Re: [algogeeks] Re: MS Question: Reverse stack using push, pop without any auxiliary data structure

[Prem Nagarajan]

I think there is a problem in this solution.
U r accessing stack elements from 1 to n in the outer loop. It is not
possible. 1st element cannot be accessed without popping first n-1 elements
out.
On Mon, Jun 18, 2012 at 11:33 AM, Rituraj  wrote:

> My  iterative approach
>
> /*code in c*/
> #include
> int main()
> {
>  int stack[]={1,2,3,4,5,6,7,8},top=7;//
>  int i,j,temp;
>
>  for(i=1;i<=top;i++)
>  {
>   temp=stack[i];
>
>   for(j=i;j>0;j--)
> stack[j]=stack[j-1];
>
>   stack[0]=temp;
>  }
>
>  for(i=0;i<=top;i++)
>printf("%d ",stack[i] );
>
>  return 0;
> }
>  /*
>
>
> Rituraj
> 2nd Yr.
> B.tech CSE
> NIT -Trichy
>
> --
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Re: [algogeeks] Re: MS Question: Reverse stack using push, pop without any auxiliary data structure

[aditya gupta]

this is not a stack at all, u have just named it as a stack. for it to be a
stack u should access only the top most element at any point of time!!!

On Mon, Jun 18, 2012 at 11:33 AM, Rituraj  wrote:

> My  iterative approach
>
> /*code in c*/
> #include
> int main()
> {
>  int stack[]={1,2,3,4,5,6,7,8},top=7;//
>  int i,j,temp;
>
>  for(i=1;i<=top;i++)
>  {
>   temp=stack[i];
>
>   for(j=i;j>0;j--)
> stack[j]=stack[j-1];
>
>   stack[0]=temp;
>  }
>
>  for(i=0;i<=top;i++)
>printf("%d ",stack[i] );
>
>  return 0;
> }
>  /*
>
>
> Rituraj
> 2nd Yr.
> B.tech CSE
> NIT -Trichy
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To view this discussion on the web visit
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>



-- 
Aditya Gupta
B.Tech III yr CSE
IITR

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Re: [algogeeks] Re: MS Question: Reverse stack using push, pop without any auxiliary data structure

[Abhishek Sharma]

In a stack, you can't access any element directly, except the top one.

On Mon, Jun 18, 2012 at 11:33 AM, Rituraj  wrote:

> My  iterative approach
>
> /*code in c*/
> #include
> int main()
> {
>  int stack[]={1,2,3,4,5,6,7,8},top=7;//
>  int i,j,temp;
>
>  for(i=1;i<=top;i++)
>  {
>   temp=stack[i];
>
>   for(j=i;j>0;j--)
> stack[j]=stack[j-1];
>
>   stack[0]=temp;
>  }
>
>  for(i=0;i<=top;i++)
>printf("%d ",stack[i] );
>
>  return 0;
> }
>  /*
>
>
> Rituraj
> 2nd Yr.
> B.tech CSE
> NIT -Trichy
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To view this discussion on the web visit
> https://groups.google.com/d/msg/algogeeks/-/n1OE58e8B7IJ.
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> http://groups.google.com/group/algogeeks?hl=en.
>



-- 
Abhishek Sharma
Under-Graduate Student,
PEC University of Technology

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Re: [algogeeks] Re: MS Question : find word in 2D array

[hary rathor]

1 search should in using KMP algo so that It can be seacrh in O(n) . let
function is   int KMP(src,trget, searchDirection )
this kmpSearch funtion should be implemented is such a fashion that is
search in both direction.
3. assume that give 2d array name is array

const int row =1;
const int col =1;
const int dig =1;

for(i=0;ihttp://groups.google.com/group/algogeeks?hl=en.

Re: [algogeeks] Re: MS Question : find word in 2D array

[Guneesh Paul Singh]

@utsav
in cases where u have repeating character in ur search string how would u
assign the numbers
eg in abaac

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Re: [algogeeks] Re: MS question : string compression

[Ashish Goel]

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


On Fri, Jun 8, 2012 at 12:54 PM, Ashish Goel  wrote:

> #include "stdafx.h"
> #include 
> using namespace std;
>
> const int len = 20;
> const int maxCount = 127;
>
> int rle(char* pStr, int length, char* pNew) {
>
> if (!pStr) return -1;
> if (length <3) return -1;
>
> int i = 0;
> int k = 0;
> char p1 = pStr[i++];
>  char p2 = pStr[i++];
> char p3 = pStr[i++];
> int pos=0;
>  int cCount = 0;
> bool verbatim = false;
> while ((p3) && (i  if (p1==p2) {
> if (p2==p3) {
> if (i == k+3) //no vRun
>  verbatim = false;//no vRun befor this cRun
> if (verbatim)
> {
>  int vEnd = (i-3)-k;;
> pNew[pos++] = vEnd;
> for (int t=k;t  pNew[pos++]=pStr[t];
> }
> }
>  cCount++;
>
if (cCount == maxCount) {
pNew[pos++] = -cCount; ///
 pNew[pos++] = p3;

p1 = pStr[i++];
if (!p1) break;
 k = 0;
p2 =  pStr[i++];
if (!p2) break;
 p3 = pStr[i++];
cCount = 0;
continue;
 } else {

> /*p1 = p2;
>  p2 = p3;  //not required*/
>  p3 = pStr[i++];
> }
> }
>  else { //run end or no run at all
> if (cCount >0) { //a run
> pNew[pos++] = -cCount; ///
>  pNew[pos++] = p2;
> p1 = p3;
> k = i-1; //p3's position
>  p2 =  pStr[i++];
> if (!p2) break;
> p3 = pStr[i++];
>  cCount = 0;
> }
> else { /*aab */
>  verbatim = true;
> p1 = p2;
> p2 = p3;
>  p3 =pStr[i++];
> }
> }
>  }
> else { //no run
> verbatim = true;
>  p1 = p2;
> p2 = p3;
> p3 =pStr[i++];
>  }
> }
> //possible run or no run here
>  if (cCount>0) {
> pNew[pos++] = -cCount;
> pNew[pos++] = p2;
>  } else {
> if (k pNew[pos++] = length-k-1;
>  for (int t=k;t pNew[pos++]=pStr[t];
> }
>  }
> }
> pNew[pos]='\0';
>  return 1;
>
> }
> void rleDecode(char *pEnc, char *pDec, char *pOrig)
> {
> int i = 0;
>  int pos =0;
> int count ;
> char character ;
>  do {
> count = pEnc[i++];
> if (count <0) {
>  count = 2-count;
> character = pEnc[i++];
> for (int j=0;j  pDec[pos++] = character;
> }
> else {
>  //pNew[pos++]=character;
> for (int j=0;j pDec[pos++]=pEnc[i++];
>  }
> }
> }while (pEnc[i]);
>  pDec[pos]='\0';
> for(int i=0;i if (pOrig[i]!=pDec[i]) cout <<"JERK, do it again!! (:" <
> }
>
>
> int _tmain(int argc, _TCHAR* argv[])
> {
> char *pStr = (char *)malloc(sizeof(char)*len);
>  pStr = "abccdddijkk"; //TRY more examples
> char *pNew = (char *)malloc(sizeof(char)*len);
>  char *pDec = (char *)malloc(sizeof(char)*len);
> //rleSimple(pStr,pNew);
> rle(pStr,len,pNew);
>  rleDecode(pNew, pDec, pStr);
> return 0;
> }
>
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
> On Fri, Jun 8, 2012 at 9:04 AM, Ashish Goel  wrote:
>
>>
>>
>> The idea here is that there will be parts of the stream which actually
>> should not be compressed. For example abcdef as well as aa do not need any
>> compression. We need to compress only if 3 characters match because for
>> compressing two chars we will take up 2 chars so no compression benefit (:
>>
>> So we need to keep a pos as a reference to say that here is the position
>> in the string i am processing now and do the compress(either verbatim or
>> real compress) when 3 same chars are found
>>
>> eg
>>
>> abcfdgffg: pos is 0 and at index 8 we get to know that there is a
>> run, so we should say 8-3+1=6 need to go verbatim so we write 6abcfdg and
>> update pos to index 6, and count to 1. Since now run flag is on, we
>> continue till we find a triplet mismatch(f==f but f!=g) which happens at g
>> (index 12)implying an end to a run, therefore now count is 4, we would
>> write 4f implying 2+4 times of next char should be expanded. now again pos
>> will be set to 12, count to 0 and three same char check should re-begin.
>> This will for sure have 2 while loops and a bit comex, and i donot think
>> this is what the interviewer should expect one to code. Kindly note that if
>> run is more than max length, we need to tweak the writing part too.
>>
>>
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>> On Thu, Jun 7, 2012 at 7:05 PM, Navin Gupta wrote:
>>
>>> If abcdef is changed to a1b1c1d1e1f1,
>>> then we need to allocate memory dynamically.
>>> Because length is increased,I think this has no practical
>>> implementation.As "abcdef " serves the same purpose.
>>>
>>>
>>> On Sunday, 3 June 2012 09:36:25 UTC+5:30, utsav sharma wrote:

 @ashish:-algo given in link wiil fail for "abcdef"

 @navin:- output of "abcdef" should be "1a1b1c1d1e1f"

 On Sun, May 27, 2012 at 3:24 PM, Ashish Goel  wrote:

> Will fail for the sing having say 257characters all same
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
> On Sat, May 26, 2012 at 12:26 PM, Navin Gupta 
> wrote:
>
>> This is called Run-Length-Encoding (RLE)  of a string.
>> Its purpose is to save space.So in case 

Re: [algogeeks] Re: MS question : string compression

[Ashish Goel]

#include "stdafx.h"
#include 
using namespace std;

const int len = 20;
const int maxCount = 127;

int rle(char* pStr, int length, char* pNew) {

if (!pStr) return -1;
if (length <3) return -1;

int i = 0;
int k = 0;
char p1 = pStr[i++];
 char p2 = pStr[i++];
char p3 = pStr[i++];
int pos=0;
 int cCount = 0;
bool verbatim = false;
while ((p3) && (i0) { //a run
pNew[pos++] = -cCount; ///
 pNew[pos++] = p2;
p1 = p3;
k = i-1; //p3's position
 p2 =  pStr[i++];
if (!p2) break;
p3 = pStr[i++];
 cCount = 0;
}
else { /*aab */
 verbatim = true;
p1 = p2;
p2 = p3;
 p3 =pStr[i++];
}
}
 }
else { //no run
verbatim = true;
 p1 = p2;
p2 = p3;
p3 =pStr[i++];
 }
}
//possible run or no run here
 if (cCount>0) {
pNew[pos++] = -cCount;
pNew[pos++] = p2;
 } else {
if (k wrote:

>
>
> The idea here is that there will be parts of the stream which actually
> should not be compressed. For example abcdef as well as aa do not need any
> compression. We need to compress only if 3 characters match because for
> compressing two chars we will take up 2 chars so no compression benefit (:
>
> So we need to keep a pos as a reference to say that here is the position
> in the string i am processing now and do the compress(either verbatim or
> real compress) when 3 same chars are found
>
> eg
>
> abcfdgffg: pos is 0 and at index 8 we get to know that there is a run,
> so we should say 8-3+1=6 need to go verbatim so we write 6abcfdg and update
> pos to index 6, and count to 1. Since now run flag is on, we continue till
> we find a triplet mismatch(f==f but f!=g) which happens at g (index
> 12)implying an end to a run, therefore now count is 4, we would write 4f
> implying 2+4 times of next char should be expanded. now again pos will be
> set to 12, count to 0 and three same char check should re-begin. This will
> for sure have 2 while loops and a bit comex, and i donot think this is what
> the interviewer should expect one to code. Kindly note that if run is more
> than max length, we need to tweak the writing part too.
>
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
> On Thu, Jun 7, 2012 at 7:05 PM, Navin Gupta wrote:
>
>> If abcdef is changed to a1b1c1d1e1f1,
>> then we need to allocate memory dynamically.
>> Because length is increased,I think this has no practical
>> implementation.As "abcdef " serves the same purpose.
>>
>>
>> On Sunday, 3 June 2012 09:36:25 UTC+5:30, utsav sharma wrote:
>>>
>>> @ashish:-algo given in link wiil fail for "abcdef"
>>>
>>> @navin:- output of "abcdef" should be "1a1b1c1d1e1f"
>>>
>>> On Sun, May 27, 2012 at 3:24 PM, Ashish Goel  wrote:
>>>
 Will fail for the sing having say 257characters all same

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


 On Sat, May 26, 2012 at 12:26 PM, Navin Gupta 
 wrote:

> This is called Run-Length-Encoding (RLE)  of a string.
> Its purpose is to save space.So in case of abcdef,I think the output
> needed is abcdef (1 is implicit).
> The added benefit is it makes the solution in-place.
>
> Approach:- (In-place and Linear Time)
> Start from the left of  string  and  PREVIOUS_CHAR = str[0]
> move forward till u match the CURRENT_CHAR with PREVIOUS_CHAR and keep
> count of PREVIOUS_CHAR
> At any point if (PREVIOUS_CHAR!=CURRENT_CHAR)
>put the count of prev_char next to the start position of the
> previous character.
>
> Below is the working code :-
> void torle(char *str)
> {   int i=0,j=0,k=0,cnt=1;
> char cur_char=str[0],num[100];
> while(str[j+1])
> {
> cnt=1;
> while(str[j+1]==cur_char && str[j]!='\0'){
> j++;
> cnt++;
> }
> str[i++]=cur_char;
> if( cnt>9 ){
> itoa(cnt,num);
> k=0;
> while(num[k]) str[i++]=num[k++];
> }
> else if( cnt>1 && cnt<10 )
> str[i++]= cnt+'0';
> j++;
> if(str[j]) cur_char=str[j];
> }
> if(i!=0){
> if(cnt==1)
> str[i++]=cur_char;
> str[i]='\0';
>
> }
> }
>
>
> On Saturday, 26 May 2012 04:32:35 UTC+5:30, utsav sharma wrote:
>>
>> Implement a method to perform basic string compression using the
>> counts of repeated characters.(inplace)
>>
>> eg:- input: "aaabcdef"
>>  output:"3a5b1c1d1e1f".
>>
>> what should be my approach to this problem
>>
>> if i calculate the size of array required to store the output string
>> and start from the last of the array then i wldn't get the right
>> answer of above input case.
>>
>> and if start from front then i wldn't get the right answer of this
>> input case
>> e

Re: [algogeeks] Re: MS question : string compression

[Ashish Goel]

The idea here is that there will be parts of the stream which actually
should not be compressed. For example abcdef as well as aa do not need any
compression. We need to compress only if 3 characters match because for
compressing two chars we will take up 2 chars so no compression benefit (:

So we need to keep a pos as a reference to say that here is the position in
the string i am processing now and do the compress(either verbatim or real
compress) when 3 same chars are found

eg

abcfdgffg: pos is 0 and at index 8 we get to know that there is a run,
so we should say 8-3+1=6 need to go verbatim so we write 6abcfdg and update
pos to index 6, and count to 1. Since now run flag is on, we continue till
we find a triplet mismatch(f==f but f!=g) which happens at g (index
12)implying an end to a run, therefore now count is 4, we would write 4f
implying 2+4 times of next char should be expanded. now again pos will be
set to 12, count to 0 and three same char check should re-begin. This will
for sure have 2 while loops and a bit comex, and i donot think this is what
the interviewer should expect one to code. Kindly note that if run is more
than max length, we need to tweak the writing part too.


