[algogeeks] Amazon Interview Q

2011-08-17 Thread Piyush Sinha
We all knw hw to find a loop in a singly linked list (Using hare and
tortoise method)..what if the linked list is a doubly linked list?can
we do it smthn better than hare n tortoise method using the advantage
of the doubly linked list?

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Re: [algogeeks] Re: Facebook Interview question at NIT Warangal

2011-07-28 Thread Piyush Sinha
@amitno it wont work..check it..

On Wed, Jul 27, 2011 at 10:07 PM, amit amit.codenam...@gmail.com wrote:

 This should be O(2^n)

 #include cstdio
 #include cstring
 using namespace std;

 int n;
 int sol[1000];
 void solve(int pos, int k) {
for(int i = 0; i  k; i++) printf(%d, , sol[i]);
putchar('\n');
for(int i = pos; i  n; i++) {
sol[k] = i+1;
solve(i+1, k+1);
}
 }

 int main() {
scanf(%d, n);
solve(0, 0);
 }


 On Jul 27, 8:33 pm, Ankur Garg ankurga...@gmail.com wrote:
  Hi
 
  The solution in the link is of complexity (n*2^n))
 
  Does anyone know any better solution ?
 
  Regards
  Ankur
  On Tue, Jul 26, 2011 at 11:10 PM, rajeev bharshetty 
 rajeevr...@gmail.comwrote:
 
 
 
   @Ankur The link does has a very good explanation. Nice solution :)
 
   On Tue, Jul 26, 2011 at 10:47 PM, Kunal Patil kp101...@gmail.com
 wrote:
 
   @Ankur Garg: Nice explanation at the link given by u...
 
   On Tue, Jul 26, 2011 at 10:35 PM, Ankur Garg ankurga...@gmail.com
 wrote:
 
   Check this
 
  
 http://codesam.blogspot.com/2011/03/find-all-subsets-of-given-set.html
 
   On Tue, Jul 26, 2011 at 9:41 PM, Vishal Thanki 
 vishaltha...@gmail.comwrote:
 
   Here is the working code..
 
   #include stdio.h
   #include stdlib.h
   int a[] = {1,2,3,4,5};
   #define ARRLEN(a) (sizeof(a)/sizeof(a[0]))
   void print_comb(int len)
   {
  int tlen = len;
  int i, j, k;
   int al = ARRLEN(a);
  for (i = 0; i  al; i++) {
  for (j=i+len-1; jal;j++) {
  for (k = i; k  i+len-1; k++) {
  printf(%d , a[k]);
  }
  printf(%d\n, a[j]);
   }
  }
   }
 
   int main(int argc, char *argv[])
   {
  int len = atoi(argv[1]);
   print_comb(len);
  return 0;
   }
 
   On Tue, Jul 26, 2011 at 5:18 PM, praneethn praneeth...@gmail.com
   wrote:
 
check this link:
 
*http://www.stefan-pochmann.info/spots/tutorials/sets_subsets/*
 
On Tue, Jul 26, 2011 at 11:59 AM, sumit sumitispar...@gmail.com
   wrote:
 
Given an array of size n, print all the possible subset of array
 of
size k.
eg:-
input:
a[]={1,2,3,4}, k = 2
output:
1,2
1,3
1,4
2,3
2,4
3,4
 
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Re: [algogeeks] Array question

2011-07-26 Thread Piyush Sinha
Use stack

On Tue, Jul 26, 2011 at 5:22 PM, Shikhar shikharko...@gmail.com wrote:

 Given an array of integers, for each element of the array, print the
 first number on the right side of the element which is smaller than
 it. print -1 is there is no such element.

 eg: 3,4,2,18,19,1,10

 Ans: 2,2,1,10,10,-1,-1
 O(n^2) solution is trivial.

 One solution could be:
 Insert the elements of the array in a binary search tree. The moment a
 node in the binary tree gets a left child, it is the element we are
 looking for. All the nodes in the right subtree of a node which has
 just received a left child can be assigned the value of the parents'
 left child as the first smaller element to the right.
 Thus, this solution is O(nlogn).

 Does this solution work for all possible cases of input?

 Is an O(n) solution possible?

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Re: [algogeeks] Array question

2011-07-26 Thread Piyush Sinha
Sorry for the above mistakeits not O(n)

On Tue, Jul 26, 2011 at 5:29 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:

 Use stack


 On Tue, Jul 26, 2011 at 5:22 PM, Shikhar shikharko...@gmail.com wrote:

 Given an array of integers, for each element of the array, print the
 first number on the right side of the element which is smaller than
 it. print -1 is there is no such element.

 eg: 3,4,2,18,19,1,10

 Ans: 2,2,1,10,10,-1,-1
 O(n^2) solution is trivial.

 One solution could be:
 Insert the elements of the array in a binary search tree. The moment a
 node in the binary tree gets a left child, it is the element we are
 looking for. All the nodes in the right subtree of a node which has
 just received a left child can be assigned the value of the parents'
 left child as the first smaller element to the right.
 Thus, this solution is O(nlogn).

 Does this solution work for all possible cases of input?

 Is an O(n) solution possible?

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Re: [algogeeks] Re: Array question

2011-07-26 Thread Piyush Sinha
@Shikhar

1) Push the first element to stack.
2) for i = 1 to n-1
a) temp =a[i]
b) while(stack not empty)
 int x = pop(stack)
 if(xtemp) print(temp);
 else
  push(x,stack)
  break;
c) push(temp,stack)

3) After the loop in step 2 is over, pop all the elements from stack and
print -1 as next element for them.


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[algogeeks] AMAZON Q

2011-07-26 Thread Piyush Sinha
You have an array like ar[]= {1,3,2,4,5,4,2}. You need to create
another array ar_low[] such that ar_low[i] = number of elements lower
than or equal to ar[i] in ar[i+1:n-1].
So the output of above should be {0,2,1,2,2,1,0}

Time complexity : O(nlogn)
use of extra space allowed.


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Re: [algogeeks] plzz explain

2011-07-26 Thread Piyush Sinha
http://www.cprogramming.com/tutorial/size_of_class_object.html

On 7/27/11, TUSHAR tusharkanta.r...@gmail.com wrote:
 1.
 #includeiostream
 using namespace std;
 class abc
 {
 int x;
 static int y;
 };
 main()
 {
 abc a;
 coutsizeof(a);
 coutsizeof(abc);
 }

 why o/p is 4 4 ???


 2.
 class abc
 { };
 main()
 {
   abc ob;
   coutsizeof(ob);
 }

 why o/p is 1 ?
 some say o/p may vary from os to os .why ?? and what may be other
 o/p s ?

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Re: [algogeeks] Adove Question.........

2011-07-22 Thread Piyush Sinha
initial call

subset(a,n,r,0,0);
printf(%d,count);//to know the value of nPr (no of subsets)
*
void subset(int a[],int n,int r,int j,int ind)
{
 int i;
 if(ind==r)
 {
   printf({);
   for(i=0;iind;i++)
printf(%d ,a[i]);
   printf(}\n);
   count++;
 }
 else if (indr)
 {
  for(i=j;in;i++)
  {
 swap(a[i],a[j]);
 subset(a,n,r,j+1,ind+1);
 swap(a[i],a[j]);
  }
 }
}*

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Re: [algogeeks] Adobe ques

2011-07-21 Thread Piyush Sinha
http://www.ideone.com/4Rq51




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[algogeeks] TREES

2011-07-21 Thread Piyush Sinha
*A preorder of a complete binary tree is given in an array , prints its
level order

*
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Re: [algogeeks] Re: Solve it

2011-07-20 Thread Piyush Sinha
ya nitish above condition will do

On 7/20/11, Nitish Garg nitishgarg1...@gmail.com wrote:
 I think:
 s[i] = max(s[i-2], s[i-2]+a[i], s[i-1], a[i]) should satisfy all the cases,
 even when all the numbers are negative.
 Pleas check.

 On Wed, Jul 20, 2011 at 12:44 AM, pnandy sayantan.nand...@gmail.com wrote:



 On Jul 19, 8:00 pm, ankit sambyal ankitsamb...@gmail.com wrote:
  @Nitish and Shubam : Since we trying to find sub sequence and not a
  sub string, so if there are negative nos. in the array, just neglect
  them.
  Piyush's algo will work perfectly..

 Piyush's algo won't work for -ve nos.
 Consider an array -12 -10 5.answer should be 5 while the algo
 gives -7 as the answer.

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Re: [algogeeks] Re: Circle Circle more Circles .........

2011-07-20 Thread Piyush Sinha
I would like to redefine my algo with cases clarified...

Create a queue that is made to contain the points...

say points queue [1000];

for i:1 to n
 for j:i+1 to n
 Calculate d (distance between the two centers)
 if (d = r0 + r1) keep them in two separate queues //the circles
don't intersect
 if(d==0 || d= abs(r0-r1))
 ignore the circle with smaller radius // one circle
wholly contains another such that  the borders do not overlap, or
overlap exactly (e.g. two identical circles)
 else
  keep both of them in one single queue


Now calculate the area of the circles in those queues which have
single element...

those with more than one element..calculate the area using simple
geometry...You can take help of this..
http://mathworld.wolfram.com/Circle-CircleIntersection.html

Hope its clear now...


On 7/20/11, SAMMM somnath.nit...@gmail.com wrote:
 I doubth .

 For (d r0 + r1) ignore the point with smaller radius as it will
 overshadowed the bigger circle completely

 There may be a case where the circle is partially overlapped by the
 other circles. Then this algo will fail .

 The area will be of like these :-

 Suppose 3 circles are there X,YZ .
 Then the area will be :-

 Case1:-  X+Y+Z
 Case2:-  X+(YUZ) == Y + Z - (YnZ) --- intersection
 case3:- There circle can overlap ... like these .

 Then Will your algo work .. I guess no .

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Re: [algogeeks] Re: Circle Circle more Circles .........

2011-07-20 Thread Piyush Sinha
@Dumanshu..i am not partitioning them into just two queues...

Moreover I just gave a raw idea...and yeah the complexity is in the order of
n^2 only.
There are many chances of improvement in it..

On Wed, Jul 20, 2011 at 5:30 PM, Dumanshu duman...@gmail.com wrote:

 @Piyush:
 Initially for partitioning the given circles into the 2 queues u r
 having an O(n^2) loop, so u are comparing each circle with every
 other.
 Now, it is possible that u have 3 or more circles A,B,C intersecting
 if i got ur algo correct, ur intersection queue will have AB, BC, CA.
 So, according to the geometry, u will find the areas. But this area
 would be different than the actual area for intersection of A,B,C.

 On Jul 20, 3:48 pm, Piyush Sinha ecstasy.piy...@gmail.com wrote:
  I would like to redefine my algo with cases clarified...
 
  Create a queue that is made to contain the points...
 
  say points queue [1000];
 
  for i:1 to n
   for j:i+1 to n
   Calculate d (distance between the two centers)
   if (d = r0 + r1) keep them in two separate queues //the circles
  don't intersect
   if(d==0 || d= abs(r0-r1))
   ignore the circle with smaller radius // one circle
  wholly contains another such that  the borders do not overlap, or
  overlap exactly (e.g. two identical circles)
   else
keep both of them in one single queue
 
  Now calculate the area of the circles in those queues which have
  single element...
 
  those with more than one element..calculate the area using simple
  geometry...You can take help of this..
 http://mathworld.wolfram.com/Circle-CircleIntersection.html
 
  Hope its clear now...
 
  On 7/20/11, SAMMM somnath.nit...@gmail.com wrote:
 
 
 
 
 
 
 
 
 
   I doubth .
 
   For (d r0 + r1) ignore the point with smaller radius as it will
   overshadowed the bigger circle completely
 
   There may be a case where the circle is partially overlapped by the
   other circles. Then this algo will fail .
 
   The area will be of like these :-
 
   Suppose 3 circles are there X,YZ .
   Then the area will be :-
 
   Case1:-  X+Y+Z
   Case2:-  X+(YUZ) == Y + Z - (YnZ) --- intersection
   case3:- There circle can overlap ... like these .
 
   Then Will your algo work .. I guess no .
 
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Re: [algogeeks] Adobe ques

2011-07-20 Thread Piyush Sinha
@Anika..can u give one example??

On 7/21/11, Anika Jain anika.jai...@gmail.com wrote:
 write a c code that takes a 32 bit ip address n prints it in dotted notation
 as a.b.c.d.
 The code shall work for big endian as well as for little endian..


 In this how can i make it common for big endian and little endian?? if in
 little endian my lower order 8 bits shall be d but for big endian they shall
 be a..

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Re: [algogeeks] Microsoft Question!

2011-07-20 Thread Piyush Sinha
unsigned int reverse_bits (unsigned int n)
{
 if(n==1) return n;
 int j,p;
 for(j = 31;j0;j--)
 if(n(1j)) break; //to find the first set bit from left
 p =0;
 while(jp)
{
 int x1 = (n (1j))j;
 int x2 = (n(1p))p;
 if(x1!=x2)
 {
   if(x1)
   {
n = ~(1j);//unset the jth bit
n |= (1p); //set the pth bit
   }
   else
   {
n = ~(1p); //unset the pth bit
n |= (1j); //set the jth bit
   }
 }
 j--;p++;
   }
   return n;
}

On 7/21/11, archita kool.arc...@gmail.com wrote:
 How to reverse the order of bits of a number in minimum space
 complexity?
 if the no is 11-1011
 the output should b 1101.

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Re: [algogeeks] Memory Allocation

2011-07-19 Thread Piyush Sinha
char a[] or any array refers to a block of memory (not a single memory
location or variable). Analogy to this can be seen in the following fact :-
The memory addresses of an array can't be changed, whereas the content of
the pointer variables, such as the base memory address of the array it
refers to or simply any variable can be changed.

Hence an undefined result persists..if we really want to return an array
from a function we use the following syntax..

*char a[] = Hello;
char *b = (char *)malloc(strlen(a)+1);
strcpy(b,a);
return b;*

here we are returning the base address of the character array, unlike as wat
you were doing previously(previously you were trying to return a block of
memory)

Hope it is clear now... :)




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Re: [algogeeks] INFINITY

2011-07-19 Thread Piyush Sinha
*printf(%c\n,236);*

On Tue, Jul 19, 2011 at 2:45 PM, SAMMM somnath.nit...@gmail.com wrote:

 Print the symbol ∞  (INFINITY) through code . Unicode ..

