Boolean Matrix
Problem Statement - link #
Given a boolean matrix of size RxC where each cell contains either 0 or 1, modify it such that if a matrix cell matrix[i][j] is 1 then all the cells in its ith row and jth column will become 1.
Your Task: You dont need to read input or print anything. Complete the function booleanMatrix() that takes the matrix as input parameter and modifies it in-place.
Expected Time Complexity: O(R * C) Expected Auxiliary Space: O(R + C)
Examples #
Example 1:
Input:
R = 2, C = 2
matrix[][] = [[1, 0], [0, 0]]
Output:
1 1
1 0
Explanation:
Only cell that has 1 is at (0,0) so all
cells in row 0 are modified to 1 and all
cells in column 0 are modified to 1.
Example 2:
Input:
R = 4, C = 3
matrix[][] = [[ 1, 0, 0], [ 1, 0, 0], [ 1, 0, 0], [ 0, 0, 0]]
Output:
1 1 1
1 1 1
1 1 1
1 0 0
Explanation:
The position of cells that have 1 in
the original matrix are (0,0), (1,0)
and (2,0). Therefore, all cells in row
0,1,2 are and column 0 are modified to 1.
Constraints #
1 ≤ R, C ≤ 10000 ≤ matrix[i][j] ≤ 1
Solutions #
class Solution{
public:
//Function to modify the matrix such that if a matrix cell matrix[i][j]
//is 1 then all the cells in its ith row and jth column will become 1.
void booleanMatrix(vector<vector<int> > &matrix) {
// code here
int m = matrix.size(), n = matrix[0].size();
int r[m] = {0} ,c[n] ={0};
for(int i=0; i<m; i++)
for(int j=0; j<n; j++)
if(matrix[i][j]){
r[i] = 1;
c[j] = 1;
}
for(int i=0; i<m; i++)
for(int j=0; j<n; j++)
if(r[i] || c[j])
matrix[i][j] = 1;
}
};