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


On Thu, Jun 7, 2012 at 7:05 PM, Navin Gupta  wrote:

> If abcdef is changed to a1b1c1d1e1f1,
> then we need to allocate memory dynamically.
> Because length is increased,I think this has no practical
> implementation.As "abcdef " serves the same purpose.
>
>
> On Sunday, 3 June 2012 09:36:25 UTC+5:30, utsav sharma wrote:
>>
>> @ashish:-algo given in link wiil fail for "abcdef"
>>
>> @navin:- output of "abcdef" should be "1a1b1c1d1e1f"
>>
>> On Sun, May 27, 2012 at 3:24 PM, Ashish Goel  wrote:
>>
>>> Will fail for the sing having say 257characters all same
>>>
>>> Best Regards
>>> Ashish Goel
>>> "Think positive and find fuel in failure"
>>> +919985813081
>>> +919966006652
>>>
>>>
>>> On Sat, May 26, 2012 at 12:26 PM, Navin Gupta wrote:
>>>
 This is called Run-Length-Encoding (RLE)  of a string.
 Its purpose is to save space.So in case of abcdef,I think the output
 needed is abcdef (1 is implicit).
 The added benefit is it makes the solution in-place.

 Approach:- (In-place and Linear Time)
 Start from the left of  string  and  PREVIOUS_CHAR = str[0]
 move forward till u match the CURRENT_CHAR with PREVIOUS_CHAR and keep
 count of PREVIOUS_CHAR
 At any point if (PREVIOUS_CHAR!=CURRENT_CHAR)
put the count of prev_char next to the start position of the
 previous character.

 Below is the working code :-
 void torle(char *str)
 {   int i=0,j=0,k=0,cnt=1;
 char cur_char=str[0],num[100];
 while(str[j+1])
 {
 cnt=1;
 while(str[j+1]==cur_char && str[j]!='\0'){
 j++;
 cnt++;
 }
 str[i++]=cur_char;
 if( cnt>9 ){
 itoa(cnt,num);
 k=0;
 while(num[k]) str[i++]=num[k++];
 }
 else if( cnt>1 && cnt<10 )
 str[i++]= cnt+'0';
 j++;
 if(str[j]) cur_char=str[j];
 }
 if(i!=0){
 if(cnt==1)
 str[i++]=cur_char;
 str[i]='\0';

 }
 }


 On Saturday, 26 May 2012 04:32:35 UTC+5:30, utsav sharma wrote:
>
> Implement a method to perform basic string compression using the
> counts of repeated characters.(inplace)
>
> eg:- input: "aaabcdef"
>  output:"3a5b1c1d1e1f".
>
> what should be my approach to this problem
>
> if i calculate the size of array required to store the output string
> and start from the last of the array then i wldn't get the right
> answer of above input case.
>
> and if start from front then i wldn't get the right answer of this
> input case
> eg:- input: "abcdef"
>  output: "1a1b1c1d1e1f"
>
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>>>
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Re: [algogeeks] Re: MS question : string compression

[Navin Gupta]

If abcdef is changed to a1b1c1d1e1f1, 
then we need to allocate memory dynamically.
Because length is increased,I think this has no practical  
implementation.As "abcdef " serves the same purpose.

On Sunday, 3 June 2012 09:36:25 UTC+5:30, utsav sharma wrote:
>
> @ashish:-algo given in link wiil fail for "abcdef"
>
> @navin:- output of "abcdef" should be "1a1b1c1d1e1f"
>
> On Sun, May 27, 2012 at 3:24 PM, Ashish Goel  wrote:
>
>> Will fail for the sing having say 257characters all same
>>
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>> On Sat, May 26, 2012 at 12:26 PM, Navin Gupta wrote:
>>
>>> This is called Run-Length-Encoding (RLE)  of a string.
>>> Its purpose is to save space.So in case of abcdef,I think the output 
>>> needed is abcdef (1 is implicit).
>>> The added benefit is it makes the solution in-place.
>>>
>>> Approach:- (In-place and Linear Time)
>>> Start from the left of  string  and  PREVIOUS_CHAR = str[0]
>>> move forward till u match the CURRENT_CHAR with PREVIOUS_CHAR and keep 
>>> count of PREVIOUS_CHAR
>>> At any point if (PREVIOUS_CHAR!=CURRENT_CHAR)
>>>put the count of prev_char next to the start position of the previous 
>>> character.
>>>
>>> Below is the working code :-
>>> void torle(char *str)
>>> {   int i=0,j=0,k=0,cnt=1;
>>> char cur_char=str[0],num[100];
>>> while(str[j+1])
>>> {
>>> cnt=1;
>>> while(str[j+1]==cur_char && str[j]!='\0'){
>>> j++;
>>> cnt++;
>>> }
>>> str[i++]=cur_char;
>>> if( cnt>9 ){
>>> itoa(cnt,num);
>>> k=0;
>>> while(num[k]) str[i++]=num[k++];
>>> }
>>> else if( cnt>1 && cnt<10 )
>>> str[i++]= cnt+'0';
>>> j++;
>>> if(str[j]) cur_char=str[j];
>>> }
>>> if(i!=0){
>>> if(cnt==1)
>>> str[i++]=cur_char;
>>> str[i]='\0';
>>>
>>> }
>>> }
>>>
>>>
>>> On Saturday, 26 May 2012 04:32:35 UTC+5:30, utsav sharma wrote:

 Implement a method to perform basic string compression using the counts 
 of repeated characters.(inplace)

 eg:- input: "aaabcdef"  
  output:"3a5b1c1d1e1f". 

 what should be my approach to this problem

 if i calculate the size of array required to store the output string 
 and start from the last of the array then i wldn't get the right answer 
 of above input case.

 and if start from front then i wldn't get the right answer of this 
 input case 
 eg:- input: "abcdef" 
  output: "1a1b1c1d1e1f"

>>>  -- 
>>> You received this message because you are subscribed to the Google 
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Re: [algogeeks] Re: MS question : string compression

[Navin Kumar]

Try this code...i think my code can be improved further.

#include 

main()
{
  int flag=0;
  int i;
  int c,prev;
  int chars[26];
  for(i=0;i<26;i++)
chars[i]=0;



 while((c=getchar())!=EOF)
 {
  if(flag==0)
  {
prev=c;
flag=1;
  }

 if(c==prev)
 {
  if(c>='a' && c<='z')
 ++chars[c-'a'];
 }
 else{
  printf("%d %c ", chars[prev-'a'],prev);
  chars[prev-'a']=0;

  if(c>='a' && c<='z')
 ++chars[c-'a'];
 }

 prev=c;

}

}


On Sun, Jun 3, 2012 at 9:36 AM, utsav sharma wrote:

> @ashish:-algo given in link wiil fail for "abcdef"
>
> @navin:- output of "abcdef" should be "1a1b1c1d1e1f"
>
>
> On Sun, May 27, 2012 at 3:24 PM, Ashish Goel  wrote:
>
>> Will fail for the sing having say 257characters all same
>>
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>> On Sat, May 26, 2012 at 12:26 PM, Navin Gupta wrote:
>>
>>> This is called Run-Length-Encoding (RLE)  of a string.
>>> Its purpose is to save space.So in case of abcdef,I think the output
>>> needed is abcdef (1 is implicit).
>>> The added benefit is it makes the solution in-place.
>>>
>>> Approach:- (In-place and Linear Time)
>>> Start from the left of  string  and  PREVIOUS_CHAR = str[0]
>>> move forward till u match the CURRENT_CHAR with PREVIOUS_CHAR and keep
>>> count of PREVIOUS_CHAR
>>> At any point if (PREVIOUS_CHAR!=CURRENT_CHAR)
>>>put the count of prev_char next to the start position of the previous
>>> character.
>>>
>>> Below is the working code :-
>>> void torle(char *str)
>>> {   int i=0,j=0,k=0,cnt=1;
>>> char cur_char=str[0],num[100];
>>> while(str[j+1])
>>> {
>>> cnt=1;
>>> while(str[j+1]==cur_char && str[j]!='\0'){
>>> j++;
>>> cnt++;
>>> }
>>> str[i++]=cur_char;
>>> if( cnt>9 ){
>>> itoa(cnt,num);
>>> k=0;
>>> while(num[k]) str[i++]=num[k++];
>>> }
>>> else if( cnt>1 && cnt<10 )
>>> str[i++]= cnt+'0';
>>> j++;
>>> if(str[j]) cur_char=str[j];
>>> }
>>> if(i!=0){
>>> if(cnt==1)
>>> str[i++]=cur_char;
>>> str[i]='\0';
>>>
>>> }
>>> }
>>>
>>>
>>> On Saturday, 26 May 2012 04:32:35 UTC+5:30, utsav sharma wrote:

