Boolean Matrix Question

Given a matrix of size n x m filled with 0′s and 1′s
e.g.:
1 1 0 1 0
0 0 0 0 0
0 1 0 0 0
1 0 1 1 0

if the matrix has 1 at (i,j), fill the column j and row i with 1′s
i.e., we get:
1 1 1 1 1
1 1 1 1 0
1 1 1 1 1
1 1 1 1 1

Method (Use two temporary arrays)
1) Create two temp array row[M] and col[N]. Initialize all values as 0.
2) Traverse the input matrix mat[M][N]. If you see an entry mat[i][j] as true, then mark row[i] and col[j] as true.
3) Traverse the input matrix mat[M][N] again. For each entry mat[i][j], check the values of row[i] and col[j]. If any of the two values (row[i] or col[j]) is true, then mark mat[i][j] as true.

Implementation:

void modifyMatrix(bool mat[R][C])
{
    bool row[R];
    bool col[C];
 
    int i, j;

    // Initialize all values of row[] and col[] as 0
    for (i = 0; i < R; i++)
    {
       row[i] = 0;
    }
 
    for (i = 0; i < C; i++)
    {
       col[i] = 0;
    }

    for (i = 0; i < R; i++)
    {
        for (j = 0; j < C; j++)
        {
            if (mat[i][j] == 1)
            {
                row[i] = 1;
                col[j] = 1;
            }
        }
    }
 
    /* Modify the input matrix mat[] using the above constructed row[] and
       col[] arrays */
    for (i = 0; i < R; i++)
    {
        for (j = 0; j < C; j++)
        {
            if ( row[i] == 1 || col[j] == 1 )
            {
                mat[i][j] = 1;
            }
        }
    }
}

Time Complexity: O(M*N)
Auxiliary Space: O(M + N)

Source: Boolean Matrix Question