http://anandtechblog.blogspot.com/2010/09/given-array-of-numbers-replace-each.html
On Sun, Jun 26, 2011 at 10:12 AM, ross <jagadish1...@gmail.com> wrote:
> @Dave: Very good solution.. I had thought an idea of using separate
> arrays to store next and previous products. And multiplying the
> elements of the 2 arrays to give B. The solution given by you
> satisfies ALL constraints inplace :)
>
> @sameer: Your solution is not O(1) in space dude..Dave's solution is
> good.
>
> On Jun 26, 9:47 pm, Ashish Goel <ashg...@gmail.com> wrote:
> > this is goog question
> > Best Regards
> > Ashish Goel
> > "Think positive and find fuel in failure"
> > +919985813081
> > +919966006652
> >
> >
> >
> >
> >
> >
> >
> > On Sun, Jun 26, 2011 at 10:04 PM, Dave <dave_and_da...@juno.com> wrote:
> > > @Ross: This satisfies your constraints...
> >
> > > B[0] = 1;
> > > for( i = 1 ; i < N ; ++i )
> > > B[i] = B[i-1] * A[i-1];
> > > int x = 1;
> > > for( i = N-1 ; i > 0 ; --i )
> > > {
> > > x *= A[i];
> > > B[i-1] *= x;
> > > }
> >
> > > Dave
> >
> > > On Jun 26, 11:08 am, ross <jagadish1...@gmail.com> wrote:
> > > > Given an array A , of N integers ( In no particular order), fill up
> an
> > > > auxilary array B such that B[i] contains the product of
> > > > all elements in A other than A[i].
> > > > Constraints:
> > > > O(n) Time,
> > > > Can this be done with O(1) space?
> > > > Division is *not* allowed .
> >
> > > > eg: A 1 2 3 4 5
> > > > B 120 60 40 30 24
> >
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