 Implement a method to perform basic string compression using the counts
 of repeated characters.(inplace)

 eg:- input: "aaabcdef"
  output:"3a5b1c1d1e1f".

 what should be my approach to this problem

 if i calculate the size of array required to store the output string
 and start from the last of the array then i wldn't get the right answer
 of above input case.

 and if start from front then i wldn't get the right answer of this
 input case
 eg:- input: "abcdef"
  output: "1a1b1c1d1e1f"

>>>  --
>>> You received this message because you are subscribed to the Google
>>> Groups "Algorithm Geeks" group.
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Re: [algogeeks] Re: MS question : string compression

[utsav sharma]

@ashish:-algo given in link wiil fail for "abcdef"

@navin:- output of "abcdef" should be "1a1b1c1d1e1f"

On Sun, May 27, 2012 at 3:24 PM, Ashish Goel  wrote:

> Will fail for the sing having say 257characters all same
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
> On Sat, May 26, 2012 at 12:26 PM, Navin Gupta wrote:
>
>> This is called Run-Length-Encoding (RLE)  of a string.
>> Its purpose is to save space.So in case of abcdef,I think the output
>> needed is abcdef (1 is implicit).
>> The added benefit is it makes the solution in-place.
>>
>> Approach:- (In-place and Linear Time)
>> Start from the left of  string  and  PREVIOUS_CHAR = str[0]
>> move forward till u match the CURRENT_CHAR with PREVIOUS_CHAR and keep
>> count of PREVIOUS_CHAR
>> At any point if (PREVIOUS_CHAR!=CURRENT_CHAR)
>>put the count of prev_char next to the start position of the previous
>> character.
>>
>> Below is the working code :-
>> void torle(char *str)
>> {   int i=0,j=0,k=0,cnt=1;
>> char cur_char=str[0],num[100];
>> while(str[j+1])
>> {
>> cnt=1;
>> while(str[j+1]==cur_char && str[j]!='\0'){
>> j++;
>> cnt++;
>> }
>> str[i++]=cur_char;
>> if( cnt>9 ){
>> itoa(cnt,num);
>> k=0;
>> while(num[k]) str[i++]=num[k++];
>> }
>> else if( cnt>1 && cnt<10 )
>> str[i++]= cnt+'0';
>> j++;
>> if(str[j]) cur_char=str[j];
>> }
>> if(i!=0){
>> if(cnt==1)
>> str[i++]=cur_char;
>> str[i]='\0';
>>
>> }
>> }
>>
>>
>> On Saturday, 26 May 2012 04:32:35 UTC+5:30, utsav sharma wrote:
>>>
>>> Implement a method to perform basic string compression using the counts
>>> of repeated characters.(inplace)
>>>
>>> eg:- input: "aaabcdef"
>>>  output:"3a5b1c1d1e1f".
>>>
>>> what should be my approach to this problem
>>>
>>> if i calculate the size of array required to store the output string
>>> and start from the last of the array then i wldn't get the right answer
>>> of above input case.
>>>
>>> and if start from front then i wldn't get the right answer of this
>>> input case
>>> eg:- input: "abcdef"
>>>  output: "1a1b1c1d1e1f"
>>>
>>  --
>> You received this message because you are subscribed to the Google Groups
>> "Algorithm Geeks" group.
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>> https://groups.google.com/d/msg/algogeeks/-/4LxWHEUJuK8J.
>>
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>
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Re: [algogeeks] Re: MS question : string compression

[Ashish Goel]

Will fail for the sing having say 257characters all same

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


On Sat, May 26, 2012 at 12:26 PM, Navin Gupta wrote:

> This is called Run-Length-Encoding (RLE)  of a string.
> Its purpose is to save space.So in case of abcdef,I think the output
> needed is abcdef (1 is implicit).
> The added benefit is it makes the solution in-place.
>
> Approach:- (In-place and Linear Time)
> Start from the left of  string  and  PREVIOUS_CHAR = str[0]
> move forward till u match the CURRENT_CHAR with PREVIOUS_CHAR and keep
> count of PREVIOUS_CHAR
> At any point if (PREVIOUS_CHAR!=CURRENT_CHAR)
>put the count of prev_char next to the start position of the previous
> character.
>
> Below is the working code :-
> void torle(char *str)
> {   int i=0,j=0,k=0,cnt=1;
> char cur_char=str[0],num[100];
> while(str[j+1])
> {
> cnt=1;
> while(str[j+1]==cur_char && str[j]!='\0'){
> j++;
> cnt++;
> }
> str[i++]=cur_char;
> if( cnt>9 ){
> itoa(cnt,num);
> k=0;
> while(num[k]) str[i++]=num[k++];
> }
> else if( cnt>1 && cnt<10 )
> str[i++]= cnt+'0';
> j++;
> if(str[j]) cur_char=str[j];
> }
> if(i!=0){
> if(cnt==1)
> str[i++]=cur_char;
> str[i]='\0';
>
> }
> }
>
>
> On Saturday, 26 May 2012 04:32:35 UTC+5:30, utsav sharma wrote:
>>
>> Implement a method to perform basic string compression using the counts
>> of repeated characters.(inplace)
>>
>> eg:- input: "aaabcdef"
>>  output:"3a5b1c1d1e1f".
>>
>> what should be my approach to this problem
>>
>> if i calculate the size of array required to store the output string
>> and start from the last of the array then i wldn't get the right answer
>> of above input case.
>>
>> and if start from front then i wldn't get the right answer of this input
>> case
>> eg:- input: "abcdef"
>>  output: "1a1b1c1d1e1f"
>>
>  --
> You received this message because you are subscribed to the Google Groups
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Re: [algogeeks] Re: MS question : string compression

[Ashish Goel]

http://michael.dipperstein.com/rle/index.html

and basic one is

http://www.fileformat.info/mirror/egff/ch09_03.htm


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


On Sat, May 26, 2012 at 1:10 PM, Hassan Monfared wrote:

> 1- try "abb"
>
>
> On Sat, May 26, 2012 at 12:07 PM, Anchal Gupta wrote:
>
>> yeah i forgot inplace so to do that we simply add count and ch in str
>> input array instead of op.
>> btw whats wrong with count it give me right answer.
>>
>> On May 26, 12:08 pm, Hassan Monfared  wrote:
>> > u forgot to do "inplace" and you have wrong conversion of count
>> >
>> > On Sat, May 26, 2012 at 11:31 AM, Anchal Gupta > >wrote:
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> > > hey, here is the function that do the compression and store the output
>> > > in an array op.
>> >
>> > > void str_comp(char *str)
>> > > {
>> > >  int count=0,j=0,i;
>> > >  char ch,op[100];
>> >
>> > >  for(i=0;i> > >  {
>> > >ch = str[i];
>> > >while(str[i] == ch)
>> > >  { count++;
>> > >i++;
>> > >  }
>> > > op[j] = count+48;
>> > > op[++j] = ch;
>> > > j++;
>> > > count=0;
>> >
>> > >   }
>> > >   cout<<"input : ";
>> > >   for(i=0;i> > >cout<> >
>> > >   cout<<"\n\noutput : ";
>> > >   for(i=0;i> > >cout<> >
>> > >  }
>> >
>> > > Best Regards
>> > > Anchal Gupta
>> > > USIT(GGSIPU), Delhi
>> > > +91-9015897983
>> >
>> > > --
>> > > You received this message because you are subscribed to the Google
>> Groups
>> > > "Algorithm Geeks" group.
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>> > >http://groups.google.com/group/algogeeks?hl=en.
>>
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Re: [algogeeks] Re: MS question : string compression