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Re: [algogeeks] Re: INFINITY

2011-07-19 Thread Piyush Sinha
In Dev C it does

On Tue, Jul 19, 2011 at 3:08 PM, SAMMM somnath.nit...@gmail.com wrote:

 It doesn't display the infinity symbol.

 On Jul 19, 2:24 pm, Piyush Sinha ecstasy.piy...@gmail.com wrote:
  *printf(%c\n,236);*
 
  On Tue, Jul 19, 2011 at 2:45 PM, SAMMM somnath.nit...@gmail.com wrote:
   Print the symbol ∞  (INFINITY) through code . Unicode ..
 
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Re: [algogeeks] Re: amazon

2011-07-19 Thread Piyush Sinha
*node *middle(node *head)
{
 node *mid;

 mid = head;
 for(int i = 2;head!=NULL;head=head-next,i++)
 {
 (i%2==1) mid = mid-next;
 }
 return mid;
}*

To find 1/3rd of a list, change (i%2==1) to (i%3==1)i.e. nth node can be
found (i%n==1) (make sure n=no. of nodes)

Now to find 3/4th of a node, we can do following

initial call *threefourth(head,3,4);*

*node *threefourth(node *head,int m,int n)
{
 node *mid,*p;
 p = head-next;
 mid = head;
 while(m)
 {
 for(int i = 2;p!=NULL;p=p-next,i++)
 {
 (i%n==1) mid = mid-next;
 }
 m--;
 if(m==0) return mid;
 else
 mid = mid-next;
 p = head-next;

 }
}*



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Re: [algogeeks] Re: amazon

2011-07-19 Thread Piyush Sinha
Ignore my previous post for find the 3/4th...its actually traversing the
list thriceso it better you traverse the list once and find the number
of nodes in it...then then traverse 3/4th times the number of nodes...

But the algo given by me is efficient if we are required to compute 1/nth of
a node...


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Re: [algogeeks] Solve it

2011-07-19 Thread Piyush Sinha
I hope it can be solved using DP...check my algo below and give any counter
case if you get it...

1. Make an array S equal to the length of the given array where
S[0] = a[0] and S[1] = max(a[0],a[1])

2. for i:2 to n-1
 S[i] = max(S[i-2]+a[i], S[i-1])

3. return S[n-1]

Hope the above algo works...

On Tue, Jul 19, 2011 at 2:59 PM, sagar pareek sagarpar...@gmail.com wrote:

 Given an array all of whose elements are positive numbesr, find the maximum
 sum of a subsequence with the constraint that no 2 numbers in the sequence
 should be adjecent in the array.

 eg:-
 3 2 7 10 should return (sum of 3 and 10)
 3 2 5 10 7 should returnn 15 (sum of 3,5,7)

 --
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 COMPUTER SCIENCE AND ENGINEERING
 NIT ALLAHABAD

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Re: [algogeeks] Solve it

2011-07-19 Thread Piyush Sinha
I think  6+4+3  6+4+2

On Tue, Jul 19, 2011 at 4:30 PM, sagar pareek sagarpar...@gmail.com wrote:

 Piyush
 sorry dude but this will not work

 say original array be
 6  8  4  1  2  3
 then ur new array be
 6  8 10 10 12 13 //but original answer is 12

 On Tue, Jul 19, 2011 at 3:49 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:

 I hope it can be solved using DP...check my algo below and give any
 counter case if you get it...

 1. Make an array S equal to the length of the given array where
 S[0] = a[0] and S[1] = max(a[0],a[1])

 2. for i:2 to n-1
  S[i] = max(S[i-2]+a[i], S[i-1])

 3. return S[n-1]

 Hope the above algo works...

 On Tue, Jul 19, 2011 at 2:59 PM, sagar pareek sagarpar...@gmail.comwrote:

 Given an array all of whose elements are positive numbesr, find the
 maximum sum of a subsequence with the constraint that no 2 numbers in the
 sequence should be adjecent in the array.

 eg:-
 3 2 7 10 should return (sum of 3 and 10)
 3 2 5 10 7 should returnn 15 (sum of 3,5,7)

 --
 **Regards
 SAGAR PAREEK
 COMPUTER SCIENCE AND ENGINEERING
 NIT ALLAHABAD

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 NIT ALLAHABAD

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Re: [algogeeks] MS

2011-07-19 Thread Piyush Sinha
are the sizes of the two arrays same??

On 7/19/11, swetha rahul swetharahu...@gmail.com wrote:
 Hi,
 Find the kth smallest element in O(logk) given 2 sorted arrays.
 Merging the arrays is not allowed. I can do it in O(k).. How to do in
 O(logk)..

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Re: [algogeeks] MS ques

2011-07-19 Thread Piyush Sinha
Divisibility of 3 of numbers in base 2 can be seen same as
divisibility of numbers by 11 in base 10...

maintain two variable even_sum  odd_sum, both initialized to 0

when an odd location in the number is set increment odd_sum
when an even location in the number is set increment even_sum

if(abs(even_sum-odd_sum)%3==0) number is divisible by 3...

Hence keep the track of even_sum and odd_sum as the bits are getting appended..

Hope I am clear... :)

On 7/19/11, sudhanshu pandey sud.nit...@gmail.com wrote:
 use automata theory. draw dfa for divisibility by 3..

 On Tue, Jul 19, 2011 at 11:23 PM, siva viknesh
 sivavikne...@gmail.comwrote:

 Given an infinite stream of bits with bits being appended at the
 highest significant position. Give an algorithm to say whether the
 number formed by sequence of bits that had been processed till then ,
 is divisible by 3 or not ?


 My sol:

 have a variable sum...find the sum of bitswhenever u add a bit
 do sum+=bit value  ... check whether sum%3==0.
 Is my solution ok?? anyother good solutions ??

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Re: [algogeeks] Re: Circle Circle more Circles .........

2011-07-19 Thread Piyush Sinha
Just a simple thoughtI am assuming all points are unique

Create a queue that is made to contain the points...

say points queue [1000];

for i:1 to n
  for j:i+1 to n
  Calculate d (distance between the two centers)
   if (d = r0 + r1) keep them in two separate queues
   if(d r0 + r1) ignore the point with smaller radius //as it
will overshadowed the bigger circle completely
keep both of them in one single queue

Now calculate the area of the circles in those queues which have
single element...

those with more than one element..calculate the area using simple geometry...

Hope the above algo is clear...

On 7/19/11, SAMMM somnath.nit...@gmail.com wrote:
 See the input will be :-

 6 No of circles

 x1 y1 R1
 x2 y2 R2
 x3 y3 R3
 x4 y4 R4
 x5 y5 R5
 x6 y6 R6

 Output:-
 Area occupied by the above circles (one line) 4 decimal points .


 On Jul 19, 9:01 pm, priyanka goel priya888g...@gmail.com wrote:
 can u pl tell wat is dis x  y coordinate?
 are dey centre coordinates or any point on circumference of circle..

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Re: [algogeeks] MICROSOFT

2011-07-17 Thread Piyush Sinha
It can be done using one extra array only that is the output array out[]

*int out = (int *)malloc(sizeof(n)); //n is the number of elements in a
int i,temp = 1;
for(i=0;in;i++)
{
 out[i] = temp;
 temp*=a[i];
}
temp =1;
for(i=n-1;i=0;i--)
{
out[i] *= temp;
temp*=a[i];
}*


On Sun, Jul 17, 2011 at 4:43 PM, manish patel manispatel...@gmail.comwrote:

 thanx the question was asked by MS today in MNNIT.

 On Sun, Jul 17, 2011 at 4:35 PM, Anurag Aggarwal 
 anurag19aggar...@gmail.com wrote:

 take two extra arrays b[] and c[]
 in b[] store the following thing
 b[0]=1;
 b[i]=b[i-1]*a[i-1];


 in c[] store following things
 c[n-1]=1;
 c[i]=c[i+1]*a[i+1]   (in-1)
 fill the c[] array in reverse order i.e. start from n-1 then go to 0;

 now output[] would be
 output[i]=b[i]*c[i];




 On Sun, Jul 17, 2011 at 4:28 PM, geek forgeek geekhori...@gmail.comwrote:

 given an array a[0..n-1]  .required to find the output array out
 [0.n-1] such that out [i] is the product of all the numbers a[0] to
 a[n-1] excluding a[i]
 for ex out[2]=a[0]*a[1]*a[3]*a[4]a[n-1]
 constraint is not using division operator

 how to do this in O(n)??

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 Anurag Aggarwal

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 BTech 2nd Year
 Computer Science And Engineering
 National Institute of Technology -Allahabad


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Re: [algogeeks] MICROSOFT

2011-07-17 Thread Piyush Sinha
Use linked list for designing  big int class

On Sun, Jul 17, 2011 at 5:17 PM, sourabh jakhar sourabhjak...@gmail.comwrote:

 10 c output question .
 questions were moderate type
 but negative marking
 +3  and -2.(30 min)

 coding round.(45 min)
 1.array simple question d.p
 2.test cases of notepad
 3.Design Big Int class in C or c++

 On Sun, Jul 17, 2011 at 5:14 PM, Harshal hc4...@gmail.com wrote:

 @manish,
 can you please tell other questions asked by ms today?


 On Sun, Jul 17, 2011 at 4:53 PM, Piyush Sinha 
 ecstasy.piy...@gmail.comwrote:

 It can be done using one extra array only that is the output array out[]

 *int out = (int *)malloc(sizeof(n)); //n is the number of elements in a
 int i,temp = 1;

 for(i=0;in;i++)
 {
  out[i] = temp;
  temp*=a[i];
 }
 temp =1;
 for(i=n-1;i=0;i--)
 {
 out[i] *= temp;
 temp*=a[i];
 }*



 On Sun, Jul 17, 2011 at 4:43 PM, manish patel 
 manispatel...@gmail.comwrote:

 thanx the question was asked by MS today in MNNIT.

 On Sun, Jul 17, 2011 at 4:35 PM, Anurag Aggarwal 
 anurag19aggar...@gmail.com wrote:

 take two extra arrays b[] and c[]
 in b[] store the following thing
 b[0]=1;
 b[i]=b[i-1]*a[i-1];


 in c[] store following things
 c[n-1]=1;
 c[i]=c[i+1]*a[i+1]   (in-1)
 fill the c[] array in reverse order i.e. start from n-1 then go to 0;

 now output[] would be
 output[i]=b[i]*c[i];




 On Sun, Jul 17, 2011 at 4:28 PM, geek forgeek 
 geekhori...@gmail.comwrote:

 given an array a[0..n-1]  .required to find the output array out
 [0.n-1] such that out [i] is the product of all the numbers a[0] 
 to
 a[n-1] excluding a[i]
 for ex out[2]=a[0]*a[1]*a[3]*a[4]a[n-1]
 constraint is not using division operator

 how to do this in O(n)??

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 Anurag Aggarwal

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Re: [algogeeks] MICROSOFT

2011-07-17 Thread Piyush Sinha
@Saurabh...ya true buddy...using malloc and realloc functions...gr88

On Sun, Jul 17, 2011 at 5:24 PM, sourabh jakhar sourabhjak...@gmail.comwrote:

 we can also use dynamically allocated array where they contain the digits
 of number and than apply the operation it is  a better option i think  as we
 can go to any index in 0(1) time.


 On Sun, Jul 17, 2011 at 5:19 PM, Harshal hc4...@gmail.com wrote:

 thanks sourabh :)

 On Sun, Jul 17, 2011 at 5:17 PM, sourabh jakhar 
 sourabhjak...@gmail.comwrote:

 10 c output question .
 questions were moderate type
 but negative marking
 +3  and -2.(30 min)

 coding round.(45 min)
 1.array simple question d.p
 2.test cases of notepad
 3.Design Big Int class in C or c++

 On Sun, Jul 17, 2011 at 5:14 PM, Harshal hc4...@gmail.com wrote:

 @manish,
 can you please tell other questions asked by ms today?


 On Sun, Jul 17, 2011 at 4:53 PM, Piyush Sinha ecstasy.piy...@gmail.com
  wrote:

 It can be done using one extra array only that is the output array
 out[]

 *int out = (int *)malloc(sizeof(n)); //n is the number of elements in
 a
 int i,temp = 1;

 for(i=0;in;i++)
 {
  out[i] = temp;
  temp*=a[i];
 }
 temp =1;
 for(i=n-1;i=0;i--)
 {
 out[i] *= temp;
 temp*=a[i];
 }*



 On Sun, Jul 17, 2011 at 4:43 PM, manish patel manispatel...@gmail.com
  wrote:

 thanx the question was asked by MS today in MNNIT.

 On Sun, Jul 17, 2011 at 4:35 PM, Anurag Aggarwal 
 anurag19aggar...@gmail.com wrote:

 take two extra arrays b[] and c[]
 in b[] store the following thing
 b[0]=1;
 b[i]=b[i-1]*a[i-1];


 in c[] store following things
 c[n-1]=1;
 c[i]=c[i+1]*a[i+1]   (in-1)
 fill the c[] array in reverse order i.e. start from n-1 then go to 0;

 now output[] would be
 output[i]=b[i]*c[i];




 On Sun, Jul 17, 2011 at 4:28 PM, geek forgeek geekhori...@gmail.com
  wrote:

 given an array a[0..n-1]  .required to find the output array out
 [0.n-1] such that out [i] is the product of all the numbers 
 a[0] to
 a[n-1] excluding a[i]
 for ex out[2]=a[0]*a[1]*a[3]*a[4]a[n-1]
 constraint is not using division operator

 how to do this in O(n)??

