Re: [algogeeks] Re: Endian-ness check

2010-06-16 Thread Sundeep Singh
@Lego: I am sorry I missed the address of operator... I wanted to type:
printf(%d, *(char *) (0x0002))

But even the above is incorrect since it is not possible to take address of
a literal in C/C++.
The best way is:
const int i=0x0002;
printf(%d, *(char *) (i));

If this print 2, then its 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 will always print 2 irrespective of the system's
 endianness!

 correct thing to do is:
 printf(%d, *(char *) (0x0002))

 --Sundeep.



 ... dereferencing a very low address pointer, are you sure?

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Re: [algogeeks] unique number in an array

2010-06-15 Thread Sundeep Singh
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



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Re: [algogeeks] Re: Endian-ness check

2010-06-15 Thread Sundeep Singh
@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 sarma.debajy...@gmail.com wrote:
  Is it possible to check endianness of a system in C without creating
  variable?
   i.e. Program should not contain any variable.

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Re: [algogeeks] Re: binary nos

2010-06-10 Thread Sundeep Singh
@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

 --
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 http://www.cse.iitb.ac.in/~rohitfeb14http://www.cse.iitb.ac.in/%7Erohitfeb14


 On Wed, Jun 9, 2010 at 9:13 PM, Rohit Saraf 
 rohit.kumar.sa...@gmail.comwrote:

 @debajyoti: read the prob before posting

 --
 Rohit Saraf
 Second Year Undergraduate,
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 http://www.cse.iitb.ac.in/~rohitfeb14http://www.cse.iitb.ac.in/%7Erohitfeb14


 On Wed, Jun 9, 2010 at 2:37 PM, Debajyoti Sarma 
 sarma.debajy...@gmail.com wrote:

 First 20 fibo no as follows with binary form
   0 = 0
   1 = 1
   1 = 1
   2 = 10
   3 = 11
   5 = 101
   8 = 1000
  13 = 1101
  21 = 10101
  34 = 100010
  55 = 110111
  89 = 1011001
 144 = 1001
 233 = 11101001
 377 = 10001
 610 = 1001100010
 987 = 011011
 1597 = 1100001
 2584 = 10111000
 4181 = 101010101

 Now please explain how fibo no is coming under consideration.Both kind of
 no is mixed here.

 On Wed, Jun 9, 2010 at 8:02 AM, Rohit Saraf rohit.kumar.sa...@gmail.com
  wrote:

 Fib comes because she wants the number of such sequences

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Re: [algogeeks] Re: binary nos

2010-06-09 Thread Sundeep Singh
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

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Re: [algogeeks] Re: divisible by 3

2010-06-07 Thread Sundeep Singh
@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 code for it.http://codepad.org/umkh3pjf
 
 
 
  On Sat, Jun 5, 2010 at 7:14 PM, Minotauraus anike...@gmail.com wrote:
   Subtract 3 from the number until either you get 0 or a negative
   number. If you get 0, its divisible, else not.
   You can probably do this by bit shifting too.
 
   On Jun 5, 11:45 am, divya sweetdivya@gmail.com wrote:
Find if a number is divisible my 3, without using %,/ or *. You can
use atoi().
 
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Re: [algogeeks] Puzzle

2010-06-06 Thread Sundeep Singh
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 can't be placed in between to cross it, as it will
 fall in water (just fitting).
 How would you cross using these L length blocks.

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Re: [algogeeks] Check if 2 linked lists are identical

2010-06-03 Thread Sundeep Singh
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 elements as well the number must
 be equal


 On Wed, Jun 2, 2010 at 5:38 PM, Antony Vincent Pandian.S. 
 sant...@gmail.com wrote:

 @Raj

 What do you mean by identical? You are just concerned about the
 presence of one element in another LL or you are concerned about the
 equality of number of elements too?

 On 6/2/10, Raj N rajn...@gmail.com wrote:
  @sharad kumar: But don't you think this'll consume a lot of space. Merge
  sort will give O(nlogn) complexity when a separate LL is used to store
 the
  sorted elements but if we disbuild the same LL it'll be O(n2). And also
 wat
  do u mean by combining LL can u explain
 
 
  On Wed, Jun 2, 2010 at 6:48 AM, sharad kumar aryansmit3...@gmail.com
 wrote:
 
  @Raj:sorting can be done in O(nlogn)..sort both and compare both.or use
 a
  hash map to store all elements of one LL and then compare it with
  otheror combine both the  LL perform merge sort and start deleting
  adjacent elements.if adjacent elements in equal count then LL are equal
  and
  at end of process we get an empty list.
 
 
  On Tue, Jun 1, 2010 at 11:55 PM, Raj N rajn...@gmail.com wrote:
 
  Hi all,
  Can someone suggest an efficient algorithm to check if 2 linked lists
  are identical. If 2 lists have to be sorted then there'll be a lot of
  space consumption for 2 separate linked lists. Can the whole process
  be done  O(n2)
 
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Re: [algogeeks] value of n

2010-05-03 Thread Sundeep Singh
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 = n = 6

--Sudneep.



 = n  8

 --Sundeep.



 On Sat, May 1, 2010 at 10:43 AM, Amit Agarwal lifea...@gmail.com wrote:

 I could not get you properly. This is an equation comes from the problem
 statement where I need to find out cut-off value of n between insertion and
 merge sort. I think equation is part of basic mathematics but I don't
 remember how do I solve it.