[Hassan Monfared]

1- try "abb"

On Sat, May 26, 2012 at 12:07 PM, Anchal Gupta wrote:

> yeah i forgot inplace so to do that we simply add count and ch in str
> input array instead of op.
> btw whats wrong with count it give me right answer.
>
> On May 26, 12:08 pm, Hassan Monfared  wrote:
> > u forgot to do "inplace" and you have wrong conversion of count
> >
> > On Sat, May 26, 2012 at 11:31 AM, Anchal Gupta  >wrote:
> >
> >
> >
> >
> >
> >
> >
> > > hey, here is the function that do the compression and store the output
> > > in an array op.
> >
> > > void str_comp(char *str)
> > > {
> > >  int count=0,j=0,i;
> > >  char ch,op[100];
> >
> > >  for(i=0;i > >  {
> > >ch = str[i];
> > >while(str[i] == ch)
> > >  { count++;
> > >i++;
> > >  }
> > > op[j] = count+48;
> > > op[++j] = ch;
> > > j++;
> > > count=0;
> >
> > >   }
> > >   cout<<"input : ";
> > >   for(i=0;i > >cout< >
> > >   cout<<"\n\noutput : ";
> > >   for(i=0;i > >cout< >
> > >  }
> >
> > > Best Regards
> > > Anchal Gupta
> > > USIT(GGSIPU), Delhi
> > > +91-9015897983
> >
> > > --
> > > You received this message because you are subscribed to the Google
> Groups
> > > "Algorithm Geeks" group.
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>
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Re: [algogeeks] Re: MS question : string compression

[Hassan Monfared]

u forgot to do "inplace" and you have wrong conversion of count

On Sat, May 26, 2012 at 11:31 AM, Anchal Gupta wrote:

> hey, here is the function that do the compression and store the output
> in an array op.
>
>
> void str_comp(char *str)
> {
>  int count=0,j=0,i;
>  char ch,op[100];
>
>  for(i=0;i  {
>ch = str[i];
>while(str[i] == ch)
>  { count++;
>i++;
>  }
> op[j] = count+48;
> op[++j] = ch;
> j++;
> count=0;
>
>   }
>   cout<<"input : ";
>   for(i=0;icout<
>   cout<<"\n\noutput : ";
>   for(i=0;icout<
>  }
>
>
>
> Best Regards
> Anchal Gupta
> USIT(GGSIPU), Delhi
> +91-9015897983
>
>
>
>
> --
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Re: [algogeeks] Re: MS Question -Reverse a Linked List in size of 2

[praveen raj]

Steps:
1)Reverse the list ...
2)Now do the swap two nodes... consecutively...

PRAVEEN RAJ
DELHI COLLEGE OF ENGINEERING

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Re: [algogeeks] Re: MS Question -Reverse a Linked List in size of 2

[Ashish Goel]

oh, a possible mistake from my side, ignore my mail please...

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


On Tue, Jan 24, 2012 at 3:08 PM, atul anand  wrote:

> @Ashish : seems exactly similar to Lucifer code  or you modified something
> in his code ?? ...
>
>
> On Tue, Jan 24, 2012 at 2:02 PM, Ashish Goel  wrote:
>
>>
>>
>>
>>
>> struct node
>> {
>>  int data;
>>  struct node *link;
>> };
>>
>> node* CreateNode(int val)
>> {
>>  node* root = (node*)malloc(sizeof(struct node));
>>  root->data = val;
>>  root->link = NULL;
>>  return root;
>> }
>>
>> node* createList(int *arr, int n)
>> {
>>  node * root = CreateNode(arr[0]);
>>  node * temp = root;
>>  for (int i =1; i < n; ++i)
>>  {
>>  temp->link = CreateNode(arr[i]);
>>  temp = temp->link;
>>  }
>>  return root;
>> }
>>
>> void deleteList(node *root)
>> {
>>  if(!root) return;
>>  deleteList(root->link);
>>  free(root);
>> }
>>
>> void printList(node *root)
>> {
>>  while(root)
>>  {
>>  printf("%d -> ", root->data);
>>  root= root->link;
>>  }
>>  printf("NULL\n");
>> }
>>
>> void reverseK(node *root, node **head, node **tail, int i, int K)
>> {
>>  if(!root->link)
>>  *head = root;
>>  else
>>  {
>>  reverseK(root->link, head, tail, (i+1)%K, K);
>>  
>>  if(i == K-1)
>>  {
>>  *tail = *head;
>>  *head = root;
>>  }
>>  else
>>  {
>>  root->link->link= root;
>>  if(i == 0) root->link = *tail;
>>  }
>>  }
>> }
>>
>> node* reverseKSize(node *root, int K)
>> {
>>  if(!root) return NULL;
>>  node *head = NULL;
>>  node *tail = NULL;
>>  reverseK(root, &head, &tail, 0, K);
>>  return head;
>> }
>>
>> int _tmain(int argc, _TCHAR* argv[])
>> {
>>  int a[11] = {1,2,3,4,5,6,7,8,9,10,11};
>>  node* root = createList(a, 11);
>>  printList(root);
>>  root = reverseKSize(root, 2);
>>  printList(root);
>>  deleteList(root);
>>
>> return 0;
>> }
>>
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>>
>> On Tue, Jan 24, 2012 at 2:30 AM, Lucifer  wrote:
>>
>>> @above
>>>
>>> attaching the file..
>>>
>>> --
>>> You received this message because you are subscribed to the Google
>>> Groups "Algorithm Geeks" group.
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>>> https://groups.google.com/d/msg/algogeeks/-/YW_phbT3me4J.
>>>
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>>>
>>
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Re: [algogeeks] Re: MS Question -Reverse a Linked List in size of 2

[DIPANKAR DUTTA]

node* prev(node *h)
{
 node *p,*q,*r;
 p=h;
 q=p->next->next;
 p->next->next=NULL;
 while(q!=NULL)
 {
   r=q->next->next;
   q->next->next=p;
   p=q;
   q=r;
 }
 return p;
}

On Tue, Jan 24, 2012 at 3:08 PM, atul anand  wrote:

> @Ashish : seems exactly similar to Lucifer code  or you modified something
> in his code ?? ...
>
>
> On Tue, Jan 24, 2012 at 2:02 PM, Ashish Goel  wrote:
>
>>
>>
>>
>>
>> struct node
>> {
>>  int data;
>>  struct node *link;
>> };
>>
>> node* CreateNode(int val)
>> {
>>  node* root = (node*)malloc(sizeof(struct node));
>>  root->data = val;
>>  root->link = NULL;
>>  return root;
>> }
>>
>> node* createList(int *arr, int n)
>> {
>>  node * root = CreateNode(arr[0]);
>>  node * temp = root;
>>  for (int i =1; i < n; ++i)
>>  {
>>  temp->link = CreateNode(arr[i]);
>>  temp = temp->link;
>>  }
>>  return root;
>> }
>>
>> void deleteList(node *root)
>> {
>>  if(!root) return;
>>  deleteList(root->link);
>>  free(root);
>> }
>>
>> void printList(node *root)
>> {
>>  while(root)
>>  {
>>  printf("%d -> ", root->data);
>>  root= root->link;
>>  }
>>  printf("NULL\n");
>> }
>>
>> void reverseK(node *root, node **head, node **tail, int i, int K)
>> {
>>  if(!root->link)
>>  *head = root;
>>  else
>>  {
>>  reverseK(root->link, head, tail, (i+1)%K, K);
>>  
>>  if(i == K-1)
>>  {
>>  *tail = *head;
>>  *head = root;
>>  }
>>  else
>>  {
>>  root->link->link= root;
>>  if(i == 0) root->link = *tail;
>>  }
>>  }
>> }
>>
>> node* reverseKSize(node *root, int K)
>> {
>>  if(!root) return NULL;
>>  node *head = NULL;
>>  node *tail = NULL;
>>  reverseK(root, &head, &tail, 0, K);
>>  return head;
>> }
>>
>> int _tmain(int argc, _TCHAR* argv[])
>> {
>>  int a[11] = {1,2,3,4,5,6,7,8,9,10,11};
>>  node* root = createList(a, 11);
>>  printList(root);
>>  root = reverseKSize(root, 2);
>>  printList(root);
>>  deleteList(root);
>>
>> return 0;
>> }
>>
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>>
>> On Tue, Jan 24, 2012 at 2:30 AM, Lucifer  wrote:
>>
>>> @above
>>>
>>> attaching the file..
>>>
>>> --
>>> You received this message because you are subscribed to the Google
>>> Groups "Algorithm Geeks" group.
>>> To view this discussion on the web visit
>>> https://groups.google.com/d/msg/algogeeks/-/YW_phbT3me4J.
>>>
>>> To post to this group, send email to algogeeks@googlegroups.com.
>>> To unsubscribe from this group, send email to
>>> algogeeks+unsubscr...@googlegroups.com.
>>> For more options, visit this group at
>>> http://groups.google.com/group/algogeeks?hl=en.
>>>
>>
>>  --
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>>
>
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-- 
Thanks and Regards,
---
*DIPANKAR DUTTA*
Software Development Engineer
Xen Server - OpenStack Development Team (DataCenter and Cloud)

