A Boolean Matrix Question
if a[i][j]=1
then fill a[i][k]=1 where k 0 to N
and fill a[k][j]=1 where k 0 to M
Method 1 (Use two temporary arrays)
1) Create two temporary arrays row[M] and col[N]. Initialize all values of row[] and col[] 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.
void
modifyMatrix(
bool
mat[R][C])
{
bool
row[R]={0};
bool
col[C]={0};
int
i, j;
/* Store the rows and columns to be marked as 1 in row[] and col[]
arrays respectively */
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;
}
}
}
}
Method 2 (A Space Optimized Version of Method 1)
This method is a space optimized version of above method 1. This method uses the first row and first column of the input matrix in place of the auxiliary arrays row[] and col[] of method 1. So what we do is: first take care of first row and column and store the info about these two in two flag variables rowFlag and colFlag. Once we have this info, we can use first row and first column as auxiliary arrays and apply method 1 for submatrix (matrix excluding first row and first column) of size (M-1)*(N-1).
1) Scan the first row and set a variable rowFlag to indicate whether we need to set all 1s in first row or not.
2) Scan the first column and set a variable colFlag to indicate whether we need to set all 1s in first column or not.
3) Use first row and first column as the auxiliary arrays row[] and col[] respectively, consider the matrix as submatrix starting from second row and second column and apply method 1.
4) Finally, using rowFlag and colFlag, update first row and first column if needed.
Time Complexity: O(M*N)
Auxiliary Space: O(1)
void
modifyMatrix(
bool
mat[R][C])
{
bool
rowFirst;
bool
colFirst;
int
i, j;
for
(i = 0; i < C; i++)
{
if(mat[0][i]){ rowFirst=1; break;
}
for
(i = 0; i < R; i++)
{
if(mat[i][o]) { colFirst=1 ; break;}
}
for
(i = 1; i < R; i++)
{
for
(j = 1; j < C; j++)
{
if
(mat[i][j] == 1)
{
mat[i][0] = 1;
mat[0][j] = 1;
}
}
}
for
(i = 1; i < R; i++)
{
for
(j = 1; j < C; j++)
{
if
( mat[i][0] == 1 || mat[0][j] == 1 )
{
mat[i][j] = 1;
}
}
}
if(rowFirst)
for
(i = 0; i < R; i++)
mat[0][i]=1;
if(colFirst)
for
(i = 0; i < C; i++)
mat[i][0]=1;
}
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