[algogeeks] Apply to US

2013-06-22 Thread pacific :-)
Hi all,

I would like to apply for US companies like google, twitter, linkedin,
facebook for next year. Any suggestions how/when should i apply ? General
suggestions on prep. Thanks in advance.

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chinna.

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Re: [algogeeks] Re: find a point closest to other points

2012-05-08 Thread pacific :-)
On Tue, May 1, 2012 at 8:44 PM, Bhupendra Dubey bhupendra@gmail.comwrote:

 Find the centroid

 X= (x1 +x2xn)/N
 Y=(y1+y2..yn)/N
 This will precisely be the point no need to calculate and check with
 distance.

@dubey : Consider this case, let there be a cluster of 4 points near the
origin and 1 point at (100,0). Then the answer would be near the origin and
not close to the centroid. median might make some sense.



 On Tue, May 1, 2012 at 1:18 PM, mohit mishra mohit7mis...@gmail.comwrote:

 Let me know if there is any bug
 /*using centroid of plane */

 struct point
 {
 int x;
 int y;
 };
 typedef struct point Point;
 int main()
 {
 int n,i;
 int d,dis;
 float sum_x=0,sum_y=0;
 Point centroid,final;
 //clrcsr();
 printf(how many points?  );
 scanf(%d,n);
 Point p[100];
 printf(enter X and Y cordinates of %d points\n,n);
 for(i=0;in;i++)
 scanf(%d%d,p[i].x,p[i].y);
 for(i=0;in;i++)
 {
 sum_x=sum_x+p[i].x;
 sum_y=sum_y+p[i].y;
 }
centroid.x=(int)sum_x/n;
centroid.y=(int)sum_y/n;
 for(i=0;in;i++)
 {
 dis=distance(centroid,p[i]);
 if(disd)
 {
  d=dis;
  final.x=p[i].x;
  final.y=p[i].y;

  }
 }
 printf(\n The point is (%3d  ,%3d),final.x,final.y);
 getch();
 return 0;
 }
 int distance(Point A,Point B)
 {
 int x,y;
 x=(A.x-B.x)*(A.x-B.x);
 y=(A.y-B.y)*(A.y-B.y);
 return sqrt(x+y);
 }


 On Mon, Apr 30, 2012 at 4:34 PM, kosgi santosh santoshko...@gmail.comwrote:

 how can we find centriod of n points in a plane?





 Regards,
 Santosh K.

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Re: [algogeeks] how to code a deterministic or Non-Deterministic finite state automata model?

2012-05-08 Thread pacific :-)
How about using function pointers ?

On Fri, Apr 6, 2012 at 11:09 PM, Doom duman...@gmail.com wrote:

 Any pointers to a code using NFA / DFA computation model?

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Re: [algogeeks] c

2011-08-20 Thread pacific :-)
Is const a compiler level construct ?
#include stdio.h

int main() {
  const int i = 0;
  int * p ;
  p = (int *)  i;
  *p = 2;
  printf((i,p): %x %x \n,i,p);
  printf((i,p): %d %d \n,i,*p);
}

I ran this program and got this output.Can somebody explain ?
(i,p): bfc3de6c bfc3de6c
(i,p): 0 2


On Sat, Aug 20, 2011 at 10:08 AM, sukran dhawan sukrandha...@gmail.comwrote:

 may be it places the  variable in read only memory


 On Thu, Aug 18, 2011 at 2:38 AM, bihari kumarvive...@gmail.com wrote:

 How to prevent the compiler to alter the value of i in statement:
 const int i=2;
 Just give the idea about the implementation of const int i=somevalue;

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Re: [algogeeks] MICROSOFT RESEARCH INTERN

2011-08-16 Thread pacific :-)
Its different.

There are separate kind of interviews for both looking for different things.

On Tue, Aug 16, 2011 at 5:18 PM, arvind kumar arvindk...@gmail.com wrote:

 Hi friends
 Can anyone please tell me about microsoft research intern-the
 selection process??
 Is it the same or is it different from microsoft itc??

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Re: [algogeeks]

2011-08-16 Thread pacific :-)
Whats the take-home salary if you work at TCS or infosys ? also work hours
expected ?

On Mon, Aug 15, 2011 at 12:01 PM, vikas singh shyguy1...@gmail.com wrote:

 TCS also have COBOL project... TCS has every language based project u can
 think of. it's your luck whether you get the latest technology to work with
 or the RETRO...


 On Mon, Aug 15, 2011 at 10:59 AM, vaibhav agrawal agrvaib...@gmail.comwrote:

 Majorly software maintainence


 On Mon, Aug 15, 2011 at 12:17 AM, siddharth srivastava 
 akssps...@gmail.com wrote:



 On 15 August 2011 00:15, sukran dhawan sukrandha...@gmail.com wrote:

 do they recruit software engineers for maintainenance?



 yes..and major chunk of projects are maintenance only



 On Mon, Aug 15, 2011 at 12:10 AM, siddharth srivastava 
 akssps...@gmail.com wrote:



 On 15 August 2011 00:03, sukran dhawan sukrandha...@gmail.com wrote:

 hmmm k

 or even purely maintenance jobs where you may not get chance to code at
 all



 On Sun, Aug 14, 2011 at 11:47 PM, rashmi i rash...@gmail.com wrote:

 Infosys and TCS are consultancies. So, the type of job is not fixed.
 It depends on the type of project a client provides, so it may involve
 variety of technologies from C to Java , Perl,etc.


 On Sun, Aug 14, 2011 at 11:31 PM, sukran dhawan 
 sukrandha...@gmail.com wrote:

  can anyone tell me what is the job provided by infosys and tcs?
 IF they do so much mass recruitment what kinda job the ppl get der?

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 DoN'T LimIt Ur cHaLlEngeS, ChAlLenGe uR LImItS.

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 Thanks and Regards
 VIKAS SINGH
 MCA- final year
 NIT DURGAPUR
 email:
  vikas.singh1...@gmail.com
  shyguy1...@gmail.com
 http://smrit.wordpress.com


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Re: [algogeeks] reason

2011-08-16 Thread pacific :-)
I believe you are assuming little endian right ?

On Mon, Aug 15, 2011 at 2:14 PM, programming love 
love.for.programm...@gmail.com wrote:

 The internal representation of array is this:

 suppose that the address starts from decimal number 10 and integer occupies
 2 bytes

 10- 0002 ( num 2 in hex)
 12- 0003 ( num 3 in hex)
 14- 0004 ( num 4 in hex)

 Now p points to address 10 and is type char. (Even after type casts) p+1
 will increment address by 1 byte (since it's char). p will now point to 11
 (int *) will say that when de-referenced 2 bytes should be extracted. So the
 2 bytes extracted are 11, 12. Numbers in these bytes are 02 and 00

 10- 00*02* ( num 2 in hex)
 12- *00*03 ( num 3 in hex)
 14- 0004 ( num 4 in hex)

 now (char *) says extract 1 byte for me. The extracted byte is 00. Hence 0
 is printed

 *Correct me if i am wrong.*


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Re: [algogeeks] c++ multi dimensional arrays

2011-08-16 Thread pacific :-)
In memory arrays are stored in row major ordering.If you dont provide the
column size then the system cannot make any access to any of the location in
the array.
when you actually look for the value at a[4][2] what the system does is that
it computes the location as  (4 * column_size + 2 ) * (int size ) and then
makes the reference a[ new_address ].So if you dont provide the column_size
it cannot perform the above operation.Try thinking how would you go about
doing that with out column_size for a[4][2] ? Not possible.

When you do int * * a , you dont need because a + 1 automatically moves to
the next row without the need of column size.Here it is because the
underlying implementation is pointer to pointer and it increments
accordingly to the next row but not next element in the same row when you do
a + 1.

Let me know if im wrong anywhere in my explanation.

On Wed, Aug 17, 2011 at 12:04 AM, priya ramesh 
love.for.programm...@gmail.com wrote:

 Why should the column size be mandatorily passed to a function which
 expects a multi dimensional array

 int arr (int marray[][5], int rows);
 int arr (int marray[][], int rows, int cols);

 1st is valid whereas second is invalid. why??

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Re: [algogeeks] Re: Probability Puzzle

2011-08-16 Thread pacific :-)
I'm little late but I too got 17/18.

On Tue, Aug 16, 2011 at 10:47 PM, Jacob Ridley jridley2...@gmail.comwrote:

 I think there is some ambiguity in the question.

  (All this time you don't know you were tossing a fair coin or not).
 1) Does the above statement mean that the thower don't know whether he
 or she threw a fair coin even after throwing? Or is the thrower not
 informed beforehand that one of them is not a fair coin?
 2) Does the coin count reduce after every throw or should it be put
 back?
 3) Depending on 1) and 2), there will be different answers.


 On Aug 9, 12:13 am, Maddy madhu.mitha...@gmail.com wrote:
  I think the answer is 17/80, because
  as you say the 5 trials are independent.. but
  the fact that a head turns up in all the 5 trials, give some
  information about our original probability of choosing the coins.
 
  in case we had obtained a tail in the first trial, we can be sure its
  the fair coin, and so the consecutive trials would become
  independent..
 
  but since that is not the case, every head is going to increase the
  chance of choosing the biased coin(initially), and hence affect the
  probability of the next head..
 
  before the first trial probability of landing a head is 3/5, but once
  u see the first head, the probability of landing a head on the second
  trial changes to 4/5*1/4+1/5, and so on..that is, there is a higher
  probability that we chose a biased coin, rather than the fair coin.
 
  hope its clear..
 
  On Aug 7, 11:36 pm, sumit gaur sumitgau...@gmail.com wrote:
 
 
 
   (3/5)
 
   On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote:
 
A bag contains 5 coins. Four of them are fair and one has heads on
both sides. You randomly pulled one coin from the bag and tossed it 5
times, heads turned up all five times. What is the probability that
you toss next time, heads turns up. (All this time you don't know you
were tossing a fair coin or not).- Hide quoted text -
 
  - Show quoted text -

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Re: [algogeeks] MS test

2011-08-06 Thread pacific :-)
I believe the answer for fifth one is o(n2)  , f(1) + f(2) + ... + f(n/2) (
increment the counter alternatively). Im not very sure.

