1. do & operation on all the values in each column and store it in the first
row of each column
2. do & operation on all the values in each row and store it in the first
column of each row.
(when writing at a[0][0] do & operation with the value computed at 1.)
3. Now to find out the value at a[i][j] ,you need to do a[i][0] & a[0][j]

On Sun, Feb 27, 2011 at 12:03 PM, Rajnish <[email protected]> wrote:

> 1.) Traverse the whole matrix and replace each 0 value with -1.
> 2.) Traverse the matrix again,all the 1 values are replaced with 0 in
> the row and column of the index where a -1 value is found.
> 3.) Set all -1 values to zero and we have the output array.
> time complexity: O(n^2)
> space complexity: O(1)
>
>
> On Feb 27, 2:29 am, gaurav gupta <[email protected]> wrote:
> > A NxN binary matrix is given. If a row contains a 0 all element in the
> > row will be set to 0 and if a column contains a 0 all element of the
> > column will be set to 0. You have to do it in O(1) space.
> >
> > example :
> >
> > input array :
> >
> > 1 0 1 1 0
> > 0 1 1 1 0
> > 1 1 1 1 1
> > 1 0 1 1 1
> > 1 1 1 1 1
> >
> > result array :
> >
> > 0 0 0 0 0
> > 0 0 0 0 0
> > 0 0 1 1 0
> > 0 0 0 0 0
> > 0 0 1 1 0
> >
> > Thanks & Regards,
> > Gaurav Gupta
>
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-- 
regards,
chinna.

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