Re: [algogeeks] Re: Microsoft interview question

2010-09-26 Thread ashita dadlani
@Mohit:
I dont think it really matters here.We just have to validate the snapshot of
the game board.Number of players should not have any relevance here.

On Sat, Sep 25, 2010 at 2:46 PM, mohit ranjan shoonya.mo...@gmail.comwrote:

 @Ashita,

 Your logic is fine for one vs one game, but as per question it's one vs
 many game
 Any idea what is that ?


 Mohit

 On Sat, Sep 25, 2010 at 1:18 PM, ashita dadlani ash@gmail.com wrote:

 1.The soldiers are initially placed at row 2 or row 7th(each-one of white
 and either of black).Also let white ones be at row 2.So they can never be at
 row 1st.Incase it is so in the game,its not a valid game.
 2.There are Bishops.Each color has one of its Bishop which moves
 diagonally on all white squares,and the other on all black squares.Incase it
 is not so,the game cannot be valid.
 3.Now suppose,no black soldier ever moved.That is,all the black soldiers
 are at row 7th.This means that the elephant(i am sorry,I generally mess up
 with their names..:P) of any other player(except horse) cannot be in any row
 but 8th one.

 I know only 3 test cases.Incase any one has more,please elaborate.
 PS:Vrinda,I also got the same question..:P


 On Sat, Sep 25, 2010 at 2:49 AM, Gene gene.ress...@gmail.com wrote:

 Valid must mean that you can get from an initial board to the the
 current game state by a series of legal moves.

 This is a classic branch and bound game tree search problem.  You
 could search either from a starting configuration and try to find
 the current game state.  Or start from the current state, use
 'backward' moves, and try to find the initial configuration.  In this
 case, you'd have to include backward moves that 'untake' pieces that
 are missing from the current state.

 Or you could do a simultaneous search from both ends, looking for
 common states in the middle.

 You'd generally use a heuristic search. Problems like this often work
 well with A-Star.  The heuristic evaluator would favor states closer
 to the desired end (either start or current).

 Gene

 On Sep 24, 6:26 am, vrinda vasishth vrindavasis...@gmail.com wrote:
  Asked in microsoft interview
 
  Given a snapshot of an ongoing chess game, which probably is a one vs
 many
  game,  identify whether it is a valid game or not.
 
  It would be great if someone would clarify on what conditions does
  validity of the game depend..
 
  --Vrinda

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Re: [algogeeks] Re: Microsoft interview question

2010-09-24 Thread ashita dadlani
1.The soldiers are initially placed at row 2 or row 7th(each-one of white
and either of black).Also let white ones be at row 2.So they can never be at
row 1st.Incase it is so in the game,its not a valid game.
2.There are Bishops.Each color has one of its Bishop which moves diagonally
on all white squares,and the other on all black squares.Incase it is not
so,the game cannot be valid.
3.Now suppose,no black soldier ever moved.That is,all the black soldiers are
at row 7th.This means that the elephant(i am sorry,I generally mess up with
their names..:P) of any other player(except horse) cannot be in any row but
8th one.

I know only 3 test cases.Incase any one has more,please elaborate.
PS:Vrinda,I also got the same question..:P

On Sat, Sep 25, 2010 at 2:49 AM, Gene gene.ress...@gmail.com wrote:

 Valid must mean that you can get from an initial board to the the
 current game state by a series of legal moves.

 This is a classic branch and bound game tree search problem.  You
 could search either from a starting configuration and try to find
 the current game state.  Or start from the current state, use
 'backward' moves, and try to find the initial configuration.  In this
 case, you'd have to include backward moves that 'untake' pieces that
 are missing from the current state.

 Or you could do a simultaneous search from both ends, looking for
 common states in the middle.

 You'd generally use a heuristic search. Problems like this often work
 well with A-Star.  The heuristic evaluator would favor states closer
 to the desired end (either start or current).

 Gene

 On Sep 24, 6:26 am, vrinda vasishth vrindavasis...@gmail.com wrote:
  Asked in microsoft interview
 
  Given a snapshot of an ongoing chess game, which probably is a one vs
 many
  game,  identify whether it is a valid game or not.
 
  It would be great if someone would clarify on what conditions does
  validity of the game depend..
 
  --Vrinda

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Re: [algogeeks] Yahoo Question:Greatest Sub-sequential sum

2010-09-11 Thread ashita dadlani
@ashish:
what if the array is {-2,3,4,17,-8,9}?