 -Regards
 Amit Agarwal
 Contact: 09765348182
 www.amitagrwal.com




 On Sat, May 1, 2010 at 9:13 AM, abhijith reddy 
 abhijith200...@gmail.comwrote:

 binary search on n

 On Fri, Apr 30, 2010 at 10:15 PM, Amit Agarwal lifea...@gmail.comwrote:

 how do I compute n from this equation.
 n  8lg(n)

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Re: [algogeeks] value of n

2010-05-03 Thread Sundeep Singh
Hi Amit,
This particular example was quite simple.. just required using calculator
couple of times.
We know log 1 =0 and log 10 = 1, so given the above equation, it was clear
that the answer had to lie within the range (1,10) and then I used the
calculator couple of times to narrow down the range.

For a more generic/complicated equation of this nature, u'll need to plot
the functions as people have suggested earlier.

Regards,
Sundeep.

On Mon, May 3, 2010 at 4:51 PM, Amit Agarwal lifea...@gmail.com wrote:

 yeah, you are right. It comes from 2 to 6. But is there any way to solve 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 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 = n = 6

 --Sudneep.



 = n  8

 --Sundeep.



 On Sat, May 1, 2010 at 10:43 AM, Amit Agarwal lifea...@gmail.comwrote:

 I could not get you properly. This is an equation comes from the problem
 statement where I need to find out cut-off value of n between insertion and
 merge sort. I think equation is part of basic mathematics but I don't
 remember how do I solve it.


 -Regards
 Amit Agarwal
 Contact: 09765348182
 www.amitagrwal.com




 On Sat, May 1, 2010 at 9:13 AM, abhijith reddy 
 abhijith200...@gmail.com wrote:

 binary search on n

 On Fri, Apr 30, 2010 at 10:15 PM, Amit Agarwal lifea...@gmail.comwrote:

 how do I compute n from this equation.
 n  8lg(n)

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Re: [algogeeks] Re: Implement a queue using a stack

2010-03-23 Thread Sundeep Singh
Thanks Dave, I didn't think about this... definitely better!

Sundeep.

On Mon, Mar 22, 2010 at 9:08 PM, Dave dave_and_da...@juno.com wrote:

 Better still.
 To enqueue: push onto stack A.
 For dequeuing: If stack B is empty, pop all items from stack A and
 push onto stack B. Then pop stack B.
 There is no need to push remaining items back to stack A.

 As every item passes through the queue, it is pushed onto stack A,
 then popped from stack A and pushed onto stack B, and finally popped
 from stack B. The time is roughly twice the time required for a direct
 implementation of a queue.

 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 Singh singh.sund...@gmail.com wrote:
  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 at 6:56 PM, Brian brianfo...@gmail.com wrote:
 How is it possible to implement a queue using a stack only?
 
   Interesting, but tricky... You need two stacks as Prakhar stated...
   In general, if you have Stack A and Stack B, you should keep all the
 items
   in stack A and then use stack B for processing.
 
   For example to insert an item:
   1. Pop all the items from A  and then push them into B (this should
 push
   the items into Stack B in reverse order)
   2. Insert the new item into A
   3. Pop all the items in B and push them back into A (again this will
 push
   them back into Stack B in reverse order)
 
   Running time should be O(1)+O(2n), which is O(n) for larger values of n
 -
   which is not good...
 
   To retrieve an item, pop the first item in stack A
 
   Hope  this helps -
   Regards
   B
 
   On 3/22/2010 4:55 AM, Prakhar Jain wrote:
 
   By a do u mean a single stack ?
   Otherwise if you use 2 its v simple
 
   Best,
   Prakhar Jain
  http://web.iiit.ac.in/~prakharjain/http://web.iiit.ac.in/%7Eprakharjain/
 http://web.iiit.ac.in/%7Eprakharjain/
 
   On Mon, Mar 22, 2010 at 12:56 AM, Pushpesh Gupta pushpes...@gmail.com
 wrote:
 
   How is it possible to implement a queue using a stack only?
 
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Re: [algogeeks] Implement a queue using a stack

2010-03-22 Thread Sundeep Singh
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 at 6:56 PM, Brian brianfo...@gmail.com wrote:

   How is it possible to implement a queue using a stack only?

 Interesting, but tricky... You need two stacks as Prakhar stated...
 In general, if you have Stack A and Stack B, you should keep all the items
 in stack A and then use stack B for processing.

 For example to insert an item:
 1. Pop all the items from A  and then push them into B (this should push
 the items into Stack B in reverse order)
 2. Insert the new item into A
 3. Pop all the items in B and push them back into A (again this will push
 them back into Stack B in reverse order)

 Running time should be O(1)+O(2n), which is O(n) for larger values of n -
 which is not good...

 To retrieve an item, pop the first item in stack A

 Hope  this helps -
 Regards
 B



 On 3/22/2010 4:55 AM, Prakhar Jain wrote:

 By a do u mean a single stack ?
 Otherwise if you use 2 its v simple

 Best,
 Prakhar Jain
 http://web.iiit.ac.in/~prakharjain/http://web.iiit.ac.in/%7Eprakharjain/


 On Mon, Mar 22, 2010 at 12:56 AM, Pushpesh Gupta pushpes...@gmail.comwrote:

 How is it possible to implement a queue using a stack only?

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