plz post a samle input output..
On Nov 4, 10:24 am, sunny agrawal sunny816.i...@gmail.com wrote:
What can be a better solution than a Brute Force O(N^2* No of iteration)
--
Sunny Aggrawal
B.Tech. V year,CSI
Indian Institute Of Technology,Roorkee
fb1_input.txt
1KViewDownload
i/p o/p files are attached in the post, see carefully :-o
On Fri, Nov 4, 2011 at 6:23 PM, Dumanshu duman...@gmail.com wrote:
plz post a samle input output..
On Nov 4, 10:24 am, sunny agrawal sunny816.i...@gmail.com wrote:
What can be a better solution than a Brute Force O(N^2* No of
Hi guys,
Social sites like facebook or linkedin must be having some vertex labelling
techniques. I cant' imagine what labelling they use since there are
trillions of nodes in there network ( coz simple
integer or alphanumeric labelling will not be efficient enough as far as i
think).
Can someone
is ur brute force O(1) for space?
On Nov 4, 10:24 am, sunny agrawal sunny816.i...@gmail.com wrote:
What can be a better solution than a Brute Force O(N^2* No of iteration)
--
Sunny Aggrawal
B.Tech. V year,CSI
Indian Institute Of Technology,Roorkee
fb1_input.txt
1KViewDownload
No,O(N^2) because all the flips happens simultaneously, it can be
reduce to O(2N) if each tile is represented using a single bit
On Fri, Nov 4, 2011 at 11:12 PM, Dumanshu duman...@gmail.com wrote:
is ur brute force O(1) for space?
On Nov 4, 10:24 am, sunny agrawal sunny816.i...@gmail.com
Here's an idea. Say we pick any element P in the 2D array A and use
it to fill in an N element array X as follows.
j = N;
for i = 1 to N do
while A(i, j) P do
j = j - 1;
end;
X(i) = j;
end;
This algorithm needs O(N) time.
The elements of X split each row with respect to P. That is,
I think this can be done in O(n) time simply by finding the median of
elements at DIAGNOL starting from right corner to left corner.
Coz as elements are sorted row-wise and column-wise , it implies that
elements below this diagonal are all greater than elements on this diagonal
and all above this
Hi,
I think the median will always lie on the diagonal
a[n][1] a[1][n]
because the elements on the LHS making the upper triangle will
always be less than or equal to the elements on the diagonal
and the RHS, elements in the lower triangle will be greater than or
equal to them.
so sort the