On Wed, Sep 8, 2010 at 8:52 AM, Anand anandut2...@gmail.com wrote:

 Maximum Value Contiguous Subsequence problem in O(n).
 http://codepad.org/BhYxSlp4


 On Tue, Sep 7, 2010 at 2:40 PM, ashish agarwal 
 ashish.cooldude...@gmail.com wrote:

 yeah..it will be i=j+1;
 it was misprinted


 On Tue, Sep 7, 2010 at 10:57 AM, Sahana Gururaj sahan...@gmail.comwrote:

 In the else if condition, the increment of the end index i should be
 i=j+1, not i=j+i; Otherwise the algo is right, follows the principles of
 Kadane's algo :
 http://en.wikipedia.org/wiki/Maximum_subarray_problem#Kadane.27s_algorithm

 On Mon, Sep 6, 2010 at 3:50 PM, ashish agarwal 
 ashish.cooldude...@gmail.com wrote:

 int max=0,sum=0,begin=0,end=0,i=0;
 for(int j=0 to n){
 sum+=a[j];
 if(maxsum){
 max=sum;
 begin=i;
 end=j;
 }
 else if(sum0){
 i=j+i;
 sum=0;
 }

 return sum;
 i will tell the starting index and j will tell ending index;
 o(n);
 correct me if i am wrong



 On Mon, Sep 6, 2010 at 1:42 PM, bittu shashank7andr...@gmail.comwrote:

 Given a sequence of integers, find a continuous subsequence which
 maximizes the sum of its elements, that is, the elements of no other
 single subsequence add up to a value larger than this one. An empty
 subsequence is considered to have the sum 0; thus if all elements are
 negative, the result must be the empty sequence.


 Solution:O(n^2)   i want O(nlogn)...???



 #include stdio.h
  #includeconio.h
 #includeiostream.h
 #includestdlib.h
 int main()
 {
int a[] = {-1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1};
int length = 11;

int begin, end, beginmax, endmax, maxsum, sum, i;

sum = 0;
beginmax = 0;
endmax = -1;
maxsum = 0;


for (begin=0; beginlength; begin++) {
sum = 0;
for(end=begin; endlength; end++) {
sum += a[end];
if(sum  maxsum) {
maxsum = sum;
beginmax = begin;
endmax = end;
}

}
 coutsum\t;
}
  coutendl;
for(i=beginmax; i=endmax; i++) {
printf(%d\n, a[i]);
}

getch();
 }


 its has time complexity O(n^2) ..does better solution exist

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Re: [algogeeks] Yahoo Question:Greatest Sub-sequential sum

2010-09-11 Thread ashita dadlani
@anand:
the maximum sum obtained from your solution is correct.
However,the subarray printed is not correct for the following case:
{-2,3,4,17,-8}
-8 is also getting printed which is not a part of thw subsequence.

On Sat, Sep 11, 2010 at 2:00 PM, ashita dadlani ash@gmail.com wrote:

 @ashish:
 what if the array is {-2,3,4,17,-8,9}?


 On Wed, Sep 8, 2010 at 8:52 AM, Anand anandut2...@gmail.com wrote:

 Maximum Value Contiguous Subsequence problem in O(n).
 http://codepad.org/BhYxSlp4


 On Tue, Sep 7, 2010 at 2:40 PM, ashish agarwal 
 ashish.cooldude...@gmail.com wrote:

 yeah..it will be i=j+1;
 it was misprinted


 On Tue, Sep 7, 2010 at 10:57 AM, Sahana Gururaj sahan...@gmail.comwrote:

 In the else if condition, the increment of the end index i should be
 i=j+1, not i=j+i; Otherwise the algo is right, follows the principles of
 Kadane's algo :
 http://en.wikipedia.org/wiki/Maximum_subarray_problem#Kadane.27s_algorithm

 On Mon, Sep 6, 2010 at 3:50 PM, ashish agarwal 
 ashish.cooldude...@gmail.com wrote:

 int max=0,sum=0,begin=0,end=0,i=0;
 for(int j=0 to n){
 sum+=a[j];
 if(maxsum){
 max=sum;
 begin=i;
 end=j;
 }
 else if(sum0){
 i=j+i;
 sum=0;
 }

 return sum;
 i will tell the starting index and j will tell ending index;
 o(n);
 correct me if i am wrong



 On Mon, Sep 6, 2010 at 1:42 PM, bittu shashank7andr...@gmail.comwrote:

 Given a sequence of integers, find a continuous subsequence which
 maximizes the sum of its elements, that is, the elements of no other
 single subsequence add up to a value larger than this one. An empty
 subsequence is considered to have the sum 0; thus if all elements are
 negative, the result must be the empty sequence.


