@brijesh ..+1
but character range.. from 0 to 255...
With regards,
Praveen Raj
DCE-IT 3rd yr
735993
praveen0...@gmail.com
On Sat, Sep 10, 2011 at 7:03 PM, Brijesh brijeshupadhyay...@gmail.comwrote:
If you are talking about character array then it can be done in space
O(128)=constant
Hi,
Actually I am not good at hash tables. can u plzz suggest me some gud link
from where I can study hash tables...
and also tell me the logic for integer arrays for which the complexity will
be o(n logn).
Thanks in advance.
--
Kind Regards
Ishan Aggarwal
Phone : +91-9654602663
On Sat, Sep
yes for integer arrays we need to sort o(nlogn) and den compare in o(n)
On Sat, Sep 10, 2011 at 7:12 PM, Ishan Aggarwal
ishan.aggarwal.1...@gmail.com wrote:
Hi,
Actually I am not good at hash tables. can u plzz suggest me some gud link
from where I can study hash tables...
and also tell me
Howz this solution??
void main()
{
int a[10] = {9,3,9,3,9,3,2};
int c[10],i,j,k,co;
k=0;
for(i=0;i7;i++)
{
co = 0;
for( j = 1+1; j7; j++)
if( a[i] == a[j])
co++ ;
if( co == 0)
{
c[k++] = a[i];
}
for (i=0;ik;i++)
printf(%d,c[i]);
}
}
On Sat, Sep 10, 2011 at 9:26 PM, sukran dhawan
@brijesh:
if we don't want to print .. and want to eliminate duplicates ...then ?
means..want to store in some place which doesn't have any duplicate for
further purpose ...
On Sun, Sep 11, 2011 at 12:08 AM, Brijesh brijeshupadhyay...@gmail.comwrote:
Watch this video for the concepts of
What would be the complexity in that case ??
On Sun, Sep 11, 2011 at 10:29 AM, bharatkumar bagana
bagana.bharatku...@gmail.com wrote:
@brijesh:
if we don't want to print .. and want to eliminate duplicates ...then ?
means..want to store in some place which doesn't have any duplicate for
Use of binary tree.. which has node(value,count)
count -number of elements having having element = value...
With regards,
Praveen Raj
DCE-IT 3rd yr
735993
praveen0...@gmail.com
On Sun, Sep 11, 2011 at 10:32 AM, Ishan Aggarwal
ishan.aggarwal.1...@gmail.com wrote:
What would be the
is there any general algo for this question,ie, upto any n(here
16) ? ?
On Sep 15, 2:46 am, Minotauraus anike...@gmail.com wrote:
I guess the problem could be modified for any 1..n where n is 2^m.
so say 132
or 1...64
Try it out. It still works! The time complexity in this case is O(log
@ash: I would treat it as a graph problem. The nodes are the numbers
from 1 to n, and the edges connect nodes that sum to a perfect square.
The problem then reduces to finding a Hamiltonian path through the
graph. This can be done with a depth-first search. Start with a node
of degree 1 if you
@Davhw can u compute in contant time where runing time dwepends
on elements of array e.g.
its has complexcity O(n) isn'tit..??
import java.util.Arrays;
import java.util.LinkedList;
class SolutionFinder
{
public static void find(final LinkedListInteger remaining, final
@Bittu: First, it makes no sense to say O(n) when n is fixed. Second,
if you know the solution, a program to print it out just consists of a
printf statement, e.g.,
printf(16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8\n);
plus whatever language stuff you need to get it to compile and
I guess the problem could be modified for any 1..n where n is 2^m.
so say 132
or 1...64
Try it out. It still works! The time complexity in this case is O(log
n).
On Sep 14, 6:02 am, Dave dave_and_da...@juno.com wrote:
@Bittu: First, it makes no sense to say O(n) when n is fixed. Second,
if
You don't really need a program. Only one number, 9, can be added to
16 to make a perfect square, so 16 must be at one end of the list.
Through a sequence of forced choices, you can build the list 16, 9, 7,
2, 14, 11, 5, 4, 12, 13, 3. At this point, you can choose either 1 or
6. Trying 1, you can
13 matches
Mail list logo