On Sat, Aug 6, 2011 at 10:39 PM, siddharam suresh
siddharam@gmail.comwrote:

 what is *Page segmentation?*
 Thank you,
 Siddharam



 On Sat, Aug 6, 2011 at 10:35 PM, ankit sambyal ankitsamb...@gmail.comwrote:

 @neha: Cud u explain how r u getting d option for ques 4 ??

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[algogeeks] interview questions

2011-08-03 Thread pacific :-)
I couldn't find the solution to this problem please help.

Find the *first unique* string from a list of strings  in *one pass.*
*
*
*unique : *It occurs only once in the list.
*one pass : *you are allowed to traverse the list only once.

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Re: [algogeeks] direct i

2011-07-27 Thread pacific :-)
O(nlogn)
1. precompute the minimum of [ i+1 N] and store in b[i]
2. Now do a binary search for a[i] in b[i+1] in the range of  b[i+1. N]

On Wed, Jul 27, 2011 at 12:27 PM, salvador_cerinza 
vishwakarma.ii...@gmail.com wrote:

 Let say stack S.
 1.insert elements in S of A[] from right to left.
 2.int val = S.top();
 3.S.pop();
 4.now check val with S.top() until u find any element smaller than val.
 5.Note down the element pop it from stack
 6.if step 4 is true , the push val in stack S and all elements which were
 popped in the order they were popped except the last matched candidate
 element.

 Yeah..dis algo is not very efficient..


 On Wed, Jul 27, 2011 at 12:20 PM, Pankaj jatka.oppimi...@gmail.comwrote:

 Can you please elaborate a little about your stack based solution. I was
 thinking of using queue but was unable to make a perfect algo.


 On Wed, Jul 27, 2011 at 12:18 PM, salvador_cerinza 
 vishwakarma.ii...@gmail.com wrote:

 i m  suggesting stack  not just for best case only .


 On Wed, Jul 27, 2011 at 12:16 PM, Pankaj jatka.oppimi...@gmail.comwrote:

 Even in array best case can be O(n). Why use stack?
 On Wed, Jul 27, 2011 at 12:14 PM, salvador_cerinza 
 vishwakarma.ii...@gmail.com wrote:

 Best case : O(n)
 Worst case : O(n^2)
 can be done using stack.

 Thinking of better solution. .


 On Wed, Jul 27, 2011 at 11:50 AM, ankit sambyal 
 ankitsamb...@gmail.com wrote:

 O(n^2) algo is trivial. Can anybody think of a better approach ???

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Re: [algogeeks] Re: AMAZON DIRECT-I !!!!!!

2011-07-25 Thread pacific :-)
+1 to the idea of a shared document.

A spreadsheet with multiple sheets for different companies should be
awesome.But the solution to the problem might as well be discussed in gmail
threads.

I find it difficult to find the best solution to a problem discussed in
gmail threads.Having a spreadsheet should help us to fill in the best
possible answer next to each question.

I personally dont like those websites like careercup etc (though they are
good) just for the reason that for each question u need to visit a diffrent
link/page ,also hard to find the best answer to a question.

What are your opinions ?

On Mon, Jul 25, 2011 at 8:04 PM, shady sinv...@gmail.com wrote:

 i would recommend you people to share questions related to each company on
 some document like this... it has access to all and anyone can edit it
 you can append questions to it in future also.. just a little more
 systematic then sharing it on thread,,, and one week later someone else will
 be asking the same question... :D

 https://docs.google.com/document/d/1GhsStHu3WaCpbGd2PzpiyTvg_IY53dxLc6g6LzhlyKo/edit?hl=en_US


 On Mon, Jul 25, 2011 at 7:52 PM, swetha rahul swetharahu...@gmail.comwrote:

 Guys pls dont reveal the placements schedule of AU here... Hope u knw dat
 v r instructed not to do so..


 On Mon, Jul 25, 2011 at 7:47 PM, Mani Bharathi 
 manibharat...@gmail.comwrote:

 ya :) :)

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Re: [algogeeks] Re: MS

2011-07-23 Thread pacific :-)
heapsort ?

On Sat, Jul 23, 2011 at 1:26 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:

 better than O(n^2)..


 On Sat, Jul 23, 2011 at 1:08 PM, Akshata Sharma akshatasharm...@gmail.com
  wrote:

 Given a string *Str of ASCII characters, write the pseudo code to remove
 the duplicate elements present in them. For example, if the given string is
 Potato, then, the output has to be Pota. Additional constraint is, the
 algorithm has to be in-place( no extra data structures allowed) . Extend
 your algorithm to remove duplicates in the string which consisted of UNICODE
 characters.



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Re: [algogeeks] Re: MS

2011-07-23 Thread pacific :-)
By heapsort i meant : After building the min heap(or max heap) ,we extract
the top of heap and put it at the end of the array constraint to
one condition that if the element already exists at a[heap-size+1] then dont
add just discard and continue.

O(nlogn) and inplace with no extra memory.But even i believe that 26 bits is
not extra memory may be bitmask is better.

On Sat, Jul 23, 2011 at 5:15 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:

 @ross, I have posted my question in a new thread, not as a reply to any
 existing thread!!!


 On Sat, Jul 23, 2011 at 5:13 PM, ross jagadish1...@gmail.com wrote:

 check for visited can also be implemented by using an integer variable
 and setting corresponding bits!

 On Jul 23, 4:39 pm, ross jagadish1...@gmail.com wrote:
  @akshata sharma:
  Kindly post a new question as a separate thread and not as a reply to
  an existing one so tat it would be noticed by many ppl!
  As akash suggestd, use a bit vector called 'visited' of 26 size for
  ASCII or of a larger size in case of Unicode ( still constant space
  and i dont think declaring 26 variables counts as an additional DS!!),
  if visited then , ignore the character while processing.
  a simple algorithm,
  int last_ptr=0;
  for ( i = 0 - N )
  {
  if(visited(a[i])) continue;
  else a[last_ptr++]=a[i];
  visited(a[i]) = true;}
 
  a[last_ptr]=NULL;
  print (%s,a) ;
 
  On Jul 23, 12:56 pm, Akshata Sharma akshatasharm...@gmail.com wrote:
 
 
 
 
 
 
 
   better than O(n^2)..
 
   On Sat, Jul 23, 2011 at 1:08 PM, Akshata Sharma
   akshatasharm...@gmail.comwrote:
 
Given a string *Str of ASCII characters, write the pseudo code to
 remove
the duplicate elements present in them. For example, if the given
 string is
Potato, then, the output has to be Pota. Additional constraint
 is, the
algorithm has to be in-place( no extra data structures allowed) .
 Extend
your algorithm to remove duplicates in the string which consisted of
 UNICODE
characters.

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Re: [algogeeks] Re: Negative index of array

2011-07-20 Thread pacific :-)
This works.
code
#include iostream
#include map

using namespace std;
int main()
{
  mapint,int a;
  a[-1] = 0;
  couta[-1]endl;
}
/code

On Wed, Jul 20, 2011 at 7:50 PM, ankit sambyal ankitsamb...@gmail.comwrote:

 You can make the following structure :

 #define MAX 100
 typedef struct
 {
  int count_positive;
  int count_negative;
 }Element;

 typedef Element Map[MAX];

 Now you can just create a map as :
 Map map;

 Now for every element read, first check whether it is +ve or -ve. Use
 the absolute value of the number read as the key.
 Increment count_positve if key is +ve and increment count -ve if key is -ve

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[algogeeks] interview questoin

2011-07-19 Thread pacific :-)
Find unique string from a list of strings in one pass.

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Re: [algogeeks] Re: interview questoin

2011-07-19 Thread pacific :-)
sorry.

On Tue, Jul 19, 2011 at 9:07 PM, Shubham Maheshwari 
shubham.veloc...@gmail.com wrote:

 what is meant by unique string ...!!

 A string which occurs only once.


 On Tue, Jul 19, 2011 at 9:04 PM, SAMMM somnath.nit...@gmail.com wrote:

 There is only one unique string  in the list of strings (words) ?

 There can be many.



 On Jul 19, 8:31 pm, pacific :-) pacific4...@gmail.com wrote:
  Find unique string from a list of strings in one pass.
 
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 ShubZz
 O.o o.O

 enJoY ...!!!

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Re: [algogeeks] MS Question

2011-07-07 Thread pacific :-)
Hi,

You may look out for plagirism detectors.

My approach would be :
1. Hash all the keywords in one file and keep the count.
2. For each keyword in the other file , check if it exists in the hash table
, decrement its count. Also increment a counter which represents the
similarity between the two docs.

For percentage you might also count the total keywords in the second doc and
do found keywords/ total keywords.

On Wed, Jul 6, 2011 at 11:41 AM, Navneet Gupta navneetn...@gmail.comwrote:

 See diff documentation. It's an application of Longest Common
 Subsequence problem.
 http://en.wikipedia.org/wiki/Diff

 On Wed, Jul 6, 2011 at 11:12 AM, priyanshu priyanshuro...@gmail.com
 wrote:
  What is the most efficient way to compare two text documents?? Also we
  need to find the percentage by which they match..
 
  Thanks,
  priyanshu
 
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Re: [algogeeks] Re: Complexity QuickSort

2011-07-07 Thread pacific :-)
Another question reg. quicksort.

If we always find the median in o(n) and use it as the pivot ,will the
quicksort by o(nlogn) ( i mean worst case also o(nlogn) )?

Since partition is anyway o(n) , im making it o(n) + o(n){for finding
median}.

On Wed, Jul 6, 2011 at 2:50 AM, Ritesh Srivastava
riteshkumar...@gmail.comwrote:

 It will be O (n*log(n)).
 Recurrence relation will be

 T(n) = T(n/4) + T(3n/4) + O(n)

 Lets just say

 n=4^p
 so p=log n in base 4
 so this will be the number of steps the array will be divided to get
 trivial constant size array.
 at every step , processing done will be O(n) because

 n --  for first step
 (n/4) +(3n/4) -- for second step == n/4 for the first partition and 3n/4
 for the second partition obtained in the first step
 ( n/4 * 1/4 + n/4 * 3/4) + ( 3n/4*1/4 + 3n/4*3/4) --third step
 ...
 ...
 so on
 for logn steps.
 Hence O(n *logn ) Omitting constant factor of base 4.