 Solution:O(n^2)   i want O(nlogn)...???



 #include stdio.h
  #includeconio.h
 #includeiostream.h
 #includestdlib.h
 int main()
 {
int a[] = {-1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1};
int length = 11;

int begin, end, beginmax, endmax, maxsum, sum, i;

sum = 0;
beginmax = 0;
endmax = -1;
maxsum = 0;


for (begin=0; beginlength; begin++) {
sum = 0;
for(end=begin; endlength; end++) {
sum += a[end];
if(sum  maxsum) {
maxsum = sum;
beginmax = begin;
endmax = end;
}

}
 coutsum\t;
}
  coutendl;
for(i=beginmax; i=endmax; i++) {
printf(%d\n, a[i]);
}

getch();
 }


 its has time complexity O(n^2) ..does better solution exist

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Re: [algogeeks] Yahoo Question:Greatest Sub-sequential sum

2010-09-11 Thread ashita dadlani
http://codepad.org/Jui20xme
http://codepad.org/Jui20xmea little modification over anand's code.

On Sat, Sep 11, 2010 at 2:33 PM, ashita dadlani ash@gmail.com wrote:

 @anand:
 the maximum sum obtained from your solution is correct.
 However,the subarray printed is not correct for the following case:
 {-2,3,4,17,-8}
 -8 is also getting printed which is not a part of thw subsequence.


 On Sat, Sep 11, 2010 at 2:00 PM, ashita dadlani ash@gmail.com wrote:

 @ashish:
 what if the array is {-2,3,4,17,-8,9}?


 On Wed, Sep 8, 2010 at 8:52 AM, Anand anandut2...@gmail.com wrote:

 Maximum Value Contiguous Subsequence problem in O(n).
 http://codepad.org/BhYxSlp4


 On Tue, Sep 7, 2010 at 2:40 PM, ashish agarwal 
 ashish.cooldude...@gmail.com wrote:

 yeah..it will be i=j+1;
 it was misprinted


 On Tue, Sep 7, 2010 at 10:57 AM, Sahana Gururaj sahan...@gmail.comwrote:

 In the else if condition, the increment of the end index i should be
 i=j+1, not i=j+i; Otherwise the algo is right, follows the principles of
 Kadane's algo :
 http://en.wikipedia.org/wiki/Maximum_subarray_problem#Kadane.27s_algorithm

 On Mon, Sep 6, 2010 at 3:50 PM, ashish agarwal 
 ashish.cooldude...@gmail.com wrote:

 int max=0,sum=0,begin=0,end=0,i=0;
 for(int j=0 to n){
 sum+=a[j];
 if(maxsum){
 max=sum;
 begin=i;
 end=j;
 }
 else if(sum0){
 i=j+i;
 sum=0;
 }

 return sum;
 i will tell the starting index and j will tell ending index;
 o(n);
 correct me if i am wrong



 On Mon, Sep 6, 2010 at 1:42 PM, bittu shashank7andr...@gmail.comwrote:

 Given a sequence of integers, find a continuous subsequence which
 maximizes the sum of its elements, that is, the elements of no other
 single subsequence add up to a value larger than this one. An empty
 subsequence is considered to have the sum 0; thus if all elements are
 negative, the result must be the empty sequence.


 Solution:O(n^2)   i want O(nlogn)...???



 #include stdio.h
  #includeconio.h
 #includeiostream.h
 #includestdlib.h
 int main()
 {
int a[] = {-1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1};
int length = 11;

int begin, end, beginmax, endmax, maxsum, sum, i;

sum = 0;
beginmax = 0;
endmax = -1;
maxsum = 0;


for (begin=0; beginlength; begin++) {
sum = 0;
for(end=begin; endlength; end++) {
sum += a[end];
if(sum  maxsum) {
maxsum = sum;
beginmax = begin;
endmax = end;
}

}
 coutsum\t;
}
  coutendl;
for(i=beginmax; i=endmax; i++) {
printf(%d\n, a[i]);
}

getch();
 }


 its has time complexity O(n^2) ..does better solution exist

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Re: [algogeeks] Re: Write a C program to sort a stack in ascending order.