Citrix R&D India Pvt Ltd
#23 Residency Road, Bangalore - 560 025
Phone: +91 8147830733
Office: Extn: 16429
Email: dipankar.du...@citrix.com

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Re: [algogeeks] Re: MS Question -Reverse a Linked List in size of 2

[atul anand]

@Ashish : seems exactly similar to Lucifer code  or you modified something
in his code ?? ...

On Tue, Jan 24, 2012 at 2:02 PM, Ashish Goel  wrote:

>
>
>
>
> struct node
> {
>   int data;
>   struct node *link;
> };
>
> node* CreateNode(int val)
> {
>   node* root = (node*)malloc(sizeof(struct node));
>   root->data = val;
>   root->link = NULL;
>   return root;
> }
>
> node* createList(int *arr, int n)
> {
>   node * root = CreateNode(arr[0]);
>   node * temp = root;
>   for (int i =1; i < n; ++i)
>   {
>   temp->link = CreateNode(arr[i]);
>   temp = temp->link;
>   }
>   return root;
> }
>
> void deleteList(node *root)
> {
>   if(!root) return;
>   deleteList(root->link);
>   free(root);
> }
>
> void printList(node *root)
> {
>   while(root)
>   {
>   printf("%d -> ", root->data);
>   root= root->link;
>   }
>   printf("NULL\n");
> }
>
> void reverseK(node *root, node **head, node **tail, int i, int K)
> {
>   if(!root->link)
>   *head = root;
>   else
>   {
>   reverseK(root->link, head, tail, (i+1)%K, K);
>   
>   if(i == K-1)
>   {
>   *tail = *head;
>   *head = root;
>   }
>   else
>   {
>   root->link->link= root;
>   if(i == 0) root->link = *tail;
>   }
>   }
> }
>
> node* reverseKSize(node *root, int K)
> {
>   if(!root) return NULL;
>   node *head = NULL;
>   node *tail = NULL;
>   reverseK(root, &head, &tail, 0, K);
>   return head;
> }
>
> int _tmain(int argc, _TCHAR* argv[])
> {
>   int a[11] = {1,2,3,4,5,6,7,8,9,10,11};
>   node* root = createList(a, 11);
>   printList(root);
>   root = reverseKSize(root, 2);
>   printList(root);
>   deleteList(root);
>
> return 0;
> }
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
>
> On Tue, Jan 24, 2012 at 2:30 AM, Lucifer  wrote:
>
>> @above
>>
>> attaching the file..
>>
>> --
>> You received this message because you are subscribed to the Google Groups
>> "Algorithm Geeks" group.
>> To view this discussion on the web visit
>> https://groups.google.com/d/msg/algogeeks/-/YW_phbT3me4J.
>>
>> To post to this group, send email to algogeeks@googlegroups.com.
>> To unsubscribe from this group, send email to
>> algogeeks+unsubscr...@googlegroups.com.
>> For more options, visit this group at
>> http://groups.google.com/group/algogeeks?hl=en.
>>
>
>  --
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>

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Re: [algogeeks] Re: MS Question -Reverse a Linked List in size of 2

[Ashish Goel]

struct node
{
int data;
struct node *link;
};

node* CreateNode(int val)
{
node* root = (node*)malloc(sizeof(struct node));
root->data = val;
root->link = NULL;
return root;
}

node* createList(int *arr, int n)
{
node * root = CreateNode(arr[0]);
node * temp = root;
for (int i =1; i < n; ++i)
{
temp->link = CreateNode(arr[i]);
temp = temp->link;
}
return root;
}

void deleteList(node *root)
{
if(!root) return;
deleteList(root->link);
free(root);
}

void printList(node *root)
{
while(root)
{
printf("%d -> ", root->data);
root= root->link;
}
printf("NULL\n");
}

void reverseK(node *root, node **head, node **tail, int i, int K)
{
if(!root->link)
*head = root;
else
{
reverseK(root->link, head, tail, (i+1)%K, K);

if(i == K-1)
{
*tail = *head;
*head = root;
}
else
{
root->link->link= root;
if(i == 0) root->link = *tail;
}
}
}

node* reverseKSize(node *root, int K)
{
if(!root) return NULL;
node *head = NULL;
node *tail = NULL;
reverseK(root, &head, &tail, 0, K);
return head;
}

int _tmain(int argc, _TCHAR* argv[])
{
int a[11] = {1,2,3,4,5,6,7,8,9,10,11};
node* root = createList(a, 11);
printList(root);
root = reverseKSize(root, 2);
printList(root);
deleteList(root);

return 0;
}

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


On Tue, Jan 24, 2012 at 2:30 AM, Lucifer  wrote:

> @above
>
> attaching the file..
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
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> https://groups.google.com/d/msg/algogeeks/-/YW_phbT3me4J.
>
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Re: [algogeeks] Re: MS question

[rahul sharma]

got it..thnx yr

On Tue, Oct 4, 2011 at 8:34 AM, rahul sharma wrote:

> so we shoul d aslo add loop at the top to find only for firrst row and
> column the initial values
>
>
> On Tue, Oct 4, 2011 at 8:30 AM, Ashish Goel  wrote:
>
>> 0 in 0th row as well as 0 in 0th col and hence true
>>
>>
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>> On Tue, Oct 4, 2011 at 8:28 AM, rahul sharma wrote:
>>
>>> row0 and col0 initilayy true coz we have 0 in 0 row???or these r default
>>> values?
>>>
>>>
>>> On Tue, Oct 4, 2011 at 8:07 AM, Ashish Goel  wrote:
>>>
 1 1 0 1
 0 1 1 1
 1 1 1 1
 1 1 1 0

 row0 is true
 col0 is true

 for (int i=1; i>>> for (int j=1;j>>> if (a[i][j] == 0) {a[i][0]=0; a[0][j]=0;}

 now after this
 1 1 0 0
 0 1 1 1
 1 1 1 1
 0 1 1 0

 for (int i=1; i>>>   if (a[i][0] ==0) for (int j=1; j>>> for (int j=1; i>>>   if (a[0][j] ==0) for (int i=1; i>>>
 after this
 1 1 0 0
 0 0 0 0
 1 1 0 0
 0 0 0 0


 because of row0 and col0 vars


 final output is
 0 0 0 0
 0 0 0 0
 0 1 0 0
 0 0 0 0

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


 On Tue, Oct 4, 2011 at 7:49 AM, rahul sharma 
 wrote:

> @ashish can u give an xample.plz...i have read a lot archives
> ...but cant find in 0(1) spaceu using 2 var only...plz give
> xample...nended urgent.thnx
>
>
> On Tue, Oct 4, 2011 at 7:26 AM, Ashish Goel  wrote:
>
>> keep two var row0 and col0 for checking if there is any 0 in row0flag
>> /col0flag
>>
>> now walk over elements from 1,1 to n,m and set corresponding entry in
>> 0th row /column if you hit a zero.
>>
>> now walk over zeroth column and rwo and set the complete row/col if a
>> 0 is there in 0th row/col.
>>
>> after this based on row0flag/col0flag, set oth col/row values to 0.
>>
>> Best Regards
>>  Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>>
>> On Mon, Oct 3, 2011 at 12:08 PM, rahul sharma <
>> rahul23111...@gmail.com> wrote:
>>
>>> yeah it is wrong..i have a solution but uses 0(n+m) space.i
>>> need it in 0(n*m) tymand o(1) space
>>>
>>>
>>> On Mon, Oct 3, 2011 at 11:55 AM, shady  wrote:
>>>
 search archives :-/


 On Mon, Oct 3, 2011 at 11:47 AM, pranav agrawal <
 pranav.is.cool.agra...@gmail.com> wrote:

> @rahul sharma, i ran this code, it is producing wrong answer :|
>
> check it,  http://codepad.org/THv1hJq1
>
> anyone with correct solution?
>
>  --
> You received this message because you are subscribed to the Google
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>
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Re: [algogeeks] Re: MS question

[rahul sharma]

so we shoul d aslo add loop at the top to find only for firrst row and
column the initial values

On Tue, Oct 4, 2011 at 8:30 AM, Ashish Goel  wrote:

> 0 in 0th row as well as 0 in 0th col and hence true
>
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
> On Tue, Oct 4, 2011 at 8:28 AM, rahul sharma wrote:
>
>> row0 and col0 initilayy true coz we have 0 in 0 row???or these r default
>> values?
>>
>>
>> On Tue, Oct 4, 2011 at 8:07 AM, Ashish Goel  wrote:
>>
>>> 1 1 0 1
>>> 0 1 1 1
>>> 1 1 1 1
>>> 1 1 1 0
>>>
>>> row0 is true
>>> col0 is true
>>>
>>> for (int i=1; i>> for (int j=1;j>> if (a[i][j] == 0) {a[i][0]=0; a[0][j]=0;}
>>>
>>> now after this
>>> 1 1 0 0
>>> 0 1 1 1
>>> 1 1 1 1
>>> 0 1 1 0
>>>
>>> for (int i=1; i>>   if (a[i][0] ==0) for (int j=1; j>> for (int j=1; i>>   if (a[0][j] ==0) for (int i=1; i>>
>>> after this
>>> 1 1 0 0
>>> 0 0 0 0
>>> 1 1 0 0
>>> 0 0 0 0
>>>
>>>
>>> because of row0 and col0 vars
>>>
>>>
>>> final output is
>>> 0 0 0 0
>>> 0 0 0 0
>>> 0 1 0 0
>>> 0 0 0 0
>>>
>>> Best Regards
>>> Ashish Goel
>>> "Think positive and find fuel in failure"
>>> +919985813081
>>> +919966006652
>>>
>>>
>>> On Tue, Oct 4, 2011 at 7:49 AM, rahul sharma wrote:
>>>
 @ashish can u give an xample.plz...i have read a lot archives ...but
 cant find in 0(1) spaceu using 2 var only...plz give xample...nended
 urgent.thnx


 On Tue, Oct 4, 2011 at 7:26 AM, Ashish Goel  wrote:

> keep two var row0 and col0 for checking if there is any 0 in row0flag
> /col0flag
>
> now walk over elements from 1,1 to n,m and set corresponding entry in
> 0th row /column if you hit a zero.
>
> now walk over zeroth column and rwo and set the complete row/col if a 0
> is there in 0th row/col.
>
> after this based on row0flag/col0flag, set oth col/row values to 0.
>
> Best Regards
>  Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
>
> On Mon, Oct 3, 2011 at 12:08 PM, rahul sharma  > wrote:
>
>> yeah it is wrong..i have a solution but uses 0(n+m) space.i
>> need it in 0(n*m) tymand o(1) space
>>
>>
>> On Mon, Oct 3, 2011 at 11:55 AM, shady  wrote:
>>
>>> search archives :-/
>>>
>>>
>>> On Mon, Oct 3, 2011 at 11:47 AM, pranav agrawal <
>>> pranav.is.cool.agra...@gmail.com> wrote:
>>>
 @rahul sharma, i ran this code, it is producing wrong answer :|

 check it,  http://codepad.org/THv1hJq1

 anyone with correct solution?

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Re: [algogeeks] Re: MS question

[Ashish Goel]

0 in 0th row as well as 0 in 0th col and hence true


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


On Tue, Oct 4, 2011 at 8:28 AM, rahul sharma wrote:

> row0 and col0 initilayy true coz we have 0 in 0 row???or these r default
> values?
>
>
> On Tue, Oct 4, 2011 at 8:07 AM, Ashish Goel  wrote:
>
>> 1 1 0 1
>> 0 1 1 1
>> 1 1 1 1
>> 1 1 1 0
>>
>> row0 is true
>> col0 is true
>>
>> for (int i=1; i> for (int j=1;j> if (a[i][j] == 0) {a[i][0]=0; a[0][j]=0;}
>>
>> now after this
>> 1 1 0 0
>> 0 1 1 1
>> 1 1 1 1
>> 0 1 1 0
>>
>> for (int i=1; i>   if (a[i][0] ==0) for (int j=1; j> for (int j=1; i>   if (a[0][j] ==0) for (int i=1; i>
>> after this
>> 1 1 0 0
>> 0 0 0 0
>> 1 1 0 0
>> 0 0 0 0
>>
>>
>> because of row0 and col0 vars
>>
>>
>> final output is
>> 0 0 0 0
>> 0 0 0 0
>> 0 1 0 0
>> 0 0 0 0
>>
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>> On Tue, Oct 4, 2011 at 7:49 AM, rahul sharma wrote:
>>
>>> @ashish can u give an xample.plz...i have read a lot archives ...but
>>> cant find in 0(1) spaceu using 2 var only...plz give xample...nended
>>> urgent.thnx
>>>
>>>
>>> On Tue, Oct 4, 2011 at 7:26 AM, Ashish Goel  wrote:
>>>
 keep two var row0 and col0 for checking if there is any 0 in row0flag
 /col0flag

 now walk over elements from 1,1 to n,m and set corresponding entry in
 0th row /column if you hit a zero.

 now walk over zeroth column and rwo and set the complete row/col if a 0
 is there in 0th row/col.

 after this based on row0flag/col0flag, set oth col/row values to 0.

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



 On Mon, Oct 3, 2011 at 12:08 PM, rahul sharma 
 wrote:

> yeah it is wrong..i have a solution but uses 0(n+m) space.i
> need it in 0(n*m) tymand o(1) space
>
>
> On Mon, Oct 3, 2011 at 11:55 AM, shady  wrote:
>
>> search archives :-/
>>
>>
>> On Mon, Oct 3, 2011 at 11:47 AM, pranav agrawal <
>> pranav.is.cool.agra...@gmail.com> wrote:
>>
>>> @rahul sharma, i ran this code, it is producing wrong answer :|
>>>
>>> check it,  http://codepad.org/THv1hJq1
>>>
>>> anyone with correct solution?
>>>
>>>  --
>>> You received this message because you are subscribed to the Google
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Re: [algogeeks] Re: MS question

[rahul sharma]

row0 and col0 initilayy true coz we have 0 in 0 row???or these r default
values?