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[algogeeks] Design questions in interviews

2011-07-06 Thread pacific :-)
Hi all ,

Can someone point to some websites where you can find  cs design questions
?

Eg. Design a data structure for sparse matrix ?

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Re: [algogeeks] Re: Yahoo Question

2011-06-30 Thread pacific :-)
Oh I got it.
If ( interview at google)
{
   Map reduce
} else if(interview at yahoo) {
   Hadoop
} else {
   Your personal preference.
}

On Thu, Jun 30, 2011 at 4:02 AM, bittu shashank7andr...@gmail.com wrote:

 1.Use Haddop  Map Reduce Framework .Obviously We Need Distributed
 Algo  we will make one computer as master  assign the job to all
   slave computer to do the crawling the web depending upon the
 geographic area ( m thinking real time problem).to crawled the
 maximum
   pages in least time we need above framework or any other
 distributed framework like google map reduce or GFS.
   computers are given for maximizing the crawling function 
 minimizing the the crawling time time..

   Algorithmically you ca think of its like a graph which has 100
 connected components in it we will bfs to traverse each computer to
 find out
   the number of pages it has been crawled  till now.

  i have given some overview hope it will help


 Thanks
 Shashank I Don't Do Code to Code But I Do Code to Build Product
 Computer Science  Engineering
 Birla Institute of Technology,Mesra

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Re: [algogeeks] Binary Tree

2011-06-30 Thread pacific :-)
My approach :

int FindPath(Node n , sum s )
{

FindPath(n-left , s ) ||
FindPath(n-right , s)  ||
OneOf { a1 =  AllSumsFromLeafToRoot( n-left )  }  + n-value +
Oneof { a2 = AllSumsFromLeafToRoot( n-right ) == s ) }
  // Use a1 and a2 to find out the actual path
}

Here all possible sums from a node to the leaves can be precomputed in O(n)
(one inorder traversal) and so lookup can be made in O(1).
For each node store  child_left , sum1,sum2,sum3 ...  ,  child_right
, sum1,sum2,sum3 ...  We may need this data to traverse the tree again
with the sum values to find the path.

Looks complicated to me too, waiting for comments.

On Thu, Jun 30, 2011 at 4:17 PM, sagar pareek sagarpar...@gmail.com wrote:

 Try traversing in LEVEL order and maintain array for each level...(Levels
 can be found by height of tree)
 then try by calculating sums starting from root(level zero) to higher
 level.
 Its typical to explain here but i found it as a solution and that will
 definitely found the solution.

 On Thu, Jun 30, 2011 at 9:08 AM, rajeev bharshetty 
 rajeevr...@gmail.comwrote:

 By traversing tree either preorder or inorder or postorder and storing the
 partial sums along the way and comparing the partial sums with the required
 sum can solve the problem


 On Wed, Jun 29, 2011 at 7:56 PM, Akshata Sharma 
 akshatasharm...@gmail.com wrote:

 How to find a path in a given binary tree which sums up to a given target
 value?
 for example if the given BT is

5
   / \
  3   2
  /
 7
 and if the target is 10, then the path is  root(5) + left node(3) +
 right node (2).

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Re: [algogeeks] Re: spoj shlights

2011-06-27 Thread pacific :-)
Can one of you provide some hints in solving this problem ?

On Sat, Jun 25, 2011 at 3:34 PM, kartik sachan kartik.sac...@gmail.comwrote:

 @jitendra that's what i am asking forwhat algo i should
 implement  to get process in 1 sec?

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Re: [algogeeks] (OOPS) Composition VS Inheritance

2011-06-26 Thread pacific :-)
The problem with inheritence is that it is compile time(i.e a class A
inheriting from class B cannot be modified again)  whereas composition can
be used to change the objects during runtime(by having a base class pointer
we can change the objects runtime).

Correct me if i'm wrong.

On Sun, Jun 26, 2011 at 7:13 AM, HowTechStuffWorks 
howtechstuffwo...@gmail.com wrote:

 Why Composition is said to be good ahead of inheritance. I am just
 learning C++ and it was said inheritance can be handled only by expert
 programmers, but I dont see anything like that. Can some one give me
 an example. Thanks in advance.

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Re: [algogeeks] Re: Sort - Consecutive Array in O(n)

2011-06-25 Thread pacific :-)
My approach :
1. Find the median.
1.5 Check if the median is min  + (max - min ) / 2.
2. Partition the array into two sub arrays using the partition function of
quicksort.
3. Check if the subarrays also satisfy the constraint.

Complexity : T(n)  = 2 T(n/2) + O(1) :: O(nlogn)




On Sat, Jun 25, 2011 at 12:15 PM, varun pahwa varunpahwa2...@gmail.comwrote:

 will this work.
 n size of array.
 cal (a[i] - min(arr) + 1).
 now cal sum of a[i], cal square sum of array as (a[i] * a[i]) , cal cube
 sum of array as (a[i] * a[i] * a[i]). now if array elements are consecutive
 then sum must be n * (n + 1) / 2. square sum must be (n * (n + 1) * (2n + 1)
 )/ 6 and cube sum must be (n * (n + 1) / 2) ^ 2.



 On Fri, Jun 24, 2011 at 11:00 PM, Adarsh s.adars...@gmail.com wrote:

 I think I got an work around for this if number of elements are
 not odd why not make them odd :)
 I variation to my prev algo

 int n = A.size();
 for (int i=0; in; i++)
total += A[i];
 findMinMax(A[1...n]); //returns first smallest (fmin), second smallest
 (smin) and largest (max) element in array

 int fmean = (max+fmin)/2;
 int smean = (max+smin)/2;
 stotal = total - fmin;
 if ((total - n*fmean) == 0)
 {
if ((stotal - n*smean) == 0)
printf(consecutive\n);
return;
 }
 printf(not consecutive\n);

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 7th Sem.
 Indian Institute of Information Technology Allahabad.
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 Official Email :: rit2008...@iiita.ac.in
 Another Email :: varunpahwa.ii...@gmail.com

 People who fail to plan are those who plan to fail.

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Re: [algogeeks] SET IN C

2011-06-24 Thread pacific :-)
Disjoint set data structure.

Refer cormen.

On Fri, Jun 24, 2011 at 10:36 PM, hary rathor harry.rat...@gmail.comwrote:


 Can anyone suggest me how to implement  a set in c .
 It means that it should take  O(1)  in insertion, deletion , finding
 element





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 Master Of Computer Application ,
 Deptt of computer Science,
 north campus , university of delhi,
 New Delhi   pin no - 110007


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[algogeeks] Writing good code in interviews

2011-06-24 Thread pacific :-)
Hey all,

I want to know a quick  list of things (positive and negative) looked  in
the code written during interviews.

I think these are a couple of things to take care.

1. Writing compilable code is definitely good.
2. Variables and methods should indicate their need.

Can you guys add more learnt from your experience ?



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Re: [algogeeks] Minimum draws for correct labels

2011-06-21 Thread pacific :-)
Is it 4 draws if we were to identify them back ?

On Tue, Jun 21, 2011 at 9:22 AM, Navneet Gupta navneetn...@gmail.comwrote:

 Nice..!

 On Tue, Jun 21, 2011 at 2:36 AM, oppilas . jatka.oppimi...@gmail.com
 wrote:
  Yes. One draw only.
  As all the box are incorrectly labeled so, box BW willl either contain
  2Black or 2 white balls.
  We pick one ball from BW box. If it's white( it contains WW ball), then
 as
  the box label WW does not contain while, it can either have BB or BW
 ball.
  But it must have BB ball inside it otherwise, we will end up getting BB
 in
  correct labeled BB box.
 
 
 
  On Tue, Jun 21, 2011 at 12:11 AM, Radhika Renganathan
  radi.coo...@gmail.com wrote:
 
  one drawing ?!
 
 
  On 6/21/11, Navneet Gupta navneetn...@gmail.com wrote:
   IMAGINE THAT YOU have three boxes, one containing two black
   marbles, one containing two white marbles, and the third, one
   black marble and one white marble. The boxes were labeled for
   their contents-BB, WW and BW-but someone has switched
   the labels so that every box is now incorrectly labeled. You are
   allowed to take one marble at a time out of any box, without
   looking inside, and by this process of sampling you are to
   determine the contents of all three boxes. What is the smallest
   number of drawings needed to do this?
  
   --Navneet
  
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Re: [algogeeks] Sum to K problem

2011-06-20 Thread pacific :-)
@vaibhav : Please note that more than two numbers can sum upto k.

On Mon, Jun 20, 2011 at 12:21 PM, vaibhav shukla vaibhav200...@gmail.comwrote:

 sort the array using merge sort : order nlogn
 take the first element,subtract it with 'k' , then search the result using
 binary search in rest of the array : order nlogn
 hence u get two elements which sum up to K in order nlogn


 On Mon, Jun 20, 2011 at 12:14 PM, Navneet Gupta navneetn...@gmail.comwrote:

 Right, in the worst case, complexity with be O(2^N).

 So what are the possible optimizations here? Would building pre-computed
 data structures with intermediate sum values help in finding such pairs in
 less complexity? I think then we can apply dynamic programming to find such
 pairs.


 On Mon, Jun 20, 2011 at 12:09 PM, oppilas . jatka.oppimi...@gmail.comwrote:

 I think its a NP problem. The solution complexity can go up O(2^N) in
 worst case.

 On Mon, Jun 20, 2011 at 11:55 AM, Navneet Gupta 
 navneetn...@gmail.comwrote:

 Given a set of integers , find a set of numbers such that they sum to a
 given number k .
 Notice the set of numbers can have 2 or more than 2 numbers.


 --Navneet

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Re: [algogeeks] Minimum Rotations

2011-06-18 Thread pacific :-)
@kk : Your approach looks like o(n^2) and only o(n) is likely to pass TLE.