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Re: [algogeeks] MS Ques

2011-07-17 Thread Piyush Sinha
check my code below...it works for all cases
*
**node *MUL(node *h1,node *h2)
{
 node *h3,*p,*r;
 h1 = reverse(h1);
 h2 = reverse(h2);
 h3 = multiply(h1,h2-data);
 h2 = h2-next;
 p = h3;
 while(h2)
 {
  r = multiply(h1,h2-data);
  p-next = add(p-next,r);
  p = p-next;
  h2 = h2-next;
 }
 h3 = reverse(h3);
 return h3;
}

**node *multiply(node *h,int x)
{
 node *head = NULL;
 node *p;
 int mul,carry=0;
 while(h)
 {
if(!head)
{
head = (node *)malloc(sizeof(node));
mul = (x*(h-data));
carry = mul/10;
mul=mul%10;
head-data=mul;
head-next = NULL;
p = head;
}
else
{
p-next = (node *)malloc(sizeof(node));
p = p-next;
p-next = NULL;
mul = (x*(h-data));
p-data = (mul%10)+carry;
carry = mul/10;
}
h = h-next;
 }
 if(carry)
 {
p-next = (node *)malloc(sizeof(node));
p = p-next;
p-next = NULL;
p-data = carry;
 }
 return head;
}

node * add(node *h1,node *h2)
{
 node *h3 = NULL;
 node *p;
 int sum,carry = 0;
 while(h1)
 {
sum = h1-data+h2-data+carry;
carry = sum/10;
sum = sum%10;
if(h3==NULL)
{
   h3 = (node *)malloc(sizeof(node));
   h3-data = sum;
   h3-next = NULL;
   p =h3;
}
else
{
   p-next = (node *)malloc(sizeof(node));
   p = p-next;
   p-next = NULL;
   p-data = sum;
}
h1 = h1-next;
h2 = h2-next;
 }
 while(h2)
 {
p-next= (node *)malloc(sizeof(node));
p = p-next;
p-next = NULL;
sum = h2-data + carry;
carry = sum/10;
sum = sum%10;
p-data = sum;
h2 = h2-next;
 }
 if(carry)
 {
p-next = (node *)malloc(sizeof(node));
p = p-next;
p-next = NULL;
p-data = carry;
 }
 return h3;
}

*
On Mon, Jul 18, 2011 at 2:34 AM, aditi garg aditi.garg.6...@gmail.comwrote:

 and 6 is carry forwarded???
 next node wud be 6*7=42+6=48
 8 and 4 carry?


 On Mon, Jul 18, 2011 at 2:28 AM, hary rathor harry.rat...@gmail.comwrote:

 sorry 7*9=63

 put 3 in list 3

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Re: [algogeeks] C output problem

2011-07-17 Thread Piyush Sinha
the output will be (b)
before looking for the reason, you should know the syntax of printf

*int printf(const char *format,)*

the function at the string after the comma operator only if the format
string aks for any format specifier like %d or %f etc...u can try it like
this also in normal main function

*printf(piyush,sinha);

*It wil compile and run successfully...and u can guess the output too :)
*
*
On Mon, Jul 18, 2011 at 2:19 AM, aditi garg aditi.garg.6...@gmail.comwrote:

 #includestdio.h
 #define FUN(arg) do\
   {\
  if(arg)\
printf(have fun...,\n);\
   } while(i--)
 void main()
 {int i=2;
 FUN(i3);
 }

 A.have fun..
   have fun..
   have fun..
 B.have fun..have fun..have fun..
 C.Error : cannot use instructions in macro
 D.No output

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Re: [algogeeks] Re: Adding w/o +

2011-07-13 Thread Piyush Sinha
we are basically assigning address value a to a char pointer p..(the
reason y char pointer is used u will get to know it at the end)

we know p[b]  means *(p+b)then p[b] means *(p+b) i.e. p+b i.e a+b

we used char pointer to increment it one by one..had we used int pointer..
it wud have incremented it +4*b.

there is another way of doing it

*printf(%d\n,printf(%*c%*c,a,,b,));
*
On Wed, Jul 13, 2011 at 1:52 PM, nicks crazy.logic.k...@gmail.com wrote:

 if someone has better idea...then please suggest that.

 plz explain how this is calculating sum --  sum= (int)p[b];


 On Wed, Jul 13, 2011 at 1:50 PM, nicks crazy.logic.k...@gmail.com wrote:

 I am looking for a code which can add without using + sign

 i searched the net and found the following code..can anyone explain me
 whats happening in this ??

 #includestdio.h
 #includeconio.h
 int main()
 {
 int a=3000,b=20,sum;
 char *p;
 p=(char *)a;
 sum= (int)p[b]; //adding a  b
 printf(Answer is :);
 printf(%d,sum);
 return 0;
 }


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Re: [algogeeks] c code bst mirror prob

2011-07-13 Thread Piyush Sinha
Its working I guess in both the cases...

On 7/14/11, Anika Jain anika.jai...@gmail.com wrote:
 can some body tell me that:

 void swap(node *p,node *q)
 {
 node *t;
 t=p;
 p=q;
 q=t;
 }

 void mirror(node *p)
 {
 if (p==NULL)
 return;

 else
 {mirror(p-r);
 mirror(p-l);
 swap(p-r,p-l);
 }
 }

 in this the swapping is occuring but if i do :

 void mirror(node *p)
 {
 if (p==NULL)
 return;

 else
 {mirror(p-r);
 mirror(p-l);
  node *t;
  t=p-r;
 p-r=p-l;
 p-l=t;
 }
 }


 here it is not getting done .. why??

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Re: [algogeeks] MS: BST

2011-07-11 Thread Piyush Sinha
If we dont want the tree back, we can convert the BST to DLL and do the
job..
On Mon, Jul 11, 2011 at 3:01 PM, aanchal goyal goyal.aanch...@gmail.comwrote:

 Given a BST and integer value K. Find all pairs of nodes (x,y), such that
 x-data + y-data = K
 Time O(n)

 Can someone provide a pseudocode/code to solve this using the concept of
  inorder and reverse inorder traversal of BST?
 PS: please don't post other solutions for this, I know this can be solved
 in other ways too. I am not able to code this using the above concept..
 --
 Regards,*
 Aanchal Goyal*.

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Re: [algogeeks] Re: GOOGLE Q

2011-07-11 Thread Piyush Sinha
@Dk..i dint frame the question buddy...:P

I found it over here
http://geeksforgeeks.org/forum/topic/google-interview-question-for-software-engineerdeveloper-about-algorithms-75

On Mon, Jul 11, 2011 at 3:14 PM, DK divyekap...@gmail.com wrote:

 @Sunny: Thanks for this counterexample. The O(N) algorithm currently seems
 unfeasible to me.
 @Piyush: Are you sure an O(N) algo exists?

 Here's an O(N log N) algo (without the N^2 memory requirements of Sunny's
 algo as it doesn't generate duplicates)

 For arrays
 A = a1 a2 ... an
 B = b1 b2  bn

 Maintain a max priority_queue ordered by Val(a,b). (Use a BST for this).
 Insert (an, bn).

 Repeat N Times {
Pop off and output the max value from the priority queue. Let that be
 (ai, bj) // O(log N)
if(i == j) {
   Insert (ai-1, bj), (ai, bj-1), (ai-1, bj-1) in the priority queue. //
 O(log n)
} else if(i  j) {
   Insert (ai-1, bj) in the priority queue. // O(log n)
} else {
   Insert (ai, bj-1) in the priority queue. // O(log n)
}
 }

 Time complexity: O(N log N)
 Space complexity: O(N) ??

 --
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 http://twitter.com/divyekapoor
 http://www.divye.in

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Re: [algogeeks] MS: BST

2011-07-11 Thread Piyush Sinha
@AanchalI think my following algo will work..its a modification of
Morris traversal...check if you can find any bug in it...I have tried my
best to make it error free..For further details regarding Morris traversal,
check out http://geeksforgeeks.org/?p=6358

*void findsum(node *T,int key)
{
 int n = countnodes(T);//returns the number of nodes in the BST
 int i=0;
 int j=n-1;
 node *cur1,*cur2,*p1,*p2;
 cur1=cur2=T;
 int flag1,flag2,sum;
 flag1=flag2=1;
 while(ij)
 {
   if(cur1-left == NULL  cur2-right == NULL)
   {
   sum = checksum(cur1,cur2,key);
   if(sum==0) //if cur1-data + cur2-data ==key
   {
   i++;j--;
   flag1=flag2=1;
   cur1 = cur1-right;
   cur2 = cur2left;
   }
   if(sum0) //if cur1-data + cur2-data  key
   {
   cur2 = cur2-left;
   flag1 = 0; //do not do anything with the left
traversal
   j--;
   }
   if(sum0) //if cur1-data + cur2-data key
   {
   cur1 = cur1-right;
   flag2 = 0;//do not do anything with the right
traversal
   i++;
   }
   }
   else
   {
   if(flag1  cur1-left!=NULL)
   {
   p1 = cur1-left;
   while(p1-right!=NULL  p1-right!=cur1)
p1 = p1-right;
   if(p1-right == NULL)
   {
p1-right = cur1;
cur1 = cur1-left;
   }
   }
   if(flag2  cur2-right!=NULL)
   {
   p2 = cur2-right;
   while(p2-left!=NULL  p2-left!=cur2)
p2 = p2-left;
   if(p2-left == NULL)
   {
p2-left = cur2;
cur2 = cur2-right;
   }
   }
   if(p1-right == cur1 || p2-left == cur2)
   {
   sum = checksum(cur1,cur2,key);
   if(sum==0) //if cur1-data + cur2-data ==key
   {
  if(cur1-left == NULL)
  {
  cur1 = cur1-right;
  flag1 = 0;
  }
  else if(p1-right == cur1)
  {
   p1-right = NULL;
   cur1 = cur1-right;
   flag1 = 1;
  }

  if(cur2-right == NULL)
  {
  cur2 = cur2-left;
  flag2 = 0;
  }
  else if(p2-left == cur2)
  {
   p2-left = NULL;
   cur2 = cur2-left;
   flag2 = 1;
  }
  i++;j--;
   }
   if (sum0) //if cur1-data + cur2-data  key
   {
  if(cur1-left == NULL)
  {
  cur1 = cur1-right;
  flag1 = 0;
  }
  else if(p1-right == cur1)
  {
   p1-right = NULL;
   cur1 = cur1-right;
   flag1 = 1;
  }
  i++;
   }
   if(sum0)//if cur1-data + cur2-data  key
   {
  if(cur2-right == NULL)
  {
  cur2 = cur2-left;
  flag2 = 0;
  }
  else if(p2-left == cur2)
  {
   p2-left = NULL;
   cur2 = cur2-left;
   flag2 = 1;
  }
  j--;
   }
   }
   }
 }

 //final correction of the links can be done again
}

int checksum ( node *c1,node *c2,int key)
{
if(c1-data+c2-data == key)
 return 0;
else if(c1-data+c2-data  key)
 return 1;
else
return -1;
}*



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Re: [algogeeks] MS: BST

2011-07-11 Thread Piyush Sinha
@Anchalthanks for pointing out the error...i found where the error
is...it is the else loop in this line in this checking...

if(p1-right == cur1 || p2-left == cur2)

actually before i have already assigned p1-right == cur1 (or
p2-left=cur2)..it shud have been in else loop..soryy for the
mistake..i will submit the modification asap..

On 7/11/11, aanchal goyal goyal.aanch...@gmail.com wrote:
 @Piyush..
 I tried your algo on BST {5,4,6,3,2,1,8,7,10,12,9,11}.
 Its only returning 1+7.
 Other pairs are 5+3, 6+2.

 On Mon, Jul 11, 2011 at 6:56 PM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 @AanchalI think my following algo will work..its a modification of
 Morris traversal...check if you can find any bug in it...I have tried my
 best to make it error free..For further details regarding Morris
 traversal,
 check out http://geeksforgeeks.org/?p=6358

 *void findsum(node *T,int key)
 {
  int n = countnodes(T);//returns the number of nodes in the BST
  int i=0;
  int j=n-1;
  node *cur1,*cur2,*p1,*p2;
  cur1=cur2=T;
  int flag1,flag2,sum;
  flag1=flag2=1;
  while(ij)
  {
if(cur1-left == NULL  cur2-right == NULL)
{
sum = checksum(cur1,cur2,key);
if(sum==0) //if cur1-data + cur2-data ==key
{
i++;j--;
flag1=flag2=1;
cur1 = cur1-right;
cur2 = cur2left;
}
if(sum0) //if cur1-data + cur2-data  key
{
cur2 = cur2-left;
flag1 = 0; //do not do anything with the left
 traversal
j--;
}
if(sum0) //if cur1-data + cur2-data key
{
cur1 = cur1-right;
flag2 = 0;//do not do anything with the right
 traversal
i++;
}
}
else
{
if(flag1  cur1-left!=NULL)
{
p1 = cur1-left;
while(p1-right!=NULL  p1-right!=cur1)
 p1 = p1-right;
if(p1-right == NULL)
{
 p1-right = cur1;
 cur1 = cur1-left;
}
}
if(flag2  cur2-right!=NULL)
{
p2 = cur2-right;
while(p2-left!=NULL  p2-left!=cur2)
 p2 = p2-left;
if(p2-left == NULL)
{
 p2-left = cur2;
 cur2 = cur2-right;
}
}
if(p1-right == cur1 || p2-left == cur2)
{
sum = checksum(cur1,cur2,key);
if(sum==0) //if cur1-data + cur2-data ==key
{
   if(cur1-left == NULL)
   {
   cur1 = cur1-right;
   flag1 = 0;
   }
   else if(p1-right == cur1)
   {
p1-right = NULL;
cur1 = cur1-right;
flag1 = 1;
   }

   if(cur2-right == NULL)
   {
   cur2 = cur2-left;
   flag2 = 0;
   }
   else if(p2-left == cur2)
   {
p2-left = NULL;
cur2 = cur2-left;
flag2 = 1;
   }
   i++;j--;
}
if (sum0) //if cur1-data + cur2-data  key
{
   if(cur1-left == NULL)
   {
   cur1 = cur1-right;
   flag1 = 0;
   }
   else if(p1-right == cur1)
   {
p1-right = NULL;
cur1 = cur1-right;
flag1 = 1;
   }
   i++;
}
if(sum0)//if cur1-data + cur2-data  key
{
   if(cur2-right == NULL)
   {
   cur2 = cur2-left;
   flag2

Re: [algogeeks] MS: BST

2011-07-11 Thread Piyush Sinha
;
else if(c1-data+c2-data  key)
 return 1;
else
return -1;
}*







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Re: [algogeeks] Re: Yahoo Question

2011-07-10 Thread Piyush Sinha
using a temp variable is considered to be the best option..

On 7/11/11, Dave dave_and_da...@juno.com wrote:
 @Vaibhav: Your method b doesn't work for floating point numbers
 because they have finite precision. E.g.,as an extreme example, try it
 on a = 1 and b = 1d-25. When you form a+b, the result is 1, not 1 +
 1d-25. Then 1 - 1d-25 gives 1 (which is correct), and 1 - 1 = 0. The
 latter should be 1d-25, so you have a total loss of precision in that
 number. More common would be just a partial loss of precision.

