little-endian else big-endian (there exists
mixed-endianness also, but lets leave that for now.)
--Sundeep.
On Tue, Jun 15, 2010 at 9:05 PM, Lego Haryanto legoharya...@gmail.comwrote:
On Mon, Jun 14, 2010 at 5:13 AM, Sundeep Singh singh.sund...@gmail.comwrote:
@saurav: your code
use a hash map
On Mon, Jun 14, 2010 at 12:14 AM, jalaj jaiswal
jalaj.jaiswa...@gmail.comwrote:
give an algo to find a unique number in an array
for eg a[]={1,3,4,1,4,5,6,1,5}
here 3 is the unique number as it occur only once... moreover array
contains only 1 unique number
--
With
@saurav: your code will always print 2 irrespective of the system's
endianness!
correct thing to do is:
printf(%d, *(char *) (0x0002))
--Sundeep.
On Mon, Jun 14, 2010 at 3:02 AM, Minotauraus anike...@gmail.com wrote:
How about a pointer? :D
On Jun 13, 5:56 am, debajyotisarma
@rohit: fibonacci sequence may be the answer to the prob, but I am curious
why? I haven't come across any such fib sequence property...
On Wed, Jun 9, 2010 at 9:16 PM, Rohit Saraf rohit.kumar.sa...@gmail.comwrote:
@junta : are fibonacci sequence is the answer of the prob, it is not used
:D
Whats the logic behind using Fibonacci in determining the number of such
sequences?
-Sundeep.
On Wed, Jun 9, 2010 at 8:02 AM, Rohit Saraf rohit.kumar.sa...@gmail.comwrote:
Fib comes because she wants the number of such sequences
--
--
Rohit
@Anand and @Minotaurus
The code seems to fail for 15.
Am I missing something?
On Mon, Jun 7, 2010 at 2:20 AM, Minotauraus anike...@gmail.com wrote:
@Anand: Thanks for the code. I knew you could do it by bit
shifting. :-)
On Jun 5, 10:21 pm, Anand anandut2...@gmail.com wrote:
Here is a
place the L block diagonally...
--Sundeep.
On Sun, Jun 6, 2010 at 7:29 PM, sharad sharad20073...@gmail.com wrote:
A square Island surrounded by bigger square, and in between there is
infinite depth water. The distance between them is L. The wooden
blocks of L are given.
The L length block
merge sort both lists: O(nlogn)
Now, for both lists to be identical, just compare the corresponding elements
in the lists i.e. L1(1) == L2(1), L1(2) == L2(2) ...
= O(n)
--Sundeep.
On Wed, Jun 2, 2010 at 10:47 PM, Raj N rajn...@gmail.com wrote:
@Antony: The 2 lists should have the same
oops
On Sat, May 1, 2010 at 5:50 PM, Sundeep Singh singh.sund...@gmail.comwrote:
Hi Amit,
here's the answer: (I am assuming in your equation lg implies log to the
base 10)
n 8 log(n)
= n/8 log(n)
= 10 ^(n/8) n
The final deduction was incorrect!!
for log base 10, the answer is:
2
it
on paper?
-Regards
Amit Agarwal
Contact: 09765348182
www.amitagrwal.com
On Mon, May 3, 2010 at 3:02 PM, Sundeep Singh singh.sund...@gmail.comwrote:
oops
On Sat, May 1, 2010 at 5:50 PM, Sundeep Singh singh.sund...@gmail.comwrote:
Hi Amit,
here's the answer: (I am assuming in your
.
There is room for a little optimization if both stacks are empty when
enquing, as you can push the item directly onto stack B. Furthermore,
when popping from stack A and pushing onto stack B, you don't need to
push the last item popped, as it is the return value.
Dave
On Mar 22, 9:29 am, Sundeep
Hey Brian,
Better still, for inserting in queue, just keep pushing onto the stack A.
You need stack B only for dequeuing: for dequeuing, push all items into
stack B, pop as many as you want from stack B and then push back all
remaining items in stack A.
Regards,
Sundeep.
On Mon, Mar 22, 2010
12 matches
Mail list logo