2010-09-11 Thread ashita dadlani
else{
ascending(s, i);
push(top);
 }
}
@swinivas:why have you used ascending(s,i) here?
On Sat, Sep 11, 2010 at 6:40 PM, Srinivas lavudyasrinivas0...@gmail.comwrote:

 reverse(stack *s){
   IsEmpty(s)
return;
   top = pop(s);
   reverse(s);
   ascending(s, top);
 }
 ascending(stack *s, int top){
  IsEmpty(s){
 push(top);
 return;
  }
  i = pop(s);
  if(i  top){
 ascending(s, top);
 push(i);
  }
  else{
 ascending(s, i);
 push(top);
  }
 }

 Please let me know if it wont work..thanks

 On Jul 18, 6:58 am, xyombie brianbl...@gmail.com wrote:
  What about a quick sort O(log n)
 
  void sort_stack(Stack *src, Stack *dst)
  {
  if(! src-IsEmpty() )
  {
  Stack smaller, larger;
  int pivot = src-Pop();
 
  while(! src-IsEmpty() )
  {
  int tmp = src-Pop();
  if(tmp  pivot)
  smaller-Push(tmp);
  else
  larger-Push(tmp);
  }
 
  sort_stack(smaller, dst);
  dst-Push(pivot);
  sort_stack(larger, dst);
  }
 
  }
 
  On Jul 17, 9:28 am, vijay auvija...@gmail.com wrote:
 
   Write a C program to sort a stack in ascending order. You should not
   make any assumptions about how the stack is implemented. The following
   are the only
   functions that should be used to write this program:
   Push | Pop | Top | IsEmpty | IsFull
The algorithm is O(N^2) and appears below.
   Do we have any other better solution which is less than O(n * n) ?
 
   void sort_stack(Stack * src, Stack * dest)
{
 while (!src-IsEmpty())
{
  Int tmp = src-Pop();
  while(!dest-IsEmpty()  dest-Top()  tmp)
 {
  src-Push(dest-Pop());
 }
  dest-Push(tmp);
 }
 
   }

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Re: [algogeeks] Re: Write a C program to sort a stack in ascending order.

2010-09-11 Thread ashita dadlani
@Srinivas:
shouldn't it be:

i = pop(s);
 if(i  top){
ascending(s, i);
push(top);
 }
 else{
ascending(s, top);
push(i);
 }

On Sat, Sep 11, 2010 at 6:52 PM, ashita dadlani ash@gmail.com wrote:


 else{
 ascending(s, i);
 push(top);
  }
 }
 @swinivas:why have you used ascending(s,i) here?
 On Sat, Sep 11, 2010 at 6:40 PM, Srinivas 
 lavudyasrinivas0...@gmail.comwrote:

 reverse(stack *s){
   IsEmpty(s)
return;
   top = pop(s);
   reverse(s);
   ascending(s, top);
 }
 ascending(stack *s, int top){
  IsEmpty(s){
 push(top);
 return;
  }
  i = pop(s);
  if(i  top){
 ascending(s, top);
 push(i);
  }
  else{
 ascending(s, i);
 push(top);
  }
 }

 Please let me know if it wont work..thanks

 On Jul 18, 6:58 am, xyombie brianbl...@gmail.com wrote:
  What about a quick sort O(log n)
 
  void sort_stack(Stack *src, Stack *dst)
  {
  if(! src-IsEmpty() )
  {
  Stack smaller, larger;
  int pivot = src-Pop();
 
  while(! src-IsEmpty() )
  {
  int tmp = src-Pop();
  if(tmp  pivot)
  smaller-Push(tmp);
  else
  larger-Push(tmp);
  }
 
  sort_stack(smaller, dst);
  dst-Push(pivot);
  sort_stack(larger, dst);
  }
 
  }
 
  On Jul 17, 9:28 am, vijay auvija...@gmail.com wrote:
 
   Write a C program to sort a stack in ascending order. You should not
   make any assumptions about how the stack is implemented. The following
   are the only
   functions that should be used to write this program:
   Push | Pop | Top | IsEmpty | IsFull
The algorithm is O(N^2) and appears below.
   Do we have any other better solution which is less than O(n * n) ?
 
   void sort_stack(Stack * src, Stack * dest)
{
 while (!src-IsEmpty())
{
  Int tmp = src-Pop();
  while(!dest-IsEmpty()  dest-Top()  tmp)
 {
  src-Push(dest-Pop());
 }
  dest-Push(tmp);
 }
 
   }

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Re: [algogeeks] Re: Write a C program to sort a stack in ascending order.