 Dave

 On Jul 10, 3:55 pm, vaibhav shukla vaibhav200...@gmail.com wrote:
 On Mon, Jul 11, 2011 at 1:51 AM, aanchal goyal
 goyal.aanch...@gmail.comwrote:

  These are the various ways to swap 2 variables
  a) Using temporary Variable

        always inefficient. using extra memory.

  b) Usnig some Arithmentic operation

        works for all numbers even floating points
       a and b are two no. to swap : a=a+b;
                                                  b=a-b;
                                                   a=a-b;
    always works

  c) Using bitwise XOR operation

    since bitwise results an error in floating point numbers so this method
 also fails.
  hence (b) is better among the three.what say ?



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[algogeeks] BIT MANIPULATION

2011-07-09 Thread Piyush Sinha
I found a good question to try for bit manipulation.Try it... :)

Given an integer x, find out the smallest integer which has same
number of set bits as x and is greater than x.

For example if the input integer is 12 (1100) then your function
should return 17(10001). If the input integer is 3(11) then your
function should return 5(101)

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Re: [algogeeks] Re: BIT MANIPULATION

2011-07-09 Thread Piyush Sinha
@Dave..canu u explain ur algo to approach this formula??

On 7/10/11, Dave dave_and_da...@juno.com wrote:
 @Piyush: See http://groups.google.com/group/algogeeks/msg/2b64c4f96fa3598e
 for a one-line algorithm.

 Dave

 On Jul 9, 4:23 am, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 I found a good question to try for bit manipulation.Try it... :)

 Given an integer x, find out the smallest integer which has same
 number of set bits as x and is greater than x.

 For example if the input integer is 12 (1100) then your function
 should return 17(10001). If the input integer is 3(11) then your
 function should return 5(101)

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Re: [algogeeks] find the integer

2011-07-08 Thread Piyush Sinha
is there anyting special about the array???
or it is aribitary array??

On 7/8/11, Dumanshu duman...@gmail.com wrote:
 given an array of intergers. find the any integer that occurs only
 once. Others might occur any no. of times.

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Re: [algogeeks] Re: find the integer

2011-07-08 Thread Piyush Sinha
try sorting then if hashing

On 7/8/11, Dumanshu duman...@gmail.com wrote:
 no constraints but try to give the optimized solution  and plz no
 hashing.

 On Jul 8, 7:02 pm, Dumanshu duman...@gmail.com wrote:
 given an array of intergers. find the any integer that occurs only
 once. Others might occur any no. of times.

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[algogeeks] GOOGLE Q

2011-07-08 Thread Piyush Sinha
Given two sorted positive integer arrays A(n) and B(n). We define a
set S = {(a,b) | a in
A and b in B}. Obviously there are n^2 elements in S. The value of such
a pair is defined as Val(a,b) = a + b. Now we want to get the n pairs
from S with largest values. The tricky part is that we need an O(n)
algorithm.


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Re: [algogeeks] Reversing the order of words in String

2011-07-07 Thread Piyush Sinha
@Navneettake a look at the solution below and tell if there is any bug
in it...

*#include string.h

typedef struct revll
{
char s[100];
struct revll *next;
}revll;

revll *rev_str(char *a)
{
  char temp[100];
  int i,len=strlen(a),j=0;
  revll *head,*p;
  head=NULL;
  for(i=0;ilen;i++)
  {
   if(a[i]!=' ')
   {
temp[j++] = a[i];
   }
   else
   {
   temp[j] = '\0';
   p = (revll *)malloc(sizeof(revll));
   p-next = head;
   strcpy(p-s,temp);
   head = p;
   j=0;
   }
  }
  /*for last word*/
  temp[j] = '\0';
  p = (revll *)malloc(sizeof(revll));
  p-next = head;
  strcpy(p-s,temp);
  head = p;
  return head;
}

main()
{
  char a[100];
  revll *head;
  gets(a);
  head = rev_str(a);
  while(head)
  {
 printf(%s-,head-s);
 head=head-next;
  }
  system(pause);
}*



On Thu, Jul 7, 2011 at 9:10 AM, Navneet Gupta navneetn...@gmail.com wrote:

 @Piyush, could you elaborate your approach with Linked List?
 From what i am getting, even with Linked List, you would need two
 traversals at least.

 On Thu, Jul 7, 2011 at 2:07 AM, Piyush Sinha ecstasy.piy...@gmail.com
 wrote:
  Can we do it using linked list if ONE TIME TRAVERSAL is a constraint??
 
  On 7/6/11, Tushar Bindal tushicom...@gmail.com wrote:
  I read that solution.
  But the same doubt as Navneet which I think you also raised i one of
 your
  posts on that thread
 
  On Wed, Jul 6, 2011 at 10:34 PM, Navneet Gupta navneetn...@gmail.com
 wrote:
 
  Saurabh,
 
  I understood your solution but wonder if it is purely single traversal
 
  In affect, you have a second traversal when you are popping the
  strings from stack to form the reverse order string.
 
  Though the second activity is less than O(n) i.e. O(#words in string)
  Nice solution, this way we can also get rid of extra spaces easily in
  the actual string if that is also to be done.
 
  On Wed, Jul 6, 2011 at 10:16 PM, saurabh singh saurab...@gmail.com
  wrote:
   I have proposed my solution in one of the previous posts.Check the
  solution
   there
  
   On Wed, Jul 6, 2011 at 10:10 PM, Tushar Bindal 
 tushicom...@gmail.com
   wrote:
  
   good job
   but how can this be done in one traversal as asked on the Adobe
  Interview
   Questions thread.
  
  
  
   On Wed, Jul 6, 2011 at 9:49 PM, Navneet Gupta 
 navneetn...@gmail.com
   wrote:
  
   I think somebody on this thread has asked this question but i am
 not
   able to find that.
  
   Question was if a string is like my name is ram, then output
 should
   be ram is name my.
  
   Wrote the code for same, so sharing.
  
   #includeiostream
   #includestring
   using namespace std;
  
   void SwapStringChars(string str, int pos1, int pos2)
   {
  char ch = str[pos1];
  str[pos1] = str[pos2];
  str[pos2] = ch;
   }
  
   void reverseString(string str, int left, int right)
   {
  for(int i = left ; i = left + (right-left)/2 ; i++)
  SwapStringChars(str, i, right + left -i));
   }
  
   void reverseWordsInString(string str)
   {
  char space = ' ';
  int len = str.length();
  int startIndex = 0, endIndex = 0;
  while(endIndex  len - 1)
  {
  while(str[endIndex] != space  endIndex 
  len)endIndex++;
  reverseString(str, startIndex, endIndex-1);
  startIndex = endIndex;
  while(str[startIndex] == space)startIndex++;
  endIndex = startIndex;
  }
   }
  
   int main()
   {
  string str;
  cout\nEnter enter the string :;
  getline(cin,str);
  
  //Reverse whole string at once
  reverseString(str, 0, str.length() - 1);
  
  //Reverse Individual words in string
  reverseWordsInString(str);
  coutstr;
  cin.get();
  return 0;
   }
  
   --
   Regards,
   Navneet
  
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Re: [algogeeks] Amazon

2011-07-07 Thread Piyush Sinha
I think for initial start it should be the minimum n values for n
milestones

On Thu, Jul 7, 2011 at 1:53 PM, Akshata Sharma akshatasharm...@gmail.comwrote:

 There is a straight roads with 'n' number of milestones. You are given an
 array with the distance between all the pairs of milestones in some random
 order. Find the position of milestones.
 Example:
 Consider a road with 4 milestones(a,b,c,d) :
 a --- 3Km ---b--- 5Km ---c--- 2Km ---d
 Distance between a and b = 3
 Distance between a and c = 8
 Distance between a and d = 10
 Distance between b and c = 5
 Distance between b and d = 7
 Distance between c and d = 2
 All the above values are given in a random order say 7, 10, 5, 2, 8, 3.
 The output must be 3,5,2 or 2,5,3

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Re: [algogeeks] Amazon

2011-07-07 Thread Piyush Sinha
But the problem arises in setting the order

On Thu, Jul 7, 2011 at 2:06 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:

 I think for initial start it should be the minimum n values for n
 milestones


 On Thu, Jul 7, 2011 at 1:53 PM, Akshata Sharma 
 akshatasharm...@gmail.comwrote:

 There is a straight roads with 'n' number of milestones. You are given an
 array with the distance between all the pairs of milestones in some random
 order. Find the position of milestones.
 Example:
 Consider a road with 4 milestones(a,b,c,d) :
 a --- 3Km ---b--- 5Km ---c--- 2Km ---d
 Distance between a and b = 3
 Distance between a and c = 8
 Distance between a and d = 10
 Distance between b and c = 5
 Distance between b and d = 7
 Distance between c and d = 2
 All the above values are given in a random order say 7, 10, 5, 2, 8, 3.
 The output must be 3,5,2 or 2,5,3

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[algogeeks] GOOGLE Q1

2011-07-07 Thread Piyush Sinha
Given an array of integers A, give an algorithm to find the longest
Arithmetic progression in it, i.e find a sequence i1  i2  …  ik,
such that

A[i1], A[i2], …, A[ik] forms an arithmetic progression, and k is the
largest possible.

The sequence S1, S2, …, Sk is called an arithmetic progression if

Sj+1 – Sj is a constant.

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[algogeeks] GOOGLE Q2

2011-07-07 Thread Piyush Sinha
Design a data structure that can be used instead of an array and
avoids the high-cost (i.e. O(n)) of initializing the array. the data
structure ought to holds n items and supports the following operations
in O(1) time:
* Init – Initializes the data structure to empty.
* Set(i, x) – Sets item x at index i in the data structure.
* Get(i) – Gets the item stored in index i (or 'empty' if nothing is there).
Remark: the data structure should use O(n) space.

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Re: [algogeeks] Merging Sorted Arrays

2011-07-07 Thread Piyush Sinha
@radha...i have an algo but its complexity is O(n^2)...check the
following and see if there is any bug as I havent tested for all
cases...also suggestions are welcomed:)

main()
{
  int a[]= {1,3,77,78,88};
  int b[]= {2,5,79,80,81,99};
  int i=sizeof(a)/sizeof(a[0]) - 1;
  int j=sizeof(b)/sizeof(b[0]) - 1;
  int temp,k,m;
  while(j=0)
  {
 if(a[i]b[j])
 {
   temp = a[i];
   k=m=i;
   while(b[j]a[k-1]) k--;
   while(i-k)
   {
a[i] = a[i-1];
i--;
   }
   a[i] = b[j];
   b[j] = temp;
   i = m;
 }
 j--;
  }
  for(k=0;ksizeof(a)/sizeof(a[0]);k++)
printf(%d ,a[k]);
  puts(\n);
  for(k=0;ksizeof(b)/sizeof(b[0]);k++)
printf(%d ,b[k]);
  puts(\n);
  system(pause);
}


On 7/8/11, radha krishnan radhakrishnance...@gmail.com wrote:
 :Given two sorted arrays a[]={1,3,77,78,90} and b[]={2,5,79,81}. Merge
 these two arrays, no extra spaces are allowed. Output has to be
 a[]={1,2,3,5,77} and b[]={78,79,81,90}.

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Re: [algogeeks] GOOGLE Q3

2011-07-07 Thread Piyush Sinha
@Rajeev...check ur logic for 4

On 7/8/11, rajeev bharshetty rajeevr...@gmail.com wrote:
 So, I think first check for excellent groups then good and then usual to
 increase the quality.
 So to get excellent groups scan the string and keep a count of the
 contagious repeating elements ,if count==3 or count==2,then put in one
 group
 The same procedure for good and usual accord to constraints .



 On Thu, Jul 7, 2011 at 11:46 PM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 You are given a String number containing the digits of a phone number
 (the number of digits, n, can be any positive integer) . To help you
 memorize the number, you want to divide it into groups of contiguous
 digits. Each group must contain exactly 2 or 3 digits. There are three
 kinds of groups:
 • Excellent: A group that contains only the same digits. For example, 000
 or 77.
 • Good: A group of 3 digits, 2 of which are the same. For example,
 030, 229 or 166.
 • Usual: A group in which all the digits are distinct. For example, 123 or
 90.
 The quality of a group assignment is defined as 2 × (number of
 excellent groups) + (number of good groups)
 Divide the number into groups such that the quality is maximized.
 Design an efficient
 algorithm to return the solution that maximizes the quality.

 --
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Re: [algogeeks] GOOGLE Q1

2011-07-07 Thread Piyush Sinha
Nopesits about finding subsequence

On 7/8/11, rajeev bharshetty rajeevr...@gmail.com wrote:
 Should the sequence beContinuos ???

 On Fri, Jul 8, 2011 at 1:18 AM, sunny agrawal
 sunny816.i...@gmail.comwrote:

 @rajiv
 if Count  = 2 means 3 elements isn't it  a,a+d,a+2d
 else according to you
 for case 10 12 14 24 26 28
 diff  2 2 10 2 2
 diff 2 has count 4 so will you say ap of 4 elements with diff 2

 On Fri, Jul 8, 2011 at 1:06 AM, rajeev bharshetty
 rajeevr...@gmail.comwrote:

 @sunny Keep count of longest repeated element in diff i.e 2 so count =2
 so
 ap of 2 elem with diff 2 .

 On Fri, Jul 8, 2011 at 1:03 AM, sunny agrawal
 sunny816.i...@gmail.comwrote:

 @rajiv
 Fails i think
 think for 10 12 24 26
 diff is 2 12  2
 so do you want to say there is an AP pf 3 elements with d = 2, i can't
 see any :P
 your solution fails because there can be many APs in the array with the
 same value of d and you will finish up by combining all those APs


 On Fri, Jul 8, 2011 at 12:55 AM, rajeev bharshetty rajeevr...@gmail.com
  wrote:


 Check the differences between the adjacent elements and store  the
 differenecs in diff[i] array
 then scan through the array .
 then keep a count for all the repeated diff elements ,the sequence of
 indexes with max count is the solution .




 On Thu, Jul 7, 2011 at 11:43 PM, Piyush Sinha ecstasy.piy...@gmail.com
  wrote:

 Given an array of integers A, give an algorithm to find the longest
 Arithmetic progression in it, i.e find a sequence i1  i2  …  ik,
 such that

 A[i1], A[i2], …, A[ik] forms an arithmetic progression, and k is the
 largest possible.

 The sequence S1, S2, …, Sk is called an arithmetic progression if

 Sj+1 – Sj is a constant.

