 2 Input matrix = [ [0, 0, 1], [1, 0, 0], [1, 0, 0] ] Output 1 Explanation Only matrix[0][2] meet the criteria. The other two 1s share the same column.

### Solution :

` ````
Solution in C++ :
int solve(vector<vector<int>>& matrix) {
vector<int> row(matrix.size(), 0), col(matrix[0].size(), 0);
for (int i = 0; i < matrix.size(); i++) {
for (int j = 0; j < matrix[i].size(); j++) {
col[j] += matrix[i][j];
row[i] += matrix[i][j];
}
}
int ans = 0;
for (int i = 0; i < matrix.size(); i++) {
for (int j = 0; j < matrix[i].size(); j++) {
ans += (matrix[i][j] == 1 and row[i] == 1 and col[j] == 1);
}
}
return ans;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] matrix) {
int count = 0;
int count2 = 0;
int ans = 0;
int index = 0;
// edge case
if (matrix.length == 1) {
int c = 0;
for (int i = 0; i < matrix[0].length; i++)
if (matrix[0][i] == 1)
c++;
if (c == 1)
return 1;
else
return 0;
}
// First we check row-wise. If only Element found then only. We have to check column wise
// from 0 - to end index of the particular column.
for (int i = 0; i < matrix.length; i++) {
count = 0;
count2 = 0;
for (int j = 0; j < matrix[i].length; j++) {
if (matrix[i][j] == 1) {
count++;
index = j;
}
}
if (count != 1)
continue;
for (int k = 0; k < matrix.length; k++)
if (matrix[k][index] == 1)
count2++;
if (count == 1 && count2 == 1)
ans++;
}
return ans;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, matrix):
if not matrix:
return 0
row = [sum(r) for r in matrix]
col = [sum(c) for c in zip(*matrix)]
m, n = len(matrix), len(matrix[0])
res = 0
for r in range(m):
for c in range(n):
if matrix[r][c] == 1 and row[r] == 1 and col[c] == 1:
res += 1
return res
```

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