2010-09-11 Thread ashita dadlani
@Srinivas:
Sorry for the confusion.
But if we have a stack {5,8,3,4,2} with 5 as the last input,ie,top,
do we have to arrange the stack such that s={2,3,4,5,8}with 2 as the top?
if I am getting it correct,then shouldn't your algo be modified slightly as
follows:?
reverse(stack *s){
  IsEmpty(s)
   return;
  top = pop(s);
  reverse(s);
  ascending(s, top);
}
ascending(stack *s, int top){
 IsEmpty(s){
push(top);
return;
 }
 i = pop(s);
 if(i  top){

push(top);
 }
 else{
ascending(s, top);
push(i);
 }
}


On Sat, Sep 11, 2010 at 7:04 PM, ashita dadlani ash@gmail.com wrote:

 @Srinivas:
 shouldn't it be:

 i = pop(s);
  if(i  top){
 ascending(s, i);
 push(top);
  }
  else{

 ascending(s, top);
 push(i);
  }


 On Sat, Sep 11, 2010 at 6:52 PM, ashita dadlani ash@gmail.com wrote:


 else{
 ascending(s, i);
 push(top);
  }
 }
 @swinivas:why have you used ascending(s,i) here?
 On Sat, Sep 11, 2010 at 6:40 PM, Srinivas 
 lavudyasrinivas0...@gmail.comwrote:

 reverse(stack *s){
   IsEmpty(s)
return;
   top = pop(s);
   reverse(s);
   ascending(s, top);
 }
 ascending(stack *s, int top){
  IsEmpty(s){
 push(top);
 return;
  }
  i = pop(s);
  if(i  top){
 ascending(s, top);
 push(i);
  }
  else{
 ascending(s, i);
 push(top);
  }
 }

 Please let me know if it wont work..thanks

 On Jul 18, 6:58 am, xyombie brianbl...@gmail.com wrote:
  What about a quick sort O(log n)
 
  void sort_stack(Stack *src, Stack *dst)
  {
  if(! src-IsEmpty() )
  {
  Stack smaller, larger;
  int pivot = src-Pop();
 
  while(! src-IsEmpty() )
  {
  int tmp = src-Pop();
  if(tmp  pivot)
  smaller-Push(tmp);
  else
  larger-Push(tmp);
  }
 
  sort_stack(smaller, dst);
  dst-Push(pivot);
  sort_stack(larger, dst);
  }
 
  }
 
  On Jul 17, 9:28 am, vijay auvija...@gmail.com wrote:
 
   Write a C program to sort a stack in ascending order. You should not
   make any assumptions about how the stack is implemented. The
 following
   are the only
   functions that should be used to write this program:
   Push | Pop | Top | IsEmpty | IsFull
The algorithm is O(N^2) and appears below.
   Do we have any other better solution which is less than O(n * n) ?
 
   void sort_stack(Stack * src, Stack * dest)
{
 while (!src-IsEmpty())
{
  Int tmp = src-Pop();
  while(!dest-IsEmpty()  dest-Top()  tmp)
 {
  src-Push(dest-Pop());
 }
  dest-Push(tmp);
 }
 
   }

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Re: [algogeeks] Re: Array Problem

2010-08-22 Thread ashita dadlani
a={1,2,2,3,4}
b={1,2,3,3,4}
in this case???
elements are not equal but they certainly consist equal set of
integers(1,2,3,4) which is what question says.

On Sun, Aug 22, 2010 at 7:53 AM, UMarius mar...@aseara.ro wrote:

 @Nikhil : I considered the array to be a linked list. xoring the
 indexes helps when you don't know how many elements you have.

 Marius.

 On Aug 22, 5:04 am, Nikhil Agarwal nikhil.bhoja...@gmail.com wrote:
  @marius Why are you xorring the indexes along with nos.?any special
 reason?
 