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Re: [algogeeks] Merging Sorted Arrays

2011-07-07 Thread Piyush Sinha
@RADHA..ya true...my bad i cudnt think dat way...but as pointed out by
Sunny the sorted output cant be brought in O(n)...In order to get the
sorted output only my algo took O(n^2)...tell me if I am missing
anything...

On 7/8/11, sunny agrawal sunny816.i...@gmail.com wrote:
 yes upto the step of swapping the elements it is O(m+n)
 but if you need final arrays also sorted (Seems like that from your first
 post)it will go nlgn


 On Fri, Jul 8, 2011 at 1:25 AM, radha krishnan radhakrishnance...@gmail.com
 wrote:

 ok ! i got a O(n lgn) finally
 i don know exact complexity
 Let N = size of first array
 Find the first N smallest elements using one pointer in each array
 now swap the list of elements  from index 0 to second-pointer in
 second array to first array
 with first_poiner+1 to N in first Array
 I think this is O(n)

 On Thu, Jul 7, 2011 at 12:53 PM, Piyush Sinha ecstasy.piy...@gmail.com
 wrote:
  @radha...i have an algo but its complexity is O(n^2)...check the
  following and see if there is any bug as I havent tested for all
  cases...also suggestions are welcomed:)
 
  main()
  {
   int a[]= {1,3,77,78,88};
   int b[]= {2,5,79,80,81,99};
   int i=sizeof(a)/sizeof(a[0]) - 1;
   int j=sizeof(b)/sizeof(b[0]) - 1;
   int temp,k,m;
   while(j=0)
   {
  if(a[i]b[j])
  {
temp = a[i];
k=m=i;
while(b[j]a[k-1]) k--;
while(i-k)
{
 a[i] = a[i-1];
 i--;
}
a[i] = b[j];
b[j] = temp;
i = m;
  }
  j--;
   }
   for(k=0;ksizeof(a)/sizeof(a[0]);k++)
 printf(%d ,a[k]);
   puts(\n);
   for(k=0;ksizeof(b)/sizeof(b[0]);k++)
 printf(%d ,b[k]);
   puts(\n);
   system(pause);
  }
 
 
  On 7/8/11, radha krishnan radhakrishnance...@gmail.com wrote:
  :Given two sorted arrays a[]={1,3,77,78,90} and b[]={2,5,79,81}. Merge
  these two arrays, no extra spaces are allowed. Output has to be
  a[]={1,2,3,5,77} and b[]={78,79,81,90}.
 
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 Indian Institute Of Technology,Roorkee

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Re: [algogeeks] GOOGLE Q3

2011-07-07 Thread Piyush Sinha
@Rajeev..there is the fault.for 4, there can be 2 excellent groups..

55
55
54

therefore, quality = 2*2 = 4

which is highest...

On 7/8/11, rajeev bharshetty rajeevr...@gmail.com wrote:
 It scans 555 count=3(excellent group)  it will put in one group than 54
 since the count is one puts that in usual group quality =2*1 =2 ,

 On Fri, Jul 8, 2011 at 1:28 AM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 @Rajeev...check ur logic for 4

 On 7/8/11, rajeev bharshetty rajeevr...@gmail.com wrote:
  So, I think first check for excellent groups then good and then usual to
  increase the quality.
  So to get excellent groups scan the string and keep a count of the
  contagious repeating elements ,if count==3 or count==2,then put in one
  group
  The same procedure for good and usual accord to constraints .
 
 
 
  On Thu, Jul 7, 2011 at 11:46 PM, Piyush Sinha
  ecstasy.piy...@gmail.comwrote:
 
  You are given a String number containing the digits of a phone number
  (the number of digits, n, can be any positive integer) . To help you
  memorize the number, you want to divide it into groups of contiguous
  digits. Each group must contain exactly 2 or 3 digits. There are three
  kinds of groups:
  • Excellent: A group that contains only the same digits. For example,
 000
  or 77.
  • Good: A group of 3 digits, 2 of which are the same. For example,
  030, 229 or 166.
  • Usual: A group in which all the digits are distinct. For example, 123
 or
  90.
  The quality of a group assignment is defined as 2 × (number of
  excellent groups) + (number of good groups)
  Divide the number into groups such that the quality is maximized.
  Design an efficient
  algorithm to return the solution that maximizes the quality.
 
  --
  *Piyush Sinha*
  *IIIT, Allahabad*
  *+91-8792136657*
  *+91-7483122727*
  *https://www.facebook.com/profile.php?id=10655377926 *
 
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Re: [algogeeks] GOOGLE Q3

2011-07-07 Thread Piyush Sinha
sorry a typing mistake...lst line contains only 4

On 7/8/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 @Rajeev..there is the fault.for 4, there can be 2 excellent
 groups..

 55
 55
 54

 therefore, quality = 2*2 = 4

 which is highest...

 On 7/8/11, rajeev bharshetty rajeevr...@gmail.com wrote:
 It scans 555 count=3(excellent group)  it will put in one group than 54
 since the count is one puts that in usual group quality =2*1 =2 ,

 On Fri, Jul 8, 2011 at 1:28 AM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 @Rajeev...check ur logic for 4

 On 7/8/11, rajeev bharshetty rajeevr...@gmail.com wrote:
  So, I think first check for excellent groups then good and then usual
  to
  increase the quality.
  So to get excellent groups scan the string and keep a count of the
  contagious repeating elements ,if count==3 or count==2,then put in one
  group
  The same procedure for good and usual accord to constraints .
 
 
 
  On Thu, Jul 7, 2011 at 11:46 PM, Piyush Sinha
  ecstasy.piy...@gmail.comwrote:
 
  You are given a String number containing the digits of a phone number
  (the number of digits, n, can be any positive integer) . To help you
  memorize the number, you want to divide it into groups of contiguous
  digits. Each group must contain exactly 2 or 3 digits. There are
  three
  kinds of groups:
  • Excellent: A group that contains only the same digits. For example,
 000
  or 77.
  • Good: A group of 3 digits, 2 of which are the same. For example,
  030, 229 or 166.
  • Usual: A group in which all the digits are distinct. For example,
  123
 or
  90.
  The quality of a group assignment is defined as 2 × (number of
  excellent groups) + (number of good groups)
  Divide the number into groups such that the quality is maximized.
  Design an efficient
  algorithm to return the solution that maximizes the quality.
 
  --
  *Piyush Sinha*
  *IIIT, Allahabad*
  *+91-8792136657*
  *+91-7483122727*
  *https://www.facebook.com/profile.php?id=10655377926 *
 
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Re: [algogeeks] GOOGLE Q3

2011-07-07 Thread Piyush Sinha
@Rajeev...ignore my previous two posts.

the grouping can be done as

55
554

quality is 2*1+1 = 3 which is the highest

On 7/8/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 sorry a typing mistake...lst line contains only 4

 On 7/8/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 @Rajeev..there is the fault.for 4, there can be 2 excellent
 groups..

 55
 55
 54

 therefore, quality = 2*2 = 4

 which is highest...

 On 7/8/11, rajeev bharshetty rajeevr...@gmail.com wrote:
 It scans 555 count=3(excellent group)  it will put in one group than 54
 since the count is one puts that in usual group quality =2*1 =2 ,

 On Fri, Jul 8, 2011 at 1:28 AM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 @Rajeev...check ur logic for 4

 On 7/8/11, rajeev bharshetty rajeevr...@gmail.com wrote:
  So, I think first check for excellent groups then good and then usual
  to
  increase the quality.
  So to get excellent groups scan the string and keep a count of the
  contagious repeating elements ,if count==3 or count==2,then put in
  one
  group
  The same procedure for good and usual accord to constraints .
 
 
 
  On Thu, Jul 7, 2011 at 11:46 PM, Piyush Sinha
  ecstasy.piy...@gmail.comwrote:
 
  You are given a String number containing the digits of a phone
  number
  (the number of digits, n, can be any positive integer) . To help you
  memorize the number, you want to divide it into groups of contiguous
  digits. Each group must contain exactly 2 or 3 digits. There are
  three
  kinds of groups:
  • Excellent: A group that contains only the same digits. For
  example,
 000
  or 77.
  • Good: A group of 3 digits, 2 of which are the same. For example,
  030, 229 or 166.
  • Usual: A group in which all the digits are distinct. For example,
  123
 or
  90.
  The quality of a group assignment is defined as 2 × (number of
  excellent groups) + (number of good groups)
  Divide the number into groups such that the quality is maximized.
  Design an efficient
  algorithm to return the solution that maximizes the quality.
 
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Re: [algogeeks] GOOGLE Q3

2011-07-07 Thread Piyush Sinha
@Sunny...can u post a definite algo for it??

On 7/8/11, Ravi Shukla shuklaravi...@gmail.com wrote:
 @sunny , yep it looks DP. more of MCM.

 solve for substrings of length 1,2,3.
 and then apply DP[i][j]=max score for a substring from i to j.
 =max(DP[i][k]+DP[k][j]) where ki  kj .

 The complexity this approach renders would be O(n^3).
 with O(n^2) space complexity.

 anyone anything better ?

 Thanks.
 Ravi Shukla
 CSE Final Year.
 BIT Mesra, Ranchi

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Re: [algogeeks] GOOGLE Q3

2011-07-07 Thread Piyush Sinha
Can u explain ur algo too??

On 7/8/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 @Sunny...nice solution but ur solution works if there can 1 to 3
 groups of digits..but in the question its mentioned the group should
 contain exactly 2 or 3 digits...

 but anyways nice solution...:)

 On 7/8/11, sunny agrawal sunny816.i...@gmail.com wrote:
 http://ideone.com/xv73J


 On Fri, Jul 8, 2011 at 2:16 AM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 @Sunny...can u post a definite algo for it??

 On 7/8/11, Ravi Shukla shuklaravi...@gmail.com wrote:
  @sunny , yep it looks DP. more of MCM.
 
  solve for substrings of length 1,2,3.
  and then apply DP[i][j]=max score for a substring from i to j.
  =max(DP[i][k]+DP[k][j]) where ki  kj .
 
  The complexity this approach renders would be O(n^3).
  with O(n^2) space complexity.
 
  anyone anything better ?
 
  Thanks.
  Ravi Shukla
  CSE Final Year.
  BIT Mesra, Ranchi
 
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[algogeeks] NVIDIA Q

2011-07-06 Thread Piyush Sinha
What is the size of an object of a class with no members in it??
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Re: [algogeeks] NVIDIA Q

2011-07-06 Thread Piyush Sinha
thanks buddy..:)

On 7/6/11, durgaprasad k durga...@gmail.com wrote:
 The size will be 1 byte as there is nothing to look into the object.

 And it is 1 instead of zero because two objects of the class will have
 different addresses by assigning each object size 1.

 Regards,
 Durga

 On Wed, Jul 6, 2011 at 2:11 PM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 What is the size of an object of a class with no members in it??
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[algogeeks]

2011-07-06 Thread Piyush Sinha
Why is it suggested not to use malloc() or calloc() in C++ for memory
allocation?


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Re: [algogeeks] Reversing the order of words in String

2011-07-06 Thread Piyush Sinha
Can we do it using linked list if ONE TIME TRAVERSAL is a constraint??

On 7/6/11, Tushar Bindal tushicom...@gmail.com wrote:
 I read that solution.
 But the same doubt as Navneet which I think you also raised i one of your
 posts on that thread

 On Wed, Jul 6, 2011 at 10:34 PM, Navneet Gupta navneetn...@gmail.comwrote:

 Saurabh,

 I understood your solution but wonder if it is purely single traversal

 In affect, you have a second traversal when you are popping the
 strings from stack to form the reverse order string.

 Though the second activity is less than O(n) i.e. O(#words in string)
 Nice solution, this way we can also get rid of extra spaces easily in
 the actual string if that is also to be done.

 On Wed, Jul 6, 2011 at 10:16 PM, saurabh singh saurab...@gmail.com
 wrote:
  I have proposed my solution in one of the previous posts.Check the
 solution
  there
 
  On Wed, Jul 6, 2011 at 10:10 PM, Tushar Bindal tushicom...@gmail.com
  wrote:
 
  good job
  but how can this be done in one traversal as asked on the Adobe
 Interview
  Questions thread.
 
 
 
  On Wed, Jul 6, 2011 at 9:49 PM, Navneet Gupta navneetn...@gmail.com
  wrote:
 
  I think somebody on this thread has asked this question but i am not
  able to find that.
 
  Question was if a string is like my name is ram, then output should
  be ram is name my.
 
  Wrote the code for same, so sharing.
 
  #includeiostream
  #includestring
  using namespace std;
 
  void SwapStringChars(string str, int pos1, int pos2)
  {
 char ch = str[pos1];
 str[pos1] = str[pos2];
 str[pos2] = ch;
  }
 
  void reverseString(string str, int left, int right)
  {
 for(int i = left ; i = left + (right-left)/2 ; i++)
 SwapStringChars(str, i, right + left -i));
  }
 
  void reverseWordsInString(string str)
  {
 char space = ' ';
 int len = str.length();
 int startIndex = 0, endIndex = 0;
 while(endIndex  len - 1)
 {
 while(str[endIndex] != space  endIndex 
 len)endIndex++;
 reverseString(str, startIndex, endIndex-1);
 startIndex = endIndex;
 while(str[startIndex] == space)startIndex++;
 endIndex = startIndex;
 }
  }
 
  int main()
  {
 string str;
 cout\nEnter enter the string :;
 getline(cin,str);
 
 //Reverse whole string at once
 reverseString(str, 0, str.length() - 1);
 
 //Reverse Individual words in string
 reverseWordsInString(str);
 coutstr;
 cin.get();
 return 0;
  }
 
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  Navneet
 
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Re: [algogeeks] MS Ques

2011-07-05 Thread Piyush Sinha
Can we do it using suffix trees and apply DFS on it??

On Tue, Jul 5, 2011 at 11:45 PM, swetha rahul swetharahu...@gmail.comwrote:

 Hi,
   Write a function which takes two char * s as inputs, one is a
 regular expression pattern and the other is a test string and check whether
 the test string is of the given regular expression pattern.
 The regular expression pattern can contain all lower-case letter, asterisk
 and question mark.
 As usual asterisk stands for 0 or more number of chars and ? for any one
 char.
 E.g. Regular Expression: a*b?c
 Test String: aaavcxbmbcxmbkc
 Result: TRUE
 Test String: abc
 Result: FALSE
 Test String: abzx
 Result: TRUE

 What is d best way 2 do this...