On Fri, Jun 17, 2011 at 5:38 PM, KK kunalkapadi...@gmail.com wrote:

 http://www.spoj.pl/problems/MINMOVE/
 This code is showing TLE after some 20th test  case what else can be
 optimized???

 try:
import psyco
psyco.full()
 except ImportError:
pass

 string = input()
 minlen = string
 length = len(string)

 string += string[:]
 #print(string)

 index = 0
 for i in range(1, length):
if string[i : i+length]  minlen:
minlen = string[i : i+length]
index = i

 print(index)


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[algogeeks] spoj problem acode

2011-06-01 Thread pacific :-)
Link : https://www.spoj.pl/problems/ACODE/

25114
BEAN’, ‘BEAAD’, ‘YAAD’, ‘YAN’, ‘YKD’ ‘BEKD’.
How many different decodings?”

My soln , but i get TLE.Please help.

#include iostream
#include cstdio
#include vector
using namespace std;

char * head;
int result[5001];
int count(char * a ,int size)
{
  if(result[a-head]!=0){
return result[a-head];
  }

  if(size==1)
return 1;
  else if (size==2)
{
  if(a[0]'2')
return 1;
  else
return 2;
}
  else
{
  int temp = count(a+1,size-1);
  if(a[0]'2' || (a[0]=='2'  a[1]'6'))
  {
result[a-head] = temp ;
return temp;
  }
  else
  {
int r = temp+count(a+2,size-2);
result[a-head] = r;
return r;
  }
}
}

int main()
{
  char ch;
  cinch;
  while(ch!='0')
{
  char input[5001];
  int index=0;
  while(ch!='\n')
{
  //input.push_back(ch-'0');
  input[index]=ch;
  index++;
  scanf(%c,ch);
}

  cinch;
  head = input ;
  for(int i=0;i=5000;i++)
result[i]=0;
  coutcount(input,index)endl;
}
return 0;
}

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Re: [algogeeks] Admin of Group

2011-05-26 Thread pacific :-)
+1

On Wed, May 25, 2011 at 11:48 PM, anuj agarwal coolbuddy...@gmail.comwrote:

 Hi,

 Are there any admins on this group? I recently joined this group and i
 liked the discussions going on here.

 But there is a problem which hurts is that unlike other groups which are
 moderated by certain admin guys (also active members of group), this one is
 not.
 So we are receiving lots of SPAMS. I guess no body wants to get a job from
 this group (specially ones which are sent by some Panzer).

 If the mods are reading this, Please take some action on that mails so that
 group will remain clean.

 Sorry for off topic.

 Anuj Agarwal
 Engineering is the art of making what you want from things you can get.

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Re: [algogeeks] Re: GOOGLE Q

2011-05-19 Thread pacific :-)
If we were given two strings and asked to check if they have the same
characters (anagrams) :

@ gene : you are sorting them both ,and trying to match.
cost : sort the first string + sort the second string + compare i.e k * k +
k * k + k .. k is the length of the string.
I presume that bubble sort is nearly optimal for smaller strings if u
consider the overall performance ( its O(n^2) but smaller constants than the
O(nlogn) with larger constants in say quicksort.

Rather i would suggest , take each character and check that in the other
string. O( k * k) ..in the average case you might do even less than nearly
O(k * k/ 2) if the two strings are not same.

On Wed, May 18, 2011 at 10:31 AM, anuj agarwal coolbuddy...@gmail.comwrote:

 Same method as Gene told.
 Only enhancement u can made is start from the word nearer to sorted string
 and compare till the nearest word of the reverse of sorted string.
 You don't need to check the whole dictionary.

 Anuj Agarwal

 Engineering is the art of making what you want from things you can get.



 On Wed, May 18, 2011 at 6:01 AM, Gene gene.ress...@gmail.com wrote:

 Sort the characters in the string. Go through the dictionary sorting the
 characters in each word in turn.  Print the words whose sorted versions
 match the sorted string.

 You can quickly print all equivalence classes of anagrams in the
 dictionary by hashing with the sorted strings as keys. It only takes a few
 seconds to get them all this way with a 2-line perl or ruby script.

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Re: [algogeeks] Re: Google Q

2011-05-15 Thread pacific :-)
perfect.

Thanks for the effort in explanation.

On Sun, May 15, 2011 at 12:20 AM, Dave dave_and_da...@juno.com wrote:

 @Pacific: A balanced binary tree is commonly defined as a binary tree
 in which the height of the two subtrees of every node never differ by
 more than 1. Thus, there could be more nodes in one subtree than in
 the other. E.g., a balanced binary tree with 11 nodes could have 7
 nodes in the left subtree and only 3 nodes in the right subtree. Thus,
 the root would not be the median.

 An additional condition is needed: the number of nodes in the left
 subtree differs by at most one from the number of nodes in the right
 subtree.

 In fact, given that the heap structure is a balanced binary tree
 structure with implicit pointers to the left and right subtrees, the
 two-heap algorithm I described results in a balanced binary tree
 satisfying this additional condition, with an implicit root node equal
 to the median.

 Dave

 On May 14, 11:55 am, pacific :-) pacific4...@gmail.com wrote:
  Will not a balanced binary tree do the job ? if we will pick the root
 each
  time for the median.
 
 
 
 
 
  On Sat, May 14, 2011 at 9:10 PM, Dave dave_and_da...@juno.com wrote:
   @Ashish: The idea is to keep two heaps, a max-heap of the smallest
   half of the elements and a min-heap of the largest elements. You
   insert incoming elements into the appropriate heap. If the result is
   that the number of elements in the two heaps differs by more than 1,
   then you move the top element from the longer heap into the other one,
   thereby equalzing the number of elements. Thus, inserting an element
   is an O(log n) operation. To get the median, it is the top element of
   the longer heap, or, if the heaps are of equal length, it is the
   average of the two top elements. This is O(1).
 
   Dave
 
   On May 14, 8:34 am, Ashish Goel ashg...@gmail.com wrote:
not clear, can u elaborate..
 
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
 
On Fri, May 13, 2011 at 7:15 PM, Bhavesh agrawal 
 agr.bhav...@gmail.com
   wrote:
 
 This problem can be solved using 2 heaps and the median can always
 be
 accessed in O(1) time ,the first node.
 
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Re: [algogeeks] Re: Google Q

2011-05-14 Thread pacific :-)
Will not a balanced binary tree do the job ? if we will pick the root each
time for the median.


On Sat, May 14, 2011 at 9:10 PM, Dave dave_and_da...@juno.com wrote:

 @Ashish: The idea is to keep two heaps, a max-heap of the smallest
 half of the elements and a min-heap of the largest elements. You
 insert incoming elements into the appropriate heap. If the result is
 that the number of elements in the two heaps differs by more than 1,
 then you move the top element from the longer heap into the other one,
 thereby equalzing the number of elements. Thus, inserting an element
 is an O(log n) operation. To get the median, it is the top element of
 the longer heap, or, if the heaps are of equal length, it is the
 average of the two top elements. This is O(1).

 Dave

 On May 14, 8:34 am, Ashish Goel ashg...@gmail.com wrote:
  not clear, can u elaborate..
 
  Best Regards
  Ashish Goel
  Think positive and find fuel in failure
  +919985813081
  +919966006652
 
  On Fri, May 13, 2011 at 7:15 PM, Bhavesh agrawal agr.bhav...@gmail.com
 wrote:
 
 
 
   This problem can be solved using 2 heaps and the median can always be
   accessed in O(1) time ,the first node.
 
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Re: [algogeeks] first fit bin packing

2011-05-14 Thread pacific :-)
i think we can use heaps for this problem..bring to the root which has
capacity to hold and pick the root each time. If the root cannot accomodate
then no other node will be able to accomodate.

On Sat, May 14, 2011 at 1:26 AM, MK stardust...@gmail.com wrote:

 Consider the first fit strategy for online bin packing.

 So if you have N bins numbered 1, 2, 3..N and you are given value V,
 you scan them from left to right and insert V into the first bin that
 currently has enough capacity.

 In the naieve case, N insertions will take O(N^2) time.

 How can this be done in NlogN time?

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Re: [algogeeks] exon chaining

2011-05-12 Thread pacific :-)
Can you provide example input and output ?

On Thu, May 12, 2011 at 9:32 AM, ganesha suresh.iyenga...@gmail.com wrote:

 Can some one help in finding out the bug in the below code.

 Input: (left,right,weight) representing intervals
 Output: maximum weight of non-overlapping intervals

 #include iostream
 #include vector
 #include math.h
 #include algorithm

 struct point
 {
int value;
bool isLeft;
long int weight;
int index;
struct point* prev;
 };

 typedef struct point* NODEPTR;

 bool compare (NODEPTR A,NODEPTR B)
 {
return (A-value  B-value);
 }

 int main()
 {

int numIntervals;
cin  numIntervals;

// read the intervals and sort the list
vectorNODEPTR points;

for (int i=0; i  numIntervals; i++)
{
NODEPTR p1 = new point;
p1-isLeft = true;
p1-weight = 0;
p1-prev = NULL;
cin  p1-value;
points.push_back(p1);

NODEPTR p2 = new point;
p2-isLeft = false;
p2-prev = p1;
cin  p2-value;
cin  p2-weight;
points.push_back(p2);
}

sort(points.begin(),points.end(),compare);


vectorNODEPTR::iterator it;
int idx=1;
for (it=points.begin(); it!=points.end(); ++it)
{
(*it)-index = idx++;
}

/*for (it=points.begin(); it!=points.end(); ++it)
{
if ((*it)-prev != NULL)
cout  (*it)-prev-value (*it)-value 
(*it)-
 weight (*it)-prev-index  endl;
}*/


int score[2*numIntervals + 1];
for (int i=0; i = 2*numIntervals; i++)
{
score[i] = 0;
}

int i = 1;
for (it=points.begin(); it!=points.end(); ++it)
{
// right end of the interval
if (false == (*it)-isLeft)
{
int j = (*it)-prev-index;

int t1 = score[j] + (*it)-weight;
int t2 = score[i-1];

if (t1  t2)
score[i] = t2;
else
score[i] = t1;
}
else
{
score[i] = score[i-1];
}

i++;
}

cout  Max weight is   score[2*numIntervals]
 2*numIntervals  endl;

return 0;
 }

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Re: [algogeeks]

2011-05-12 Thread pacific :-)
Just a try for :

You have a billion urls, where each has a huge page. How to you detect the
duplicate documents?

