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 :



title-img




                        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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