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Re: [algogeeks] Flatten a BST to produce inorder traversal

2011-07-04 Thread Piyush Sinha
Use the concept used in Morris traversal (same as TBT concept)...



On 7/4/11, sunny agrawal sunny816.i...@gmail.com wrote:
 I think Threaded Binary Tree solves your Problem
 see this http://en.wikipedia.org/wiki/Threaded_binary_tree

 On Mon, Jul 4, 2011 at 11:34 PM, Navneet Gupta navneetn...@gmail.comwrote:

 Tree has an extra pointer next apart from left and right. Objective
 is to set next pointer to point to node successor in the tree.
 Following the next pointer, we would be able to produce sorted list.

 Looking for both a recursive and non-recursive approach.

 --Navneet

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Re: [algogeeks] Flatten a BST to produce inorder traversal

2011-07-04 Thread Piyush Sinha
http://geeksforgeeks.org/?p=6358

On 7/4/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 Use the concept used in Morris traversal (same as TBT concept)...



 On 7/4/11, sunny agrawal sunny816.i...@gmail.com wrote:
 I think Threaded Binary Tree solves your Problem
 see this http://en.wikipedia.org/wiki/Threaded_binary_tree

 On Mon, Jul 4, 2011 at 11:34 PM, Navneet Gupta
 navneetn...@gmail.comwrote:

 Tree has an extra pointer next apart from left and right. Objective
 is to set next pointer to point to node successor in the tree.
 Following the next pointer, we would be able to produce sorted list.

 Looking for both a recursive and non-recursive approach.

 --Navneet

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Re: [algogeeks] Flatten a BST to produce inorder traversal

2011-07-04 Thread Piyush Sinha
I know its not about the traversali just suggested that one can
use the trick used by Morris traversal to locate the next node of the
inorder traversal...

On 7/4/11, Navneet Gupta navneetn...@gmail.com wrote:
 @Piyush, it is not about the traversal, you actually have to establish
 those links such that once they are established, inorder traversal
 would be just like traversing a list.

 @Sunny - thanks, exactly what i was looking for

 On Mon, Jul 4, 2011 at 11:45 PM, Piyush Sinha ecstasy.piy...@gmail.com
 wrote:
 http://geeksforgeeks.org/?p=6358

 On 7/4/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 Use the concept used in Morris traversal (same as TBT concept)...



 On 7/4/11, sunny agrawal sunny816.i...@gmail.com wrote:
 I think Threaded Binary Tree solves your Problem
 see this http://en.wikipedia.org/wiki/Threaded_binary_tree

 On Mon, Jul 4, 2011 at 11:34 PM, Navneet Gupta
 navneetn...@gmail.comwrote:

 Tree has an extra pointer next apart from left and right. Objective
 is to set next pointer to point to node successor in the tree.
 Following the next pointer, we would be able to produce sorted list.

 Looking for both a recursive and non-recursive approach.

 --Navneet

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Re: [algogeeks] Flatten a BST to produce inorder traversal

2011-07-04 Thread Piyush Sinha
U only mentioned in ur question that we have to use next pointer to
connect the nodes...while TBT used the left and right pointers

On 7/5/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 I know its not about the traversali just suggested that one can
 use the trick used by Morris traversal to locate the next node of the
 inorder traversal...

 On 7/4/11, Navneet Gupta navneetn...@gmail.com wrote:
 @Piyush, it is not about the traversal, you actually have to establish
 those links such that once they are established, inorder traversal
 would be just like traversing a list.

 @Sunny - thanks, exactly what i was looking for

 On Mon, Jul 4, 2011 at 11:45 PM, Piyush Sinha ecstasy.piy...@gmail.com
 wrote:
 http://geeksforgeeks.org/?p=6358

 On 7/4/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 Use the concept used in Morris traversal (same as TBT concept)...



 On 7/4/11, sunny agrawal sunny816.i...@gmail.com wrote:
 I think Threaded Binary Tree solves your Problem
 see this http://en.wikipedia.org/wiki/Threaded_binary_tree

 On Mon, Jul 4, 2011 at 11:34 PM, Navneet Gupta
 navneetn...@gmail.comwrote:

 Tree has an extra pointer next apart from left and right. Objective
 is to set next pointer to point to node successor in the tree.
 Following the next pointer, we would be able to produce sorted list.

 Looking for both a recursive and non-recursive approach.

 --Navneet

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Re: [algogeeks] output plzzzz

2011-07-02 Thread Piyush Sinha
i)  [Linker error] undefined reference to `a'

ii) 4

On 7/3/11, HARSH PAHUJA hpahuja.mn...@gmail.com wrote:
 1)
 #includestdio.h
 int main()
 {
 extern int a;
 a=20;
 printf(%d,sizeof(a));
 return 0;
 }

 2)
 #includestdio.h
 int main()
 {
 extern int a;
 printf(%d,sizeof(a));
 return 0;
 }



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Re: [algogeeks] query

2011-06-29 Thread Piyush Sinha
Its working fine..

On Wed, Jun 29, 2011 at 5:26 PM, sunny agrawal sunny816.i...@gmail.comwrote:

 the error is in   quotes, just rewrite them

 On Wed, Jun 29, 2011 at 5:24 PM, Anika Jain anika.jai...@gmail.comwrote:

 hey, printf(%d ya %u in both same error cming..


 On Wed, Jun 29, 2011 at 5:10 PM, sunny agrawal 
 sunny816.i...@gmail.comwrote:

 Because u copied the code i think :P
 try writing printf statement yourself again :)

   On Wed, Jun 29, 2011 at 5:05 PM, Anika Jain anika.jai...@gmail.comwrote:

  int main()
 {
 int I =10, j=2;
 int *ip = I ,*jp =j;
 int k = *ip/ *jp;
 printf(“%u”,k);

 return 0;
 }


 in this code error is coming why??

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Re: [algogeeks] query

2011-06-29 Thread Piyush Sinha
ya noticed just now when sunny agrawal pointed it out...


On Wed, Jun 29, 2011 at 5:38 PM, oppilas . jatka.oppimi...@gmail.comwrote:

 Piyush the original code has double apostrophe( ' ) instead of inverted
 comma's.
 http://ideone.com/zPpGA


 On Wed, Jun 29, 2011 at 5:30 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:

 Its working fine..


 On Wed, Jun 29, 2011 at 5:26 PM, sunny agrawal 
 sunny816.i...@gmail.comwrote:

 the error is in   quotes, just rewrite them

 On Wed, Jun 29, 2011 at 5:24 PM, Anika Jain anika.jai...@gmail.comwrote:

 hey, printf(%d ya %u in both same error cming..


 On Wed, Jun 29, 2011 at 5:10 PM, sunny agrawal sunny816.i...@gmail.com
  wrote:

 Because u copied the code i think :P
 try writing printf statement yourself again :)

   On Wed, Jun 29, 2011 at 5:05 PM, Anika Jain 
 anika.jai...@gmail.comwrote:

  int main()
 {
 int I =10, j=2;
 int *ip = I ,*jp =j;
 int k = *ip/ *jp;
 printf(“%u”,k);

 return 0;
 }


 in this code error is coming why??

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Re: [algogeeks] linked list

2011-06-29 Thread Piyush Sinha
node *segregate(node *head)
{
node *even,*odd,*even1,*odd1;
even=odd=NULL;
while(head)
{
if((head-data)%2)  
{
if(!odd)
{
odd = head;
odd1 = odd;
}
else
{
odd1-next = head;
odd1 = odd1-next;
}
}
else
{
if(!even)
{
even = head;
even1 = even;
}
else
{
even1-next = head;
even1 = even1-next;
}
}
head=head-next;
}
odd1-next=NULL;
even1-next = odd;
return (even);
}

On 6/29/11, Nishant Mittal mittal.nishan...@gmail.com wrote:
 segregate even and odd nodes in a singly linked list.Order of even and
 odd numbers must be same...
 e.g:-
 i/p list is 4-1-3-6-12-8-7-NULL
 o/p list  4-6-12-8-1-3-7-NULL

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Re: [algogeeks] Binary Tree

2011-06-29 Thread Piyush Sinha
7+3 also give the sum to be 10???

On 6/29/11, 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] Amazon - max substring

2011-06-29 Thread Piyush Sinha
i think suffix tres will do the job if i have not misunderstood the question...

On 6/29/11, Swathi chukka.swa...@gmail.com wrote:
 Given a string (assume there no spaces or punctuations), write a code that
 returns the max. length of the string that has repeated more than once.

 Thanks,
 Swathi

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Re: [algogeeks] Amazon - max substring

2011-06-29 Thread Piyush Sinha
I dnt think any company is gonna ask u to code suffix tree..:P :P

On 6/29/11, Swathi chukka.swa...@gmail.com wrote:
 It does but i am asked to code.. if you know the code for suffix tree then
 please provide..

 On Wed, Jun 29, 2011 at 10:30 PM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 i think suffix tres will do the job if i have not misunderstood the
 question...

 On 6/29/11, Swathi chukka.swa...@gmail.com wrote:
  Given a string (assume there no spaces or punctuations), write a code
 that
  returns the max. length of the string that has repeated more than once.
 
  Thanks,
  Swathi
 
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Re: [algogeeks] Amazon - max substring

2011-06-29 Thread Piyush Sinha
jokes apart...then i think it can be done using dynamic programming
using the similar approach of LCS but the time and space complexity
both will be N^2.i am working on the pseudocode and post it within
a few moments if it works..

On 6/29/11, Swathi chukka.swa...@gmail.com wrote:
 I dont know why people reply in plain words.. I personally had this
 experience and i was asked to code but i couldn't

 On Wed, Jun 29, 2011 at 10:53 PM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 I dnt think any company is gonna ask u to code suffix tree..:P :P

 On 6/29/11, Swathi chukka.swa...@gmail.com wrote:
  It does but i am asked to code.. if you know the code for suffix tree
 then
  please provide..
 
  On Wed, Jun 29, 2011 at 10:30 PM, Piyush Sinha
  ecstasy.piy...@gmail.comwrote:
 
  i think suffix tres will do the job if i have not misunderstood the
  question...
 
  On 6/29/11, Swathi chukka.swa...@gmail.com wrote:
   Given a string (assume there no spaces or punctuations), write a code
  that
   returns the max. length of the string that has repeated more than
 once.
  
   Thanks,
   Swathi
  
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Re: [algogeeks] 2 D array(dynamic allocation)

2011-06-29 Thread Piyush Sinha
int **p;
p = (int **)malloc(sizeof(int *)*row);
for(i = 0;irow;i++)
   p[i] = (int *)malloc(sizeof(int)*column);

On 6/30/11, Apoorve Mohan apoorvemo...@gmail.com wrote:
 though thankx :)

 On Thu, Jun 30, 2011 at 12:44 AM, Apoorve Mohan
 apoorvemo...@gmail.comwrote:

 @above: man i need a 2d array not a 1d array...


 On Thu, Jun 30, 2011 at 12:38 AM, hary rathor
 harry.rat...@gmail.comwrote:


 #includestdlib.h

 int main ()
 {
 int *mat;
 int i,j;
 int ROW=4;
 int COL=3;
 int k=0;
 mat=(int *)malloc(ROW*COL*sizeof(int));

for(i=0;iROW;i++)
for(j=0;jCOL;j++)
mat[i*COL+j]=++k;

for(i=0;iROW;i++)
for(j=0;jCOL;j++)
printf(%d,,mat[i*COL+j]);

 return 0;

 }



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Re: [algogeeks] 2 D array(dynamic allocation)

2011-06-29 Thread Piyush Sinha
ohh sorrymy bad...i didnt read the whole question..i just read the
subject...:P

i think its not possible if u want other than hary's solution...

On 6/30/11, Apoorve Mohan apoorvemo...@gmail.com wrote:
 @piyush: only one call to malloc...ur sol has 2

 On Thu, Jun 30, 2011 at 12:58 AM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 int **p;
 p = (int **)malloc(sizeof(int *)*row);
 for(i = 0;irow;i++)
   p[i] = (int *)malloc(sizeof(int)*column);

 On 6/30/11, Apoorve Mohan apoorvemo...@gmail.com wrote:
  though thankx :)
 
  On Thu, Jun 30, 2011 at 12:44 AM, Apoorve Mohan
  apoorvemo...@gmail.comwrote:
 
  @above: man i need a 2d array not a 1d array...
 
 
  On Thu, Jun 30, 2011 at 12:38 AM, hary rathor
  harry.rat...@gmail.comwrote:
 
 
  #includestdlib.h
 
  int main ()
  {
  int *mat;
  int i,j;
  int ROW=4;
  int COL=3;
  int k=0;
  mat=(int *)malloc(ROW*COL*sizeof(int));
 
 for(i=0;iROW;i++)
 for(j=0;jCOL;j++)
 mat[i*COL+j]=++k;
 
 for(i=0;iROW;i++)
 for(j=0;jCOL;j++)
 printf(%d,,mat[i*COL+j]);
 
  return 0;
 
  }
 
 
 
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Re: [algogeeks] Printing unique rows.

2011-06-28 Thread Piyush Sinha
In any case, I don't think the complexity can be improved from O(n^2)
because even in creating hash function every column element of every row
which itself is N^2 only..

On Tue, Jun 28, 2011 at 11:38 AM, Ankit Agrawal 
ankitmnnit.agra...@gmail.com wrote:

 u can use hashing ...
 hash fun can b base2 to base10

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Re: [algogeeks] given a bst and a value x.find pair of nodes in the tree that sum upto x

2011-06-27 Thread Piyush Sinha
Instead of using an array...we can do this by converting the BST into a
doubly linked list...but this will definitely modify the BST..


On Mon, Jun 27, 2011 at 3:01 PM, varun pahwa varunpahwa2...@gmail.comwrote:

 @ankit: no need to use hash table for that.
 simply run two pointers one from 0 and second from n - 1.

 On Mon, Jun 27, 2011 at 2:51 PM, ankit sambyal ankitsamb...@gmail.comwrote:

 @Bharath : Cud u plz explain how r u searching the elements in O(n) time?
 Because if we use binary search, it will have O(n*log n )  worst case
 time complexity. One way in which I think it cud be made O(n) is that
 we can use a hash table, with a good hash function apart frm the
 array. And then for each element 'm' in the array, we cud search if
 there is an element (sum - m) in O(1) time by using hash table. Still
 we can't assure O(n) time complexity. Because coming up with a good
 hash function is not easy. Again, hash table takes more space




 On Mon, Jun 27, 2011 at 1:40 AM, Nishant Mittal
 mittal.nishan...@gmail.com wrote:
  do inorder traversal of tree and store values in an array.
  Now find pairs by applying binary search on array..
 