1. Compute a simple hash for each of the document.
2. And for those docs where the hash matches ,
3.  use a different hash function and go back to step 1..

you might want to stop when u after a certain similarity.

On Thu, May 5, 2011 at 7:33 PM, sourabh jakhar sourabhjak...@gmail.comwrote:

 You have a billion urls, where each has a huge page. How to you detect the
 duplicate documents?

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 best way.

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Re: [algogeeks] Finding subarray

2011-05-12 Thread pacific :-)
myapproach( set S , sum s ) :
Let all elements be set  S  and we need to find sum , s
1. for each element taken from the set ,e
2. now call the function recursively with   myapproach( S - { e } , s - e )
and myapproach( S - {e } , s )

On Mon, Apr 11, 2011 at 3:17 PM, Subhransu subhransu.panigr...@gmail.comwrote:

 I didnt get the step 3. Could you please elaborate this(dry run be good to
 understand and bringing it for generic solution).
 Also how best the complexity will be .

 *Subhransu Panigrahi
 *
 *Mobile:* *+91-9840931538*
  *Email:* subhransu.panigr...@gmail.com



 On Mon, Apr 11, 2011 at 12:27 AM, ArPiT BhAtNaGaR 
 arpitbhatnagarm...@gmail.com wrote:

 well i hav deviced a approach :


 well say we hav sorted array  {-1 -3 2 3 3 5 9 13 24}

 say we hav to seach 6

 take sum of all neg no   store it   -4   means we can atmost reduce any no
 by 4 units means in any case we cant take no greater than 10-4=6 for finding
 6.

 now locate the upto position just less than we r searching for here 9

 now sum up all positive upto 9 3+2+3+3+5 ie 16   now

 WE hav 3 sets

 (1)   negative no sum
 (2) postive no less than requred sum
 (3)  greater no set  we can easily check here since only of
 this set less than 10 is useful so we hav 9 check for 9  -1  -3 here
 we get 6.

 either way we hav 16  -4 need some thing done here




 NOT PURE SOLN JUST AN IDEA THOUGH GOOD TO BE SHARE






 On Thu, Mar 31, 2011 at 1:06 AM, Subhransupanigrahi 
 subhransu.panigr...@gmail.com wrote:

 could you please point to some similar
 I just want to validate the approach which I am thinking of.


 Sent from my iPhone

 On Mar 31, 2011, at 0:59, hammett hamm...@gmail.com wrote:

  We did :-)  Dynamic programming seems as the best way to approach this
  kind of problem.
 
  On Wed, Mar 30, 2011 at 12:56 AM, Subhransu
  subhransu.panigr...@gmail.com wrote:
  For this X = {-1,4,2,3}
  Nearest will be 4 then remain is 1 but the there is no 1 so again in
  recursion nearest number is 2 then it search for suitable number where
 the
  sum is zero so 3 is not suitable then it will pick the next closest
 number
  to 2 which is one and make it (5= 4 + 2 -1 )
 
  I just propose one approach you guys also can think a better one.
 
  Subhransu Panigrahi
 
  Mobile: +91-9840931538
 
  Email: subhransu.panigr...@gmail.com
 
 
  On Wed, Mar 30, 2011 at 12:55 PM, Saikat Debnath 
 crazysai...@gmail.com
  wrote:
 
  @ Subhranshu : Your approach will fail for a case let  X = {-1,4,2,3}
 and
  your sum is 5. You will get nearest element 4, but there is no 1 in
 the set.
  A DP solution will be like this :
  Let a[i][j] be the 1 if using the first i elements we can get sum of
 j,
  and 0 otherwise. Initialise a[0][j] = 0, a[i][0] = 1;
  Now a[i][j] = a[i-1][j] | a[i-1][ j - X[i] ], where X[i] is ith
 element of
  set X.
 
  On Wed, Mar 30, 2011 at 12:35 PM, hammett hamm...@gmail.com wrote:
 
  I'd try a tabular approach. Check dynamic programming. Samples like
  LCS may give you a click.
 
 
  On Tue, Mar 29, 2011 at 11:46 PM, Subhransu
  subhransu.panigr...@gmail.com wrote:
  set of integers in an array X that the sum equals a given number N
 
  Eg. X = {-1, -3, 5, 13, 2, 24, 9, 3, 3}
 
  Lets say the number 5 which can be formed in the following manner
 5= 2
  + 3
  (the values from array). If there is no match it has to return
  invalid numbers.
 
  We also have to see the complexity of the solution.
 
  I thought of one approach but not sure if there is more approach
 where
  the
  complexity will be minimal.
  Lets say sort the array and now take number and find the closest
 number
  for
  that.
  Then in a recursion manner search for another( Lets say number '5',
 it
  search the list and found closest match
  will be 3. Then recursively search for 3 now where closest match is
 2)
 
  Any algo with better complexity will be appreciated.
 
  Subhransu Panigrahi
 
  Email: subhransu.panigr...@gmail.com
 
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Re: [algogeeks] Re: Need help on Divide and Conquer Algorithm

2011-05-12 Thread pacific :-)
a sort and another traversal would also do the same job in o( nlogn + n )
??

On Fri, Apr 15, 2011 at 12:49 AM, vishwakarma
vishwakarma.ii...@gmail.comwrote:

 complexity : O(n) + O(nlogn)

 Sweety wrote:
  Question :Let A[1..n] be an array of integers. Design an efficient
  divide and conquer algorithm to determine if A contains a majority
  element, i.e an element appears more than n/2 times in A. What is the
  time complexity of your algorithm?
 
  Answer:
  a[1..n] is an array
  int majorityElement(int a[], int first, int last)
  {
   If (first = = last)
  {
 return a[first]; // Array has one element and its count =
 1
  and it is major element
   }
  mid= (first+last)/2;
 
 (majorL,countL)= majorityElement(a,first,mid);
 (majorR,countR)= majorityElement(a,mid
  +1,last);
  n = total elements in an array;
If(majorL==majorR)
  return(countL+countR);
   else
   {
 If(countLcountR)
  return(majorL,countL);
elseif(countL countR)
  return(majorR,countR);
else
 return(majorL,majorR);
}
   if(countLn/2)
  temp1=majorL;
if(countRn/2)
   temp2=majorR;
 
 If(temp1 = = temp2)
return temp1;
elseif(countLcountR)
   return temp1;
   else (countRcountL)
  return temp2;
  else
return -1;
  }
 
  int main()
  {
int a[8] = {2,3,2,2,4,2,2,2};
int first =1;
int last=8;   //change the value of last when the array
  increases or decreases in size
int x = majorityElement(a,first,last);
if(x= = -1)
  printf(“No Majority Element”)
else
Majority element = x;
   }

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Re: [algogeeks] Re: Amazon Interview Question

2011-05-10 Thread pacific :-)
My approach :

 Have a pointer to the start (smallest of the array) of each of the N
arrays.
 Until all pointers reach end of respective arrays :
take the smallest value from all of the pointers
and compute the difference between the smallest and the current pointers
of each of the arrays
let newdiff be the smallest among all the diffs
if newdiff  olddiff :
   olddiff = newdiff
   only advance the smallest number pointer

Correctness :

I believe that if the smallest difference occurs between a and b ,(a is
smallest) you will come across that comparison and find the least.

Complexity : 0(kn) , it should be the best because you atleast need to read
all of the input.

Please correct me if i'm wrong.


On Mon, May 9, 2011 at 8:55 PM, bittu shashank7andr...@gmail.com wrote:

 see here  let me know if anything wrong..??


 http://shashank7s.blogspot.com/2011/05/wap-to-find-minimum-difference-between.html



 Thanks  Regrads
 Shashank   the Best Way to Escape From The problem is to Solve it
 Computer Science  Engg.
 BIT Mesra

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Re: [algogeeks] Need the algorithm or idea

2011-05-10 Thread pacific :-)
You may have to use a trie and also the edit distance for this problem.

Firstly , walk down the trie as you can keep matching the alphabets.
When you encounter a first mismatch , findout the edit distance for the rest
of the substring of the input with all of the strings possible from that
node down to the leaves.
Now output the one with the least edit distance as the correct spelling.
You might want to keep a bound on the max edit distance and say no
suggestions if edit distance exceeds that.


On Tue, May 3, 2011 at 9:47 AM, Sathaiah Dontula don.sat...@gmail.comwrote:

 is it not EDIT DISTANCE  (DP) problem ?

 Thanks  regards,
 Sathaiah Dontula


 On Tue, May 3, 2011 at 9:03 AM, lichenga2404 lichenga2...@gmail.comwrote:

 The question in an interview. And I got lost with this one.

 Could you guys give some algorithm or idea on  this ?


 Write a program that reads a large list of English words (e.g. from /
 usr/share/dict/words on a unix system) into memory, and then reads
 words from stdin, and prints either the best spelling suggestion, or
 NO SUGGESTION if no suggestion can be found. The program should
 print  as a prompt before reading each word, and should loop until
 killed.

 Your solution should be faster than O(n)

 For example:

  shep
 sheep
  peepple
 people
  sheeple
 NO SUGGESTION
 The class of spelling mistakes to be corrected is as follows:

 Case (upper/lower) errors: inSIDE = inside
 Repeated letters: jjoobbb = job
 Incorrect vowels: weke = wake
 Any combination of the above types of error in a single word should be
 corrected (e.g. CUNsperrICY = conspiracy).

 If there are many possible corrections of an input word, your program
 can choose one in any way you like. It just has to be an English word
 that is a spelling correction of the input by the above rules.

 Final step: Write a second program that *generates* words with
 spelling mistakes of the above form, starting with correctly spelled
 English words. Pipe its output into the first program and verify that
 there are no occurrences of NO SUGGESTION in the output

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Re: [algogeeks] Output of the code

2011-03-29 Thread pacific :-)
ruby :

putsIndia will win the worldcup 2011


On Mon, Mar 28, 2011 at 9:04 PM, shady sinv...@gmail.com wrote:

 python 2.6

 print 'India will win the World Cup 2010'


 On Mon, Mar 28, 2011 at 8:45 PM, Praveen Kumar praveen97...@gmail.comwrote:

 Here is the correct program :

 #includeiostream
 using namespace std;

 int main()
 {
while(true)
{
  coutIndia will win the World Cup 2010endl;
}

 return 0;


 }


 On Mon, Mar 28, 2011 at 8:42 PM, Praveen Kumar praveen97...@gmail.comwrote:

 true is a keyword representing 1 and false as 0. The program will print
 the line a single time.

