@Icy: I forgot about a[0][0]. So I need to add a few lines at the end
of my code, so that it becomes:
for( i = 1 ; i < n ; ++i )
for( j = 1 ; j < m ; ++j )
if( a[i][j] != 0 )
a[i][0] = a[0][j] = 1;
for( i = 1 ; i < n ; ++i )
for( j = 1 ; j < m ; ++j )
if( a[i][0] + a[0][j] != 0 )
a[i][j] = 1;
// the following added to propogate a[0][0], if necessary.
if( a[0][0] != 0 )
{
for( i = 1 ; i < n ; ++i )
a[i][0] = 1;
for( j = 1 ; j < m ; ++j )
a[0][j] = 1;
}
Dave
On Aug 31, 5:12 pm, "icy`" <[email protected]> wrote:
> Dave has a nice idea but I cant get it to work =/
> [[1, 0, 0, 1], [0, 0, 1, 0], [0, 0, 0, 0]] original matrix
>
> [[1, 0, 1, 1], [1, 1, 1, 1], [0, 0, 1, 1]] dave's
> [[1, 1, 1, 1], [1, 1, 1, 1], [1, 0, 1, 1]] expected
>
> Maybe I converted it wrong. My method was basically the same as
> Anup's --
> 1st pass fill rows and convert 1's to 2's.
> 2nd pass check for 2's and fill those columns.
>
> But complexity seems to be n*m*m + n*m*n = nm^2 + mn^2
> which is about O(n^2 m^2) ? =/
>
> I would like to get Dave's to work =P
>
> On Aug 31, 1:47 pm, siva viknesh <[email protected]> wrote:
>
>
>
> > @dave...additionally u ve to do this...checking the 1st row nd 1st
> > column...
>
> > if(a[0][0])
> > set both first row and first column;
> > else
> > for(i=1;i<n;i++)
> > if(a[0][i])
> > set first row;
> > else
> > set first column;
>
> > On Aug 31, 10:34 pm, siva viknesh <[email protected]> wrote:
>
> > > dave s algo is nice :)
>
> > > On Aug 31, 10:09 pm, Dave <[email protected]> wrote:
>
> > > > @Ashima: Scan all but the first row and the first column. If there is
> > > > a 1 in a row, set the first element of that row to 1. If there is a 1
> > > > in a column, set the first element of that column to zero. Now, set
> > > > any element in all but the first row and the first column of the
> > > > matrix that has a 1 it the first element of its row or a 1 in its
> > > > first element of its colunn to 1.
>
> > > > Dave
>
> > > > On Aug 31, 12:02 pm, "Ashima ." <[email protected]> wrote:
>
> > > > > @dave wats d logic behind ur code
>
> > > > > Ashima
> > > > > M.Sc.(Tech)Information Systems
> > > > > 4th year
> > > > > BITS Pilani
> > > > > Rajasthan
>
> > > > > On Wed, Aug 31, 2011 at 9:05 AM, Dave <[email protected]> wrote:
> > > > > > @Manish:
>
> > > > > > for( i = 1 ; i < n ; ++i )
> > > > > > for( j = 1 ; j < m ; ++j )
> > > > > > if( a[i][j] != 0 )
> > > > > > a[i][0] = a[0][j] = 1;
> > > > > > for( i = 1 ; i < n ; ++i )
> > > > > > for( j = 1 ; j < m ; ++j )
> > > > > > if( a[i][0] + a[0][j] != 0 )
> > > > > > a[i][j] = 1;
>
> > > > > > Dave
>
> > > > > > On Aug 31, 8:40 am, manish kapur <[email protected]> wrote:
> > > > > > > Input is a matrix of size n x m of 0s and 1s.
>
> > > > > > > eg:
> > > > > > > 1 0 0 1
> > > > > > > 0 0 1 0
> > > > > > > 0 0 0 0
>
> > > > > > > If a location has 1; make all the elements of that row and column
> > > > > > > = 1. eg
>
> > > > > > > 1 1 1 1
> > > > > > > 1 1 1 1
> > > > > > > 1 0 1 1
>
> > > > > > > Solution should be with Time complexity = O(n*m) and O(1) extra
> > > > > > > space
>
> > > > > > --
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