  On 6/27/11, manish kapur manishkapur.n...@gmail.com wrote:
 
 
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Re: [algogeeks] Re: given a bst and a value x.find pair of nodes in the tree that sum upto x

2011-06-27 Thread Piyush Sinha
how about converting the BST into a doubly linked list...but this will
definitely modify the BST..

On Mon, Jun 27, 2011 at 11:32 PM, Swathi chukka.swa...@gmail.com wrote:

 Dave,

 I am unable to write code for this so i am asking your help.

 Thanks,
 Swathi


 On Mon, Jun 27, 2011 at 11:28 PM, Dave dave_and_da...@juno.com wrote:

 @Swathi: No. I think the high level description should be adequate for
 you to write your own code or pseudocode, albeit recognizing that you
 may have to look up how to do an inorder traversal using a stack
 instead of recursion.

 Dave

 On Jun 27, 9:34 am, Swathi chukka.swa...@gmail.com wrote:
  Dave,
 
  Can you provide the psuedo code for this..
 
  Thanks,
  Swathi
 
 
 
  On Mon, Jun 27, 2011 at 7:30 PM, Dave dave_and_da...@juno.com wrote:
   @Sunny. Mea culpa. You are correct. Revised (and correct) algorithm.
   Do two inorder traversals, one in the usual (descend to the left
   before descendung to the right) direction and the other in the
   reversed(descend to the right before descending to the left)
   direction. Let u and r be the current nodee of the two traversals,
   respectively. If u + r  x, then advance the usual traversal and
   repeat the comparison. If u + r  x, advance the reverse traversal and
   repeat the comparison. If u + r = x, and if u != r, then terminate
   with success. If u = r, then terminate with failure.
 
   Dave
 
   On Jun 27, 7:53 am, sunny agrawal sunny816.i...@gmail.com wrote:
@Dave
i think your solution won't work
consider inorder traversal of a BST is 1 6 7 8 15 and x = 14
initially both u,v (1,1)
according to u your algorithm will proceed like
(1,1) - (1,6) - (1,7) - (1,8) - (1,15) - (6,15)  -
   (15,15)
 
and clearly in second step of your solution if (u+v)  x after
 advancing
   u
still u+v will be greater than x
so something is wrong
I think your solution will work in case we need to find 2 nodes with
difference x.
 
correct me if i am wrong.!!
 
On Mon, Jun 27, 2011 at 6:13 PM, Dave dave_and_da...@juno.com
 wrote:
 @Nishant: No need to store the data in an array. Do two inorder
 traversals simultaneously. Let u and v be the current nodes of the
 two
 traversals, respectively. If u + v  x, then advance the v
 traversal. If u + v  x, advance the u traversal.
 
 Dave
 
 On Jun 27, 3:40 am, Nishant Mittal mittal.nishan...@gmail.com
 wrote:
  do inorder traversal of tree and store values in an array.
  Now find pairs by applying binary search on array..
 
  On 6/27/11, manish kapur manishkapur.n...@gmail.com wrote:
 
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Re: [algogeeks] novel written test

2011-06-27 Thread Piyush Sinha
1,6,6

On Tue, Jun 28, 2011 at 2:07 AM, amit the cool amitthecoo...@gmail.comwrote:

 Conversation between two mathematicians: first : I have three
 children. The product of their ages is 36 . If you sum their ages . it
 is exactly same as my neighbor's door number on my left. The second
 mathematician verifies the door number and says that the not
 sufficient . Then the first says  o.k one more clue is that my
 youngest is the youngest Immediately the second mathematician
 answers . Can you answer the question asked by the first
 mathematician? What are the children ages?

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Re: [algogeeks] novel written test

2011-06-27 Thread Piyush Sinha
We just need to look the 3 set of number that multiply to 36...

the set can be as
(1,1,36),(1,2,18),(1,4,9)..(1,6,6),(1,3,12)..(2,2,9)

The catch here is we need not find the whole set...the second mathematician
can't guess their ages through their sum that means there are at least two
possible combination that sum to the same value..

we have 1,6,6 and 2,2,9

second clue says my youngest is youngest...

if we look into the combination, we find the first set have unique age for
the youngest son..hence the ages..

and 2 sons can have same age without being twins...one born in the month of
jan and the other born in the month of december...:P :P
moreover its no where mentioned there can't be any twins

On Tue, Jun 28, 2011 at 2:24 AM, amit kumar amitthecoo...@gmail.com wrote:

 @piyush..will u xplain plz.
 by d way 2 sons cant hav d same age(hope they r nt twins)

 On Tue, Jun 28, 2011 at 2:14 AM, Piyush Sinha ecstasy.piy...@gmail.comwrote:

 1,6,6


 On Tue, Jun 28, 2011 at 2:07 AM, amit the cool 
 amitthecoo...@gmail.comwrote:

 Conversation between two mathematicians: first : I have three
 children. The product of their ages is 36 . If you sum their ages . it
 is exactly same as my neighbor's door number on my left. The second
 mathematician verifies the door number and says that the not
 sufficient . Then the first says  o.k one more clue is that my
 youngest is the youngest Immediately the second mathematician
 answers . Can you answer the question asked by the first
 mathematician? What are the children ages?

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Re: [algogeeks] Queue to support insert , delete, find max in o(1)

2011-06-24 Thread Piyush Sinha
Can we use circular linked list with each new inserted node keeping track of
the minimum before it??


On Fri, Jun 24, 2011 at 3:20 PM, ross jagadish1...@gmail.com wrote:

 Hi,
 I know that a stack can be modified with another stack to support push
 pop min in const time.
 Design a FIFO data structure to support ins, del, and find min in
 O(1). Extra space allowed.

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Re: [algogeeks] Re: Queue to support insert , delete, find max in o(1)

2011-06-24 Thread Piyush Sinha
ohh sorry, my bad..I missed that issue...then we can use the same logic
of using one more stack that we use for implementing modified stack keeping
track of the min()..I hope this will solve the issue

On Fri, Jun 24, 2011 at 3:57 PM, ross jagadish1...@gmail.com wrote:

 @piyush,
 Dude, how will that make findmin() to be O(1) because, once the
 minimum element is
 deleted, u would require changes in the others .. Correct me if i am
 wrong..
 Eg:
 consider inserting, 1 5 6 7 9 in order into the circular LL.
 When u make each node keep track of the minm before it, all will
 retain 1 as minm..

 Now consider deleting 1. how will that affect the rest of the list?

 On Jun 24, 3:07 pm, Piyush Sinha ecstasy.piy...@gmail.com wrote:
  Can we use circular linked list with each new inserted node keeping track
 of
  the minimum before it??
 
 
 
 
 
 
 
 
 
  On Fri, Jun 24, 2011 at 3:20 PM, ross jagadish1...@gmail.com wrote:
   Hi,
   I know that a stack can be modified with another stack to support push
   pop min in const time.
   Design a FIFO data structure to support ins, del, and find min in
   O(1). Extra space allowed.
 
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Re: [algogeeks] Re: Sort - Consecutive Array in O(n)

2011-06-24 Thread Piyush Sinha
I have a solution that will do the job in O(n) time but will require three
variablesbut this solution won't work if the array contains -ve numbers.
*
int findrepeat(int a[],int n)
{
 int i,xor = 0;
 int min = findmin(a,n);
 int max = findmax(a,n);
 if((max-min+1)!=n)
 return 0;
 for(i = 0 ;in;i++)
  xor^=a[i];
 for(i=min;i=max;i++)
 xor^=i;
 if(xor)
 return 0;
 else
  return 1;
}*

Please let me know if there is any counter example..


On Sat, Jun 25, 2011 at 1:17 AM, ross jagadish1...@gmail.com wrote:

 One solution would be to : check if max-min = N and
 that all elements are unique in the array.
 However, this may require space complexity.. Looking for a
 better solution.

 On Jun 25, 12:44 am, ross jagadish1...@gmail.com wrote:
  Given an array, A, find if all elements in the sorted version of A are
  consecutive in less than O(nlogn).
  eg: A: 5 4 1 2 3 on sorting 1 2 3 4 5 all elements are consecutive --
  true
  A: 1 9 2 22 on sorting 1 2 9 22 all elements are NOT consecutive -
  false

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Re: [algogeeks] Question on Combination

2011-06-23 Thread Piyush Sinha
sorry a lil bit modification in the above answer..


*int ref[] = {1,2,3};*

*void printcombination(int n,int index,int i)*

*{*

*static int a[100];*

*int j;*

*if (n == 0)*

*   {*

*  for(j=0;jindex;j++)*

*  printf(%d ,a[j]);*

*   printf(\n);*

*}*

*   else if(n0)*

*   {*

*   for(j=i;j3;j++)*

*  {*

*a[index]=ref[j];*

*  printcombination(n-ref[j],index+1,j);*

*   }*

*  }*

*} *

*main()*

*{*

*  int n;*

*   printf(enter value of n :);*

*  scanf(%d,n);*

*  printcombination(n,0,0);*

*}*

On Thu, Jun 23, 2011 at 11:17 AM, Piyush Sinha ecstasy.piy...@gmail.comwrote:

 pass one more argument to the function *int index* and instead of
 starting the loop from *i = 0 to N, *make it start from *i = index to N *and
 then call

 *printcombinations(a,sum-a[i],level+1,index+1);
 *I think it will work then...
   On Thu, Jun 23, 2011 at 10:48 AM, ross jagadish1...@gmail.com wrote:

 Given an array and a sum S output all combinations of elements that
 sum to S.
 eg: 1 2 3

 sum = 3
 1+1+1,
 2+1
 3

 I came up with the foll algorithm, but it outputs 2+1 and 1+2 again.
 (does not handle repetitions)

 printcombinations(int a[],int sum,int level) {
 if(sum==0) { print array}
 else if (sum0) {
 for ( i = 0 to N ) {
 array[level]=a[i];
 printcombinations(a,sum-a[i],level+1);
 }
 }
 }

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Re: [algogeeks] Please explain the output

2011-06-23 Thread Piyush Sinha
printf(%d,*power(432)) will expand as

*printf(%d, *432)*

432 represents here a string and *432 is pointing to the first string
literal i.e 4 whose ascii value is 52..hence the output is 52


On Thu, Jun 23, 2011 at 4:02 PM, Shachindra A C sachindr...@gmail.comwrote:

  #includestdio.h
 #define power(a) #a
 int main()
 {
  printf(%d,*power(432));
   return 0;
 }

 the printf statement, after preprocessing, will look like
 printf(%d,*432);

 so, when u print the value at the first position of the string, 52, which
 is the ascii value of 4, will be printed.

 On Thu, Jun 23, 2011 at 3:40 PM, vaibhav shukla 
 vaibhav200...@gmail.comwrote:

 #includestdio.h
 #define power(a) #a
 int main()
 {
  printf(%d,*power(432));
   return 0;
 }


 ans is 52 on gcc. Explain plss

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Re: [algogeeks] c query

2011-06-23 Thread Piyush Sinha
the comma operator evaluates its first operand and discards the
result, and then evaluates the second operand and returns this value.
instead of 3, whatever u write. the result will depend on the second
operand i.e x1 i.e 2 (5/2 = 2)

On 6/23/11, Anika Jain anika.jai...@gmail.com wrote:
 #includestdio.h
 typedef struct
 {
   char *name;
double salary;
 }job;

 main()
 {
 static job a={tcs,15000.0};
 static job a={ibm,25000.0};
 static job a={google,35000.0};
 int x=5;
 job *arr[3]={a,b,c};
  printf(%s %f\t,(3,x1)[*arr]);
 }

 output: google 35000.00

 what this printf statement means plz explain...

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Re: [algogeeks] c query

2011-06-23 Thread Piyush Sinha
for further details.

http://en.wikipedia.org/wiki/Comma_operator

On 6/23/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 the comma operator evaluates its first operand and discards the
 result, and then evaluates the second operand and returns this value.
 instead of 3, whatever u write. the result will depend on the second
 operand i.e x1 i.e 2 (5/2 = 2)

 On 6/23/11, Anika Jain anika.jai...@gmail.com wrote:
 #includestdio.h
 typedef struct
 {
   char *name;
double salary;
 }job;

 main()
 {
 static job a={tcs,15000.0};
 static job a={ibm,25000.0};
 static job a={google,35000.0};
 int x=5;
 job *arr[3]={a,b,c};
  printf(%s %f\t,(3,x1)[*arr]);
 }

 output: google 35000.00

 what this printf statement means plz explain...

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Re: [algogeeks] c doubt again

2011-06-23 Thread Piyush Sinha
An asterisk indicates that the data is to be retrieved from the use
but ignored, i.e. it is not stored in the corresponding
argument...hence the third value entered gets stored for b and for c
the output comes to garbage value

One beautiful application of such type of implementation is in finding
the sum of 2 variables without using + operator..:)

On 6/23/11, Anika Jain anika.jai...@gmail.com wrote:
 int main()
 {
int a,b, c;
scanf(%d%*d%d,a,b,c);
printf(%d %d %d,a,b,c);
 }

 output: 25 35 garbage

 how is it happening??

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Re: [algogeeks] c query

2011-06-23 Thread Piyush Sinha
I had a doubt which I wanted to ask...for me output is coming as follows

google 0.00

r u sure ur output is coming to be
google 35000.00

On 6/23/11, Anika Jain anika.jai...@gmail.com wrote:
 ya *arr[2] is fine but in printing a struct element do we need to give all
 necessary format specifiers for printing??

 On Thu, Jun 23, 2011 at 7:15 PM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 for further details.

 http://en.wikipedia.org/wiki/Comma_operator

 On 6/23/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
  the comma operator evaluates its first operand and discards the
  result, and then evaluates the second operand and returns this value.
  instead of 3, whatever u write. the result will depend on the second
  operand i.e x1 i.e 2 (5/2 = 2)
 
  On 6/23/11, Anika Jain anika.jai...@gmail.com wrote:
  #includestdio.h
  typedef struct
  {
char *name;
 double salary;
  }job;
 
  main()
  {
  static job a={tcs,15000.0};
  static job a={ibm,25000.0};
  static job a={google,35000.0};
  int x=5;
  job *arr[3]={a,b,c};
   printf(%s %f\t,(3,x1)[*arr]);
  }
 
  output: google 35000.00
 
  what this printf statement means plz explain...
 