 On Mon, Mar 28, 2011 at 11:13 AM, balaji a peshwa.bal...@gmail.comwrote:

 the code will give error as there is nothing called true defined.

 On Sun, Mar 27, 2011 at 10:38 PM, Umer Farooq the.um...@gmail.comwrote:

 Hi,

 Can anyone tell me the output of the following code?

 #include iostream.h

 int main()
 {
 .. if (true)
 .. cout  Pakistan will win the WorldCup 2011\n;
 return 0;
 }

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Re: [algogeeks] RR Scheduling

2011-03-28 Thread pacific :-)
 You can solve this problem using Binary Indexed Trees.
(copied from spoj forum)

On Mon, Mar 21, 2011 at 3:09 PM, Terence technic@gmail.com wrote:

  In this problem, sum can be as large as 5*10^9.
 Try breaking the whole interval into several stages (no more than 2*N),
 each with a fixed amount of tasks.
 Then in each stage, the schedule is a simple loop: A B C D E A B C D E A B
 ...
 Process each stage in O(1) time, then the total time complexity is O(N).



 On 2011-3-21 17:32, saurabh singh wrote:

 using scanf and printf and still tle,I am not pretty sure how malloc or new
 arrays can speed up execution?

 On Mon, Mar 21, 2011 at 2:25 PM, sanchit mittal sm14it...@gmail.comwrote:

 use scanf n printf instead of cin n cout,
 malloc array of structures after reading value of n if working in c else
 use new in cpp
 rest i guess...is ok
   On Sun, Mar 20, 2011 at 11:53 PM, ankit sambyal ankitsamb...@gmail.com
  wrote:

 I worked on this problem but cud not get a more efficient algo than
 yours.
 Plz get back 2 me if u find a better algo.


 On Sun, Mar 20, 2011 at 3:24 AM, Akshata Sharma 
 akshatasharm...@gmail.com wrote:

 I tried to solve this problem
 https://www.spoj.pl/problems/RRSCHED/

 I am getting TLE!! How can I improve my code??

 #includeiostream
 #includestdio.h

 using namespace std;

 struct process
 {
 long time;
 int finished;
 long elapsed_time;
 };

 int main()
 {
 long n,sum=0;
 cinn;
 struct process prss[5];
 for(long i=0;in;i++)
 {
 scanf(%ld,prss[i].time);
 prss[i].finished=0;
 sum+=prss[i].time;
 }
 long index=0;
  for(long k=1;k=sum;k++)
 {
   while(prss[index].finished==1)
   index++;

   prss[index].time--;

   if(prss[index].time==0)
   {
prss[index].finished=1;
prss[index].elapsed_time=k;
   }

   index++;
   if(index==n)
   index=0;
 }

 for(long i=0;in;i++)
 printf(%ld\n,prss[i].elapsed_time);
 return 0;
 }


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[algogeeks] Reg. file I/O in unix

2011-03-27 Thread pacific pacific
I have two questions :

1. How many files can a unix process open ( the same file and different
files )?

2. How many processes can open the same file (in unix) ?

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Re: [algogeeks] Binary Tree to BST

2011-03-25 Thread pacific pacific
I dont see any property of binary tree which  u can take advantage of  to
convert the  binary tree to a BST.

On Tue, Mar 22, 2011 at 9:31 PM, Decipher ankurseth...@gmail.com wrote:

 Convert Binary tree to BST in most efficient way ?

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Re: [algogeeks] Re: Print Hello

2011-03-17 Thread pacific pacific
Does  C struct have constructor ?

On Fri, Mar 18, 2011 at 12:55 AM, Don dondod...@gmail.com wrote:

 int n = printf(Hello\n);

 int main()
 {

 }

 On Mar 17, 8:08 am, himanshu kansal himanshukansal...@gmail.com
 wrote:
  is there any way to print hello in c also wdout writing anythng in
  main()
 
  On Thu, Mar 17, 2011 at 6:34 PM, kunal srivastav 
 kunal.shrivas...@gmail.com
 
   wrote:
   n...c does not support classes
 
   On Thu, Mar 17, 2011 at 6:10 PM, himanshu kansal 
   himanshukansal...@gmail.com wrote:
 
   is this possible in c also
 
   On Wed, Mar 16, 2011 at 11:57 PM, Manikanta 
 manikantabab...@gmail.comwrote:
 
   Thats right in C++. How about in C.
 
   On Mar 16, 9:44 pm, kumar anurag anurag.it.jo...@gmail.com wrote:
@anurag good.
On Wed, Mar 16, 2011 at 9:28 PM, Anurag atri 
 anu.anurag@gmail.com
   wrote:
 
 #includeiostream
 using namespace std ;
 class a
 {
 public :
 a() { couthello;}
 }a1;
 int main()
 {
 }
 
 On Wed, Mar 16, 2011 at 8:25 PM, himanshu kansal 
 himanshukansal...@gmail.com wrote:
 
 How can u print hello in a c/c++ program without writing a
 single
 word in main() function
 
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Re: [algogeeks] what to learn python or perl

2011-03-15 Thread pacific pacific
I vote for python.

On Wed, Mar 16, 2011 at 10:03 AM, kracekumar ramaraju 
kracethekingma...@gmail.com wrote:

 Hello
  Python vs Perl  been battle for more than 20 years.Perl is been around 23+
 years(not sure,people say 25 years pls check) and python for 21 years.

 Python would be my choice

 1.Python achieves code readability.

 2.Python can do what perl can do.

 more on this fight you can find here
 http://infohost.nmt.edu/~tcc/help/lang/python/vsperl.html
 http://www.linuxjournal.com/article/3882

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Re: [algogeeks] Re: Given an integer n , you have to print all the ways in which n can be represented as sum of positive integers

2011-03-14 Thread pacific pacific
My approach :
partition(n) =
1 + partition(n-1)
2+partition(n-2)
3+partition(n-3)
..
..
n-1+partition(1)
n

Assuming 1+2 and 2+1 as different.


On Mon, Mar 14, 2011 at 11:53 PM, Aviral Gupta aviral@gmail.com wrote:

 you can use coin change problem as one of the solutions.

 Regards
 Aviral Gupta
 http://coders-stop.blogspot.com/

 On Mar 14, 8:43 pm, Ralph Boland rpbol...@gmail.com wrote:
  On Mar 11, 9:33 am, saurabh agrawal saurabh...@gmail.com wrote:
 
   Given an integer n , you have to print all the ways in which n can be
   represented as sum of positive integers
 
  I suggest you
 1)  generate the numeric partitions of  n.
  That's the lists of numbers in sorted order whose sum is n.
   e.g. The numeric partitions of  3 are: {(1,1,1), (1,2), 3}
 2)  For each partition generate its multiset permutations.
 
  Note: there is a formula for how many of sums of positive integers to
  n
  there are but I don't what it is.
 
  Regards,
 
  Ralph Boland

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Re: [algogeeks] Re: An interesting question

2011-02-27 Thread pacific pacific
1. do  operation on all the values in each column and store it in the first
row of each column
2. do  operation on all the values in each row and store it in the first
column of each row.
(when writing at a[0][0] do  operation with the value computed at 1.)
3. Now to find out the value at a[i][j] ,you need to do a[i][0]  a[0][j]

On Sun, Feb 27, 2011 at 12:03 PM, Rajnish rajnish.i...@gmail.com wrote:

 1.) Traverse the whole matrix and replace each 0 value with -1.
 2.) Traverse the matrix again,all the 1 values are replaced with 0 in
 the row and column of the index where a -1 value is found.
 3.) Set all -1 values to zero and we have the output array.
 time complexity: O(n^2)
 space complexity: O(1)


 On Feb 27, 2:29 am, gaurav gupta 1989.gau...@googlemail.com wrote:
  A NxN binary matrix is given. If a row contains a 0 all element in the
  row will be set to 0 and if a column contains a 0 all element of the
  column will be set to 0. You have to do it in O(1) space.
 
  example :
 
  input array :
 
  1 0 1 1 0
  0 1 1 1 0
  1 1 1 1 1
  1 0 1 1 1
  1 1 1 1 1
 
  result array :
 
  0 0 0 0 0
  0 0 0 0 0
  0 0 1 1 0
  0 0 0 0 0
  0 0 1 1 0
 
  Thanks  Regards,
  Gaurav Gupta

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Re: [algogeeks] GOOG: Design a circular buffer and write the code for the reader and writer routines

2011-02-15 Thread pacific pacific
here is my approach.

shared variables : in, out
shared memory : buf[N]

writer
while(1)
{
while(in+1 % N ==out)
   { // waiting
}
   buf[in+1]= input;
   in = in+1 % N
}


reader
while(1)
{
while(out == in)
   { // waiting
 }
   output = buf[out]
   out = out+1 % N
}


On Tue, Feb 15, 2011 at 10:35 PM, Ashish Goel ashg...@gmail.com wrote:

 as i see this would need synchronization...
 anyone tried it so far?

 Best Regards
 Ashish Goel
 Think positive and find fuel in failure
 +919985813081
 +919966006652

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Re: [algogeeks] Re: google paper/...plz help..its urgent

2011-01-17 Thread pacific pacific
thanks very much.

On Sun, Jan 16, 2011 at 5:04 PM, Lakhan Arya lakhan.a...@gmail.com wrote:

 @pacific

 Sets of size 2 can have 2 elements common with set of size greater
 than 2. for example if set is (1,2) than it is adjacent to sets like
 (1,2,3) (1,2,4), (1,2,3,4...n) etc.
 So (1,2) is adjacent to (1,2,3), (1,2,4) etc.

 On Jan 16, 1:04 pm, pacific pacific pacific4...@gmail.com wrote:
  @Lakhan
  Why are you not considering sets of size 2 ? Because two sets of size two
  cannot have both of the elements as same.
 
