@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 <sivavikne...@gmail.com> wrote:
> dave s algo is nice :)
>
> On Aug 31, 10:09 pm, Dave <dave_and_da...@juno.com> 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 ." <ashima.b...@gmail.com> 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 <dave_and_da...@juno.com> 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 <manishkapur.n...@gmail.com> 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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