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Re: [algogeeks] c doubt again

2011-06-23 Thread Piyush Sinha
sorry by mistake i added it in scanf situation..
actually this type of specifier can be used with printf statement for
finding the sum...

look at the code below

main()
{
 int a=9;
 int b=3;
 printf(%d\n,printf(%*s%*s,a,,b,));
 system(pause);
}

On 6/23/11, Anika Jain anika.jai...@gmail.com wrote:
 thanx .. can u explain me how this is used in finding sum of 2 vars without
 using + ??


 On Thu, Jun 23, 2011 at 7:20 PM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 An asterisk indicates that the data is to be retrieved from the use
 but ignored, i.e. it is not stored in the corresponding
 argument...hence the third value entered gets stored for b and for c
 the output comes to garbage value

 One beautiful application of such type of implementation is in finding
 the sum of 2 variables without using + operator..:)

 On 6/23/11, Anika Jain anika.jai...@gmail.com wrote:
  int main()
  {
 int a,b, c;
 scanf(%d%*d%d,a,b,c);
 printf(%d %d %d,a,b,c);
  }
 
  output: 25 35 garbage
 
  how is it happening??
 
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Re: [algogeeks] c doubt again

2011-06-23 Thread Piyush Sinha
there is no as such logic behind it..its just the format specifier...

u must be knowing printf returns the number of values it has printed(u can
check that)

now, in printf if u write like *printf(%7s,a), *it will create 7
columns for the output and print a in the last column and the returned value
of this printf will be 7..(u can check it)

now if u write *printf(%*s,7,a)* then u r giving additional information
of format specifier i.e 7..returned value of this printf is also 7.

Hence the above logic..hope I am able to clarify it...:)

On Thu, Jun 23, 2011 at 8:06 PM, Anika Jain anika.jai...@gmail.com wrote:

 i mean how it working actually?


 On Thu, Jun 23, 2011 at 8:06 PM, Anika Jain anika.jai...@gmail.comwrote:

 hey ya its working :) but whats the logic behind it??


 On Thu, Jun 23, 2011 at 7:52 PM, Piyush Sinha 
 ecstasy.piy...@gmail.comwrote:

 sorry by mistake i added it in scanf situation..
 actually this type of specifier can be used with printf statement for
 finding the sum...

 look at the code below

 main()
 {
 int a=9;
 int b=3;
 printf(%d\n,printf(%*s%*s,a,,b,));
 system(pause);
 }

 On 6/23/11, Anika Jain anika.jai...@gmail.com wrote:
  thanx .. can u explain me how this is used in finding sum of 2 vars
 without
  using + ??
 
 
  On Thu, Jun 23, 2011 at 7:20 PM, Piyush Sinha
  ecstasy.piy...@gmail.comwrote:
 
  An asterisk indicates that the data is to be retrieved from the use
  but ignored, i.e. it is not stored in the corresponding
  argument...hence the third value entered gets stored for b and for c
  the output comes to garbage value
 
  One beautiful application of such type of implementation is in finding
  the sum of 2 variables without using + operator..:)
 
  On 6/23/11, Anika Jain anika.jai...@gmail.com wrote:
   int main()
   {
  int a,b, c;
  scanf(%d%*d%d,a,b,c);
  printf(%d %d %d,a,b,c);
   }
  
   output: 25 35 garbage
  
   how is it happening??
  
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Re: [algogeeks] c query

2011-06-23 Thread Piyush Sinha
even I am getting output as google 0.0

On 6/23/11, Bhavesh agrawal agr.bhav...@gmail.com wrote:
 i got (null) 0.0 on my gcc compiler , is there any syntax error

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Re: [algogeeks] c doubt again

2011-06-23 Thread Piyush Sinha
or u cud consult ANSI C by Balaguruswamy in chapter of Console I/Ps and O/Ps

On 6/23/11, harshit pahuja hpahuja.mn...@gmail.com wrote:
 @rajeev

 http://www.cplusplus.com/reference/clibrary/cstdio/scanf/
 http://www.cplusplus.com/reference/clibrary/cstdio/printf/

 On Thu, Jun 23, 2011 at 9:39 PM, rajeev bharshetty
 rajeevr...@gmail.comwrote:

 @ Piyush Could u provide the link to some source , because i am still
 unclear about the above concept .

 Regards
 Rajeev N B




 On Thu, Jun 23, 2011 at 8:32 PM, Piyush Sinha
 ecstasy.piy...@gmail.comwrote:

 there is no as such logic behind it..its just the format specifier...

 u must be knowing printf returns the number of values it has printed(u
 can
 check that)

 now, in printf if u write like *printf(%7s,a), *it will create 7
 columns for the output and print a in the last column and the returned
 value
 of this printf will be 7..(u can check it)

 now if u write *printf(%*s,7,a)* then u r giving additional
 information of format specifier i.e 7..returned value of this printf is
 also
 7.

 Hence the above logic..hope I am able to clarify it...:)


 On Thu, Jun 23, 2011 at 8:06 PM, Anika Jain
 anika.jai...@gmail.comwrote:

 i mean how it working actually?


 On Thu, Jun 23, 2011 at 8:06 PM, Anika Jain
 anika.jai...@gmail.comwrote:

 hey ya its working :) but whats the logic behind it??


 On Thu, Jun 23, 2011 at 7:52 PM, Piyush Sinha ecstasy.piy...@gmail.com
  wrote:

 sorry by mistake i added it in scanf situation..
 actually this type of specifier can be used with printf statement for
 finding the sum...

 look at the code below

 main()
 {
 int a=9;
 int b=3;
 printf(%d\n,printf(%*s%*s,a,,b,));
 system(pause);
 }

 On 6/23/11, Anika Jain anika.jai...@gmail.com wrote:
  thanx .. can u explain me how this is used in finding sum of 2 vars
 without
  using + ??
 
 
  On Thu, Jun 23, 2011 at 7:20 PM, Piyush Sinha
  ecstasy.piy...@gmail.comwrote:
 
  An asterisk indicates that the data is to be retrieved from the use
  but ignored, i.e. it is not stored in the corresponding
  argument...hence the third value entered gets stored for b and for
  c
  the output comes to garbage value
 
  One beautiful application of such type of implementation is in
 finding
  the sum of 2 variables without using + operator..:)
 
  On 6/23/11, Anika Jain anika.jai...@gmail.com wrote:
   int main()
   {
  int a,b, c;
  scanf(%d%*d%d,a,b,c);
  printf(%d %d %d,a,b,c);
   }
  
   output: 25 35 garbage
  
   how is it happening??
  
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Re: [algogeeks] Time Complexity of Merge Sort(Linked list)

2011-06-23 Thread Piyush Sinha
Merge sort is often the best choice for sorting a linked list. Reason
because it requires only Θ(1) extra space only and generally beats
other algorithms like quicksort,heapsort for sorting linked lists. The
overall complexity remain O(N log N) even in the worst case

On 6/23/11, Anantha Krishnan ananthakrishnan@gmail.com wrote:
 Hi All,

 Can someone explain about the time complexity of Merge sort(Linked list with
 billions of node)?

 There is no way to find the middle of sub-list without traversing
 completely.

 Please clear my doubts.

 Thanks  Regards,
 Anantha Krishnan

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Re: [algogeeks] Time Complexity of Merge Sort(Linked list)

2011-06-23 Thread Piyush Sinha
I googled a better explaination for this and found this...
hope this will be useful...

Sorting a LinkedList is different from sorting an Array as you can not
make index based references to any arbitrary item in the List. Instead
you need to remember the items during each recursive step and then
pass them on to the merge function. For each recursion step you only
need to remember the first item of each list half. If you do not
remember these items you will quickly end up with indexes, but this
leads you to the problem that in your merge-function you need to
traverse the whole list with for-loops to find the items to merge.
That in turn means you get a complexity of O(n^2).

Another important point is the step of recursing into the list and
dividing the list in two halves. You can do this step in the recursive
part by using for-loops. Contrary to the merge-part at this step the
for-loops will only yield a complexity of O(log(n) * n/2) and this is
still below the overall O(n*log(n)) complexity. Here is why:

   1. You always need to find the first item of each half of list part.
   2.  In the first recursion step you need to pass the first item and
the item at position n/2. This takes n/2 steps to find.
   3.  In each following step you need to find the middle item for
each of the two halves of the list which gives us n/4 to find the item
in the first halve and n/4 in the other halve. In total thats n/2.
   4.  In each following recursive step the amount of list parts
double and the lengths is divided by two:
  * 4 * n/8 in the 3rd recursion depth
  * 8 * n/16 in the 4th recursion depth, and so on...
   5. The recursion depth is log(n) and in each step we perform n/2
steps. This equals O(log(n)*n/2)


On 6/24/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 Merge sort is often the best choice for sorting a linked list. Reason
 because it requires only Θ(1) extra space only and generally beats
 other algorithms like quicksort,heapsort for sorting linked lists. The
 overall complexity remain O(N log N) even in the worst case

 On 6/23/11, Anantha Krishnan ananthakrishnan@gmail.com wrote:
 Hi All,

 Can someone explain about the time complexity of Merge sort(Linked list
 with
 billions of node)?

 There is no way to find the middle of sub-list without traversing
 completely.

 Please clear my doubts.

 Thanks  Regards,
 Anantha Krishnan

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 *+91-7483122727*
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Re: [algogeeks] Time Complexity of Merge Sort(Linked list)

2011-06-23 Thread Piyush Sinha
Also u can refer
http://www.chiark.greenend.org.uk/~sgtatham/algorithms/listsort.html

On 6/24/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 I googled a better explaination for this and found this...
 hope this will be useful...

 Sorting a LinkedList is different from sorting an Array as you can not
 make index based references to any arbitrary item in the List. Instead
 you need to remember the items during each recursive step and then
 pass them on to the merge function. For each recursion step you only
 need to remember the first item of each list half. If you do not
 remember these items you will quickly end up with indexes, but this
 leads you to the problem that in your merge-function you need to
 traverse the whole list with for-loops to find the items to merge.
 That in turn means you get a complexity of O(n^2).

 Another important point is the step of recursing into the list and
 dividing the list in two halves. You can do this step in the recursive
 part by using for-loops. Contrary to the merge-part at this step the
 for-loops will only yield a complexity of O(log(n) * n/2) and this is
 still below the overall O(n*log(n)) complexity. Here is why:

1. You always need to find the first item of each half of list part.
2.  In the first recursion step you need to pass the first item and
 the item at position n/2. This takes n/2 steps to find.
3.  In each following step you need to find the middle item for
 each of the two halves of the list which gives us n/4 to find the item
 in the first halve and n/4 in the other halve. In total thats n/2.
4.  In each following recursive step the amount of list parts
 double and the lengths is divided by two:
   * 4 * n/8 in the 3rd recursion depth
   * 8 * n/16 in the 4th recursion depth, and so on...
5. The recursion depth is log(n) and in each step we perform n/2
 steps. This equals O(log(n)*n/2)


 On 6/24/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 Merge sort is often the best choice for sorting a linked list. Reason
 because it requires only Θ(1) extra space only and generally beats
 other algorithms like quicksort,heapsort for sorting linked lists. The
 overall complexity remain O(N log N) even in the worst case

 On 6/23/11, Anantha Krishnan ananthakrishnan@gmail.com wrote:
 Hi All,

 Can someone explain about the time complexity of Merge sort(Linked list
 with
 billions of node)?

 There is no way to find the middle of sub-list without traversing
 completely.

 Please clear my doubts.

 Thanks  Regards,
 Anantha Krishnan

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Re: [algogeeks] Google Question

2011-06-22 Thread Piyush Sinha
This question has been discussed over here once...It was concluded
that this can be solved in O(n) if we know there is a fixed range up
to which the elements keep on increasing and decreasing..for example
in an array of 12 elements, we know 3 elements keep on increasing
monotonically, then 3 elements keep on decreasing monotonically and so
on

On 6/22/11, chirag ahuja sparkle.chi...@gmail.com wrote:
 Given an array of size n wherein elements keep on increasing
 monotonically upto a certain location after which they keep on
 decreasing monotonically, then again keep on increasing, then
 decreasing again and so on. Sort the array in O(n) and O(1).

 I didn't understand the question, any array of n elements will be like
 this except when first there is a decrese from index 0 to a higher
 index. Any ideas about how to solve it in given constraints??

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Re: [algogeeks] Explain this.....

2011-06-22 Thread Piyush Sinha
r u sure the last output is also 100..for me its coming 99

On 6/22/11, udit sharma sharmaudit...@gmail.com wrote:
 #includestdio.h
 int main()
 {
 void print(int *,int *,int *,int *,int *);
 static int arr[]={97,98,99,100,101,102,103,104};
 int *ptr=arr+1;
 print(++ptr,ptr--,ptr,ptr++,++ptr);
 return 0;
 }
 void print(int *a,int *b,int *c,int *d,int *e)
 {
 printf(%d\t%d\t%d\t%d\t%d\n,*a,*b,*c,*d,*e);
 }

 Why the output is:
 10010010099100


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  DU- MCA

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Re: [algogeeks] Explain this.....

2011-06-22 Thread Piyush Sinha
the arguments are passed from right to left in a function...

initially ptr is pointing to location of 98 (i =1)

the last argument ++ptr makes it point to 99 therefore output of *e = 99
 the second last argument passes pointer to 99 only and then
increments its location to i=3 i.e 100...therefore output of *d = 99
and *c = 100
the second argument is pointing to location of i=3 only and then
decrements to point to location of i=2..therefore output of *a =100
and *b =100..

hope you understood the sequence of outputs...:) :)

On 6/22/11, Piyush Sinha ecstasy.piy...@gmail.com wrote:
 r u sure the last output is also 100..for me its coming 99

 On 6/22/11, udit sharma sharmaudit...@gmail.com wrote:
 #includestdio.h
 int main()
 {
 void print(int *,int *,int *,int *,int *);
 static int arr[]={97,98,99,100,101,102,103,104};
 int *ptr=arr+1;
 print(++ptr,ptr--,ptr,ptr++,++ptr);
 return 0;
 }
 void print(int *a,int *b,int *c,int *d,int *e)
 {
 printf(%d\t%d\t%d\t%d\t%d\n,*a,*b,*c,*d,*e);
 }

 Why the output is:
 10010010099100


 --
 Regards
  UDIT
  DU- MCA

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