 
 
 
 
 
 
 
 
  On Sat, Jan 15, 2011 at 9:39 PM, Lakhan Arya lakhan.a...@gmail.com
 wrote:
   @bittu
   I don't think answer of 6th question to be a)
   No. of vertices of degree 0 will be those who didnot intersect with
   any set i exactly 2 points. All sets of size greater than equal 2 must
   intersect with any other set having exactly 2 common elements between
   them in exactly 2 points. e.g if a set is (1,2) then it will be
   adjacent to (1,2,3) , (1,2,3,4) etc..
   The sets of size 0 and 1 cannot intersect in 2 points so they all will
   be of degree 0.
   Number of Sets of size 0 --- 1
   Number of Sets of size 1 --- n
   so Total number of vertices n+1.
 
   In the similar way number of connected components will be n+2.
 
   On Jan 15, 8:44 pm, bittu shashank7andr...@gmail.com wrote:
1.c U Can verify by putting n =I where I is positive integer value
 say
n=5  try it out its so easy
 
2 a...what i have understood.
as we know that  formal grammar is defined as (N, Σ, P, S)
so  For instance, the grammar G with N = {S, A}, Σ = {a, b}, P
with start symbol S and rules
 
S → aA
A → Sb
S → ε
 
  generates { a^ib^i : =0}   so answer is A.
 
3  expected value doe discrete distributional is defined as
E(i)=sum(pi * xi);  so from my points of view ans is 1/n ...Really
 Gud
Question one has think..still thinking
 
4.b -Explaination
 
Informally the NP-complete problems are the toughest problems in NP
in the sense that they are the ones most likely not to be in P. NP-
complete problems are a set of problems that any other NP-problem can
be reduced to in polynomial time, but retain the ability to have
 their
solution verified in polynomial time. In comparison, NP-hard problems
are those at least as hard as NP-complete problems, meaning all NP-
problems can be reduced to them, but not all NP-hard problems are in
NP, meaning not all of them have solutions verifiable in polynomial
time.
 
(A) is incorrect because set NP includes both P(Polynomial time
solvable) and NP-Complete .
(B) is incorrect because X may belong to P (same reason as (A))
(C) is correct because NP-Complete set is intersection of NP and NP-
Hard sets.
(D) is incorrect because all NP problems are decidable in finite set
of operations.
 
5. The Most Typical..Still Need Time
6 a   zero degree means vertex is not connected from any other vertex
in graph
7.a
8.No Answer  Answer Comes to Be 252
15c10,14c9,10c5,10*9*8*7*6 all are greater then from output so
 say
No Answer
 
  Correct Me if I am Wrong
 
  Next Time I will Try to provide the solution of 2nd, 5th
problem ..explanations from-others are appreciated
 
Thanks  Regards
Shashank Mani Don't B Evil U Can Earn while U Learn
Computer Science  Engg.
BIT Mesra
 
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Re: [algogeeks] Re: google paper/...plz help..its urgent

2011-01-16 Thread pacific pacific
@Lakhan
Why are you not considering sets of size 2 ? Because two sets of size two
cannot have both of the elements as same.




On Sat, Jan 15, 2011 at 9:39 PM, Lakhan Arya lakhan.a...@gmail.com wrote:

 @bittu
 I don't think answer of 6th question to be a)
 No. of vertices of degree 0 will be those who didnot intersect with
 any set i exactly 2 points. All sets of size greater than equal 2 must
 intersect with any other set having exactly 2 common elements between
 them in exactly 2 points. e.g if a set is (1,2) then it will be
 adjacent to (1,2,3) , (1,2,3,4) etc..
 The sets of size 0 and 1 cannot intersect in 2 points so they all will
 be of degree 0.
 Number of Sets of size 0 --- 1
 Number of Sets of size 1 --- n
 so Total number of vertices n+1.

 In the similar way number of connected components will be n+2.


 On Jan 15, 8:44 pm, bittu shashank7andr...@gmail.com wrote:
  1.c U Can verify by putting n =I where I is positive integer value say
  n=5  try it out its so easy
 
  2 a...what i have understood.
  as we know that  formal grammar is defined as (N, Σ, P, S)
  so  For instance, the grammar G with N = {S, A}, Σ = {a, b}, P
  with start symbol S and rules
 
  S → aA
  A → Sb
  S → ε
 
generates { a^ib^i : =0}   so answer is A.
 
  3  expected value doe discrete distributional is defined as
  E(i)=sum(pi * xi);  so from my points of view ans is 1/n ...Really Gud
  Question one has think..still thinking
 
  4.b -Explaination
 
  Informally the NP-complete problems are the toughest problems in NP
  in the sense that they are the ones most likely not to be in P. NP-
  complete problems are a set of problems that any other NP-problem can
  be reduced to in polynomial time, but retain the ability to have their
  solution verified in polynomial time. In comparison, NP-hard problems
  are those at least as hard as NP-complete problems, meaning all NP-
  problems can be reduced to them, but not all NP-hard problems are in
  NP, meaning not all of them have solutions verifiable in polynomial
  time.
 
  (A) is incorrect because set NP includes both P(Polynomial time
  solvable) and NP-Complete .
  (B) is incorrect because X may belong to P (same reason as (A))
  (C) is correct because NP-Complete set is intersection of NP and NP-
  Hard sets.
  (D) is incorrect because all NP problems are decidable in finite set
  of operations.
 
  5. The Most Typical..Still Need Time
  6 a   zero degree means vertex is not connected from any other vertex
  in graph
  7.a
  8.No Answer  Answer Comes to Be 252
  15c10,14c9,10c5,10*9*8*7*6 all are greater then from output so say
  No Answer
 
Correct Me if I am Wrong
 
Next Time I will Try to provide the solution of 2nd, 5th
  problem ..explanations from-others are appreciated
 
  Thanks  Regards
  Shashank Mani Don't B Evil U Can Earn while U Learn
  Computer Science  Engg.
  BIT Mesra

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Re: [algogeeks] Explain anybody solution of Soduo Game is unique or Not......

2011-01-15 Thread pacific pacific
On Sat, Jan 15, 2011 at 6:58 PM, UMESH KUMAR kumar.umesh...@gmail.comwrote:

 Hello..


Explain and write Algorithm for* Soduo Game* and also describe
Solution is unique or Not.

 I think using a search algorithm like A* would work.

  if Anybody do you have C/C++ code please send to me.

I dont have.



 Thanks and Regards

   Umesh kumar


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Re: [algogeeks] Recursive Function For insertion in BST

2011-01-15 Thread pacific pacific
#include stdio.h
#include stdlib.h
struct node {
  int a ;
};

int insert(struct node * head  )
{
  struct node * temp;
  temp= (struct node *) malloc(sizeof(struct node ));
  printf(temp in function  %d \n,temp);
  head = temp;
}

int main()
{
  struct node * temp;
  temp= (struct node *) malloc(sizeof(struct node ));
  printf( temp before function  %d\n ,temp);
  insert(temp);
  printf( temp after function %d\n ,temp);
}

Output :
temp before function  167862280
 temp in function  167862296
 temp after function 167862280

The assignment in the function insert to the head doesn't affect in the
caller.I hope this example clears the doubt.

I suggest an implementation like this ,
struct node * insert(struct node * head , int data )  which returns the head
of the tree to the caller.

Let me know if i'm wrong anywhere.

On Sun, Jan 16, 2011 at 12:34 AM, juver++ avpostni...@gmail.com wrote:

 @above
 Of course, NO.

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Re: [algogeeks] Adobe Question

2011-01-14 Thread pacific pacific
Integers larger than which fit in 32/64 bits.


On Thu, Jan 13, 2011 at 8:58 PM, bittu shashank7andr...@gmail.com wrote:

 1. 10 test cases for entering 3 values representing sides of a
 triangle and the program giving output as scalene, isosceles or
 equilateral--Means  At Least 10



 2 .Question on a program that calculates P=R/I where R, I are integer
 inputs and P a
 floating point output. Write 10 test cases for thisAt Least 10


 Regards
 Shashank

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Re: [algogeeks] google paper/...plz help..its urgent

2011-01-14 Thread pacific pacific
On Wed, Jan 12, 2011 at 2:44 PM, snehal jain learner@gmail.com wrote:

 1. Quick-sort is run on two inputs shown below to sort in ascending
 order
 (i) 1,2,3, ….,n
 (ii) n, n - 1, n - 2, …., 2, 1
 Let C1 and C2 be the number of comparisons made for the inputs (i) and
 (ii) respectively. Then,
 a) C1  C2
 b) C1  C2
 c) C1 = C2
 d) We cannot say anything for arbitrary n
 2. Which of the following languages over {0, 1} is regular?
 a) 0i1j such that i ≤ j
 b) 0iw1j such that w ∈ {0, 1}∗ and i ≥ 0
 c) All strings of 0s and 1s such that every pth character is 0 where p
 is prime
 d) None of the above
 3. We are given a set X = {x1, x2, ..., xn} where xi = 2i. A sample S
 (which is a subset of X) is
 drawn by selecting each xi independently with probability pi = 1 / 2.
 The expected value of the
 smallest number in sample S is:
 a) 1 / n
 b) 2
 c) sqrt(n)
 d) n
 4. Let S be an NP-complete problem and Q and R be two other problems
 not known to be in
 NP. Q is polynomial time reducible to S and S is polynomial-time
 reducible to R. Which one of
 the following statements is true?
 a) R is NP-complete
 b) R is NP-hard
 c) Q is NP-complete
 d) Q is NP-hard
 5. For any string s ∈ (0 + 1)*, let d(s) denote the decimal value of s
 (eg: d(101) = 5, d(011) = 3).
 Let L = {s ∈ (0+1)* | d(s) mod 5 = 2 and d(s) mod 7 = 4}. Which of the
 following statements is
 true?
 a) L is recursively enumerable, but not recursive
 b) L is is recursive, but not context-free
 c) L is context-free, but not regular
 d) L is regular
 Common data for questions 6 and 7
 The 2n vertices of a graph G corresponds to all subsets of a set of
 size n. Two vertices of G are
 adjacent if and only if the corresponding sets intersect in exactly 2
 elements
 6. The number of vertices of degree zero in G is:
 a) 1
 b) n
 c) 2n - 1
 d) None


Nodes which dont have any edges are :
subset of the given set of size  0   --- 1
subsets of the given set of size  1 --- n
subsets of the given set of size  2 --- nc2
so answer is 1 + n + nc2.
Since all subsets of size greater than 2 share atleast two elements with
some other node.