 
 
 
 
 
 
 
 
  On Sun, Aug 22, 2010 at 7:19 AM, UMarius mar...@aseara.ro wrote:
   @Dave: I read the question correctly and for A = { 1 , 2} B = { 2 , 1}
   the output is correct.
   Maybe I didn't explain the steps correctly. This is the code:
 
   for(int i = 0 ; i  arr1.Length ; i++)
  {
  arr1XOR ^= arr1[i];
  arr1XOR ^= i;
 
  arr1SUM += arr1[i];
  arr1MUL *= arr1[i];
  }
 
  for (int i = 0; i  arr2.Length; i++)
  {
  arr2XOR ^= arr2[i];
  arr2XOR ^= i;
 
  arr2SUM += arr2[i];
  arr2MUL *= arr2[i];
  }
 
  if(arr1XOR == arr2XOR  arr1SUM == arr2SUM  arr1MUL ==
   arr2MUL)
  {
  //SAME VALUES - IDENTICAL ARRAYS
  }
  else
  {
  //NOT IDENTICAL
  }
 
   Please correct me if I'm wrong.
 
   Marius.
 
   On Aug 22, 3:45 am, Dave dave_and_da...@juno.com wrote:
@UMarius: A = {1,2}, B = {2,1} fails your test. If you reread the
original problem, you see that the question is not whether the arrays
are identical (which is easily determined by simply comparing them
element-by-element in O(n)), but whether they contain the same
 values,
possibly in a different order.
 
Dave
 
On Aug 21, 11:01 am, UMarius mar...@aseara.ro wrote:
 
 What about this?
 
 1. xor all elements of each array and their corresponding indexes 
 sum all the elements of each array  mul all elements of each array
 2. if all results are the same then the arrays are identical
 
 Nice to meet you all, I just joined and this is my first post :)
 ...
 
  Given two arrays of numbers, find if each of the two arrays have
 the
  same set of integers ? Suggest an algo which can run faster than
   NlogN
  without extra space?- Hide quoted text -
 
 - Show quoted text -
 
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  Thanks  Regards
  Nikhil Agarwal
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  Computer Science  Engineering,
  National Institute Of Technology,
 Durgapur,Indiahttp://tech-nikk.blogspot.comhttp://
 beta.freshersworld.com/communities/nitd

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[algogeeks]

2010-08-16 Thread ashita dadlani
You have a string say foobarfoo in which fo and oo aree repeated
twice.You have to find all such repeated pairs in O(n) time,The string can
only have alphanumeric elements in it.

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[algogeeks]

2010-08-16 Thread ashita dadlani
write the test cases for rectangle overlap.

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Re: [algogeeks] Re: Modified Binary Search

2010-08-08 Thread ashita dadlani
its possible.first apply binary search to find the index where the array is
rotated.
eg.a=[8,9,1,2,3,4,56,7]
first apply binary search to find the value i=2.
now we have 2 sub sorted arrays,ie, from i=0 to i=1 and from i=2 to i=8.
apply binary search to these sub sorted arrays each of which will take time
logn.
hence total time:O(3logn)=O(logn)

On Sun, Aug 8, 2010 at 12:04 PM, Manjunath Manohar manjunath.n...@gmail.com
 wrote:

 hey wait a sec,.. we wont be given the k value..

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Re: [algogeeks] Re: Modified Binary Search

2010-08-08 Thread ashita dadlani
first of all Gene..this is not he but she :P
and secondly,can you please mention the details you are talking about so
that we can reach to a solution?

On Sun, Aug 8, 2010 at 7:48 PM, Gene gene.ress...@gmail.com wrote:

 Well, then, you must say that in the problem!  To to find k, ashita
 dadlani's solution works fine, except he has left out important
 details.  This binary search is a bit tricker that the one to find a
 value.

 On Aug 8, 2:34 am, Manjunath Manohar manjunath.n...@gmail.com wrote:
  hey wait a sec,.. we wont be given the k value..

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Re: [algogeeks] Adobe Ques

2010-07-14 Thread ashita dadlani
@ashish:yours is not a generalized solution.Here you are assuming that u
know r=3.
What if we need to generalize the solution?

On Wed, Jul 14, 2010 at 11:58 PM, Ashish Goel ashg...@gmail.com wrote:

 for (int i=0;in-r+1;i++)
 printf(%c%c%c , a[i],a[i+1],a[i+2]);


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



 On Wed, Jul 14, 2010 at 6:23 PM, ankit mahendru ankit.mahend...@gmail.com
  wrote:

 Write the code to print combinations of a given array (n r were given)
 e.g for abcde .r=3 ...print  abc, bcd' cde etc

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