 7. The number of connected components in G is:
 a) 2n
 b) n + 2
 c) n C 2
 d) None


I doubt if it is number of nodes of degree 0 + 1.




 8. There are 5 nested loops written as follows,
 int counter = 0;
 for (int loop_1=0; loop_1  10; loop_1++) {
 for (int loop_2=loop_1 + 1; loop_2  10; loop_2++) {
 for (int loop_3=loop_2 + 1; loop_3  10; loop_3++) {
 for (int loop_4=loop_3 + 1; loop_4  10; loop_4++) {
 for (int loop_5=loop_4 + 1; loop_5  10; loop_5++) {
 counter++;
 }
 }
 }
 }
 }
 What will be the value of counter in the end (after all the loops
 finished running)?
 a) 15C5
 b) 14C5
 c) 10C5
 d) 10 * 9 * 8 * 7 * 6 * 5

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Re: [algogeeks] Re: Amazon Interview - Algorithms

2011-01-14 Thread pacific pacific
At each location if the value is k  ,
find the largest value in the next k elements and jump there.

This greedy approach works in 0(n^2) and i believe it works. If not can
someone give me a counter example ?

On Sat, Jan 15, 2011 at 3:30 AM, Avi Dullu avi.du...@gmail.com wrote:

 @jammy Even I felt the same, but the greedy 'algo' u suggest is actually
 IMHO not a greedy approach. You just take each arr[i] and jump *without
 deciding a locally optimal policy* . SO, if u were to *see* arr[i] and
 *decide* on the optimal policy I feel one would follow d same steps as in a
 DP solution. Its only just that the implementation would be O(n^2). Just to
 add, this is the greedy approach I feel:

 greedy_min_steps[n]
 for i = 0; i  n; i++:
   for (j = 0; j  input[i]; j++)
 greedy_min_steps[ i + j ] = min(greedy_min_step[ i + j ],
 greedy_min_steps[ i ] + 1)

 this is the greedy approach I build and I see this being exactly similar to
 my DP approach. There are instances of greedy approach based algorithms
 which have *optimized* DP counter parts. I feel this problem is one of them.
 More ideas ?




 Programmers should realize their critical importance and responsibility in
 a world gone digital. They are in many ways similar to the priests and monks
 of Europe's Dark Ages; they are the only ones with the training and insight
 to read and interpret the scripture of this age.


 On Sat, Jan 15, 2011 at 2:14 AM, Jammy xujiayiy...@gmail.com wrote:

 @Avi Greedy approach doesn't work since you can't ensure the choice is
 locally optimum. Consider 3,9,2,1,8,3.  Using greedy alg. would give
 you 3,1,8,3 while otherwise DP would give you 3,9,3.

 On Jan 14, 6:11 am, Avi Dullu avi.du...@gmail.com wrote:
  I guess u got confused with the comment I wrote, I have added 2 print
  statements and now I guess it should be clear to you as to why the code
 is
  O(n). The comment means that each element of the min_steps_dp will be
  ACCESSED only ONCE over the execution of the entire program. Hence the
 outer
  loop still remains O(n). The next_vacat variable if u notice is always
  incremental, never reset to a previous value.
 
  #includestdio.h
  #includestdlib.h
 
  #define MAX 0x7fff
 
  inline int min(int a, int b) {
return a = b ? b : a;
 
  }
 
  int find_min_steps(int const * const input, const int n) {
int min_steps_dp[n], i, temp, next_vacant;
for (i = 0; i  n; min_steps_dp[i++] = MAX);
 
min_steps_dp[0] = 0;
next_vacant = 1; // Is the index in the array whose min_steps needs to
 be
  updated
 // in the next iteration.
for (i = 0; i  n  min_steps_dp[n - 1] == MAX; i++) {
  temp = i + input[i];
  if (temp = n) {
min_steps_dp[n - 1] = min(min_steps_dp[n - 1], min_steps_dp[i] +
 1);
temp = n - 1;
  } else {
printf(Updating min[%d] to %d \n, i + input[i], min_steps_dp[i]
 +
  1);
min_steps_dp[temp] = min(min_steps_dp[temp], min_steps_dp[i] + 1);
  }
  if (temp  next_vacant) {
printf(i: %d \n, i);
for (; next_vacant  temp; next_vacant++) {
printf(next_vacant: %d \n, next_vacant);
  min_steps_dp[next_vacant]
= min(min_steps_dp[temp], min_steps_dp[next_vacant]);
}
  }
}
for (i=0;in;printf(%d ,min_steps_dp[i++]));printf(\n);
return min_steps_dp[n-1];
 
  }
 
  int main() {
int n, *input, i;
scanf(%d,n);
if ((input = (int *)malloc(n * sizeof(int))) == NULL) {
  return -1;
}
for (i = 0;i  n; scanf(%d,input[i++]));
printf(Minimum steps: %d\n,find_min_steps(input, n));
return 0;
 
  }
 
  Programmers should realize their critical importance and responsibility
 in a
  world gone digital. They are in many ways similar to the priests and
 monks
  of Europe's Dark Ages; they are the only ones with the training and
 insight
  to read and interpret the scripture of this age.
 
  On Fri, Jan 14, 2011 at 1:49 PM, Decipher ankurseth...@gmail.com
 wrote:
   I don't think the inner loop is executing only once . Kindly check it
 for
   this test case {1,3,5,8,9,2,6,7,6,8,9} . And try to print i in inner
 loop
   you will find that for same values of i(Outer index) inner loop is
 called.
   Its an O(n2) solution .
 
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Re: [algogeeks] Re: Tejas Networks - Written Test

2011-01-12 Thread pacific pacific
2. can this be done in the following manner ?
print the preorder and inorder traversal of the tree and then you can build
it back but for binary trees

On Tue, Jan 11, 2011 at 1:12 PM, juver++ avpostni...@gmail.com wrote:

 1. Any shortest path algorithms on graphs (Dijkstra, Bellman-Ford, Floyd).
 2. Store it's edges while doing DFS.

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Re: [algogeeks] Re: Amazon Interview - Algorithms

2011-01-12 Thread pacific pacific
doesn't greedy approach work here ?

On Sun, Jan 9, 2011 at 8:00 PM, juver++ avpostni...@gmail.com wrote:

 It doesn't matter how to make transitions: from current position make all
 necessary moves,
 or determine all positions from which we can achieve i-th position.
 So your method is only the one of possible.

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Re: [algogeeks] BT

2011-01-12 Thread pacific pacific
Here is my pseudo code :
check( node t1 , node t2 )
{
if ( t1-data == t2-data)
{
return check( t1-left , t2-left )  check(t1-right, t2-right) ;
}
else
{
return check(t1-left , t2) || check(t1-right , t2);
}
}

time complexity : o(n) because each node in t1 needs to be visited once.let
me know if this works.


On Sun, Jan 9, 2011 at 1:30 PM, Harshal hc4...@gmail.com wrote:

 @Nishaanth
 T1 has millions of nodes. Suppose all the nodes of T1 are equal to root of
 T2. Then u will have to check every where in T1. Putting height as
 constraint, u will have to check only those nodes whose height is equal to
 T2. It will reduce time complexity.

 Well m not able to think of better time complexity, another way would be:
 Find Height of T2 ... O(k)  //k is no. of nodes in T2
 Find Height of each node in T1... O(N)  //N is no. of nodes in T1

 now if p nodes in T1 have height same as T2, then we can find if a subtree
 rooted at any of those p nodes are identical to T2 in O(pk) time.

 Thus total time complexity: O(N) + O(k) + O(pk).
 correct me if I am wrong..

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Re: [algogeeks] Google Interview Question

2010-12-27 Thread pacific pacific
here is my approach :

Starting from the root node ,
if root node need to have a 1 ...
if root is an and gate :
 flips  = minflips for left child to have value 1 + minflips for the
right child to have value 1
else
 flips = minimum of ( minflips for left child to have value 1 , minflips
for right child to have value 1)

For  a leaf node , return 0 if the leaf has the value needed else return
INFINITY.Also if at any internal node if it is not possible return INFINITY.

On Tue, Dec 28, 2010 at 10:06 AM, Anand anandut2...@gmail.com wrote:

 @Terence.

 Could please elaborate your answer. Bottom up level order traversal helps
 to get the final root value but how to use it to find minimum flips needed
 to Obtain the desired root value.




 On Fri, Dec 24, 2010 at 1:56 AM, Terence technic@gmail.com wrote:

 Using the same level order traversal (bottom up), calculating the minimum
 flips to turn each internal node to given value (0/1).



 On 2010-12-24 17:19, bittu wrote:


 Boolean tree problem:

 Each leaf node has a boolean value associated with it, 1 or 0. In
 addition, each interior node has either an AND or an OR gate
 associated with it. The value of an AND gate node is given by the
 logical AND of its two children's values. The value of an OR gate
 likewise is given by the logical OR of its two children's values. The
 value of all of the leaf nodes will be given as input so that the
 value of all nodes can be calculated up the tree.
 It's easy to find the actual value at the root using level order
 traversal and a stack(internal if used recursion).

 Given V as the desired result i.e we want the value calculated at the
 root to be V(0 or 1) what is the minimum number of gates flip i.e. AND
 to OR or OR to AND be required at internal nodes to achieve that?

 Also for each internal node you have boolean associated which tells
 whether the node can be flipped or not. You are not supposed to flip a
 non flippable internal node.


 Regards
 Shashank Mani Narayan
 BIT Mesra


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Re: [algogeeks] palindrome

2010-12-22 Thread pacific pacific
On Wed, Dec 22, 2010 at 12:11 PM, mo...@ismu mohan...@gmail.com wrote:



 if x is a  32 bit number
 if((x0x)==((x16)0x)))
x's  bit pattern is a polyndrome

 @snehal :Do you want to consider binary representation of 5 as 101 or
..0101 ?
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