On Tue, Oct 4, 2011 at 8:07 AM, Ashish Goel  wrote:

> 1 1 0 1
> 0 1 1 1
> 1 1 1 1
> 1 1 1 0
>
> row0 is true
> col0 is true
>
> for (int i=1; i for (int j=1;j if (a[i][j] == 0) {a[i][0]=0; a[0][j]=0;}
>
> now after this
> 1 1 0 0
> 0 1 1 1
> 1 1 1 1
> 0 1 1 0
>
> for (int i=1; i   if (a[i][0] ==0) for (int j=1; j for (int j=1; i   if (a[0][j] ==0) for (int i=1; i
> after this
> 1 1 0 0
> 0 0 0 0
> 1 1 0 0
> 0 0 0 0
>
>
> because of row0 and col0 vars
>
>
> final output is
> 0 0 0 0
> 0 0 0 0
> 0 1 0 0
> 0 0 0 0
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
> On Tue, Oct 4, 2011 at 7:49 AM, rahul sharma wrote:
>
>> @ashish can u give an xample.plz...i have read a lot archives ...but
>> cant find in 0(1) spaceu using 2 var only...plz give xample...nended
>> urgent.thnx
>>
>>
>> On Tue, Oct 4, 2011 at 7:26 AM, Ashish Goel  wrote:
>>
>>> keep two var row0 and col0 for checking if there is any 0 in row0flag
>>> /col0flag
>>>
>>> now walk over elements from 1,1 to n,m and set corresponding entry in 0th
>>> row /column if you hit a zero.
>>>
>>> now walk over zeroth column and rwo and set the complete row/col if a 0
>>> is there in 0th row/col.
>>>
>>> after this based on row0flag/col0flag, set oth col/row values to 0.
>>>
>>> Best Regards
>>>  Ashish Goel
>>> "Think positive and find fuel in failure"
>>> +919985813081
>>> +919966006652
>>>
>>>
>>>
>>> On Mon, Oct 3, 2011 at 12:08 PM, rahul sharma 
>>> wrote:
>>>
 yeah it is wrong..i have a solution but uses 0(n+m) space.i need
 it in 0(n*m) tymand o(1) space


 On Mon, Oct 3, 2011 at 11:55 AM, shady  wrote:

> search archives :-/
>
>
> On Mon, Oct 3, 2011 at 11:47 AM, pranav agrawal <
> pranav.is.cool.agra...@gmail.com> wrote:
>
>> @rahul sharma, i ran this code, it is producing wrong answer :|
>>
>> check it,  http://codepad.org/THv1hJq1
>>
>> anyone with correct solution?
>>
>>  --
>> You received this message because you are subscribed to the Google
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Re: [algogeeks] Re: MS question

[Ashish Goel]

1 1 0 1
0 1 1 1
1 1 1 1
1 1 1 0

row0 is true
col0 is true

for (int i=1; iwrote:

> @ashish can u give an xample.plz...i have read a lot archives ...but
> cant find in 0(1) spaceu using 2 var only...plz give xample...nended
> urgent.thnx
>
>
> On Tue, Oct 4, 2011 at 7:26 AM, Ashish Goel  wrote:
>
>> keep two var row0 and col0 for checking if there is any 0 in row0flag
>> /col0flag
>>
>> now walk over elements from 1,1 to n,m and set corresponding entry in 0th
>> row /column if you hit a zero.
>>
>> now walk over zeroth column and rwo and set the complete row/col if a 0 is
>> there in 0th row/col.
>>
>> after this based on row0flag/col0flag, set oth col/row values to 0.
>>
>> Best Regards
>>  Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>>
>> On Mon, Oct 3, 2011 at 12:08 PM, rahul sharma wrote:
>>
>>> yeah it is wrong..i have a solution but uses 0(n+m) space.i need
>>> it in 0(n*m) tymand o(1) space
>>>
>>>
>>> On Mon, Oct 3, 2011 at 11:55 AM, shady  wrote:
>>>
 search archives :-/


 On Mon, Oct 3, 2011 at 11:47 AM, pranav agrawal <
 pranav.is.cool.agra...@gmail.com> wrote:

> @rahul sharma, i ran this code, it is producing wrong answer :|
>
> check it,  http://codepad.org/THv1hJq1
>
> anyone with correct solution?
>
>  --
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Re: [algogeeks] Re: MS question

[rahul sharma]

@ashish can u give an xample.plz...i have read a lot archives ...but
cant find in 0(1) spaceu using 2 var only...plz give xample...nended
urgent.thnx

On Tue, Oct 4, 2011 at 7:26 AM, Ashish Goel  wrote:

> keep two var row0 and col0 for checking if there is any 0 in row0flag
> /col0flag
>
> now walk over elements from 1,1 to n,m and set corresponding entry in 0th
> row /column if you hit a zero.
>
> now walk over zeroth column and rwo and set the complete row/col if a 0 is
> there in 0th row/col.
>
> after this based on row0flag/col0flag, set oth col/row values to 0.
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
>
> On Mon, Oct 3, 2011 at 12:08 PM, rahul sharma wrote:
>
>> yeah it is wrong..i have a solution but uses 0(n+m) space.i need
>> it in 0(n*m) tymand o(1) space
>>
>>
>> On Mon, Oct 3, 2011 at 11:55 AM, shady  wrote:
>>
>>> search archives :-/
>>>
>>>
>>> On Mon, Oct 3, 2011 at 11:47 AM, pranav agrawal <
>>> pranav.is.cool.agra...@gmail.com> wrote:
>>>
 @rahul sharma, i ran this code, it is producing wrong answer :|

 check it,  http://codepad.org/THv1hJq1

 anyone with correct solution?

  --
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Re: [algogeeks] Re: MS question

[Ashish Goel]

keep two var row0 and col0 for checking if there is any 0 in row0flag
/col0flag

now walk over elements from 1,1 to n,m and set corresponding entry in 0th
row /column if you hit a zero.

now walk over zeroth column and rwo and set the complete row/col if a 0 is
there in 0th row/col.

after this based on row0flag/col0flag, set oth col/row values to 0.

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


On Mon, Oct 3, 2011 at 12:08 PM, rahul sharma wrote:

> yeah it is wrong..i have a solution but uses 0(n+m) space.i need it
> in 0(n*m) tymand o(1) space
>
>
> On Mon, Oct 3, 2011 at 11:55 AM, shady  wrote:
>
>> search archives :-/
>>
>>
>> On Mon, Oct 3, 2011 at 11:47 AM, pranav agrawal <
>> pranav.is.cool.agra...@gmail.com> wrote:
>>
>>> @rahul sharma, i ran this code, it is producing wrong answer :|
>>>
>>> check it,  http://codepad.org/THv1hJq1
>>>
>>> anyone with correct solution?
>>>
>>>  --
>>> You received this message because you are subscribed to the Google Groups
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Re: [algogeeks] Re: MS question

[rahul sharma]

yeah it is wrong..i have a solution but uses 0(n+m) space.i need it
in 0(n*m) tymand o(1) space

On Mon, Oct 3, 2011 at 11:55 AM, shady  wrote:

> search archives :-/
>
>
> On Mon, Oct 3, 2011 at 11:47 AM, pranav agrawal <
> pranav.is.cool.agra...@gmail.com> wrote:
>
>> @rahul sharma, i ran this code, it is producing wrong answer :|
>>
>> check it,  http://codepad.org/THv1hJq1
>>
>> anyone with correct solution?
>>
>>  --
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Re: [algogeeks] Re: MS question

[shady]

search archives :-/

On Mon, Oct 3, 2011 at 11:47 AM, pranav agrawal <
pranav.is.cool.agra...@gmail.com> wrote:

> @rahul sharma, i ran this code, it is producing wrong answer :|
>
> check it,  http://codepad.org/THv1hJq1
>
> anyone with correct solution?
>
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Re: [algogeeks] Re: MS Question -> Median of a BST without using extra space and in O(n)

[anshu mishra]

*@all to median of BST time O(n) space O(1) (modified code of nitin to
get median)


medianBST*(node, n)
  int x = 0;
  *while* hasleftchild(node) *do*
node = node.left
  *do*

x++;
if (x == n/2) return node->val;
*if* (hasrightchild(node)) *then*
  node = node.right
  *while* hasleftchild(node) *do*
node = node.left
*else*

  *while* node.parent ≠ *null* *and* node == node.parent.right *do*
node = node.parent
  node = node.parent
  *while* node ≠ *null*


@dheeraj u

U  can get the number of elements by just traversing the who tree by above
method

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Re: [algogeeks] Re: MS Question -> Median of a BST without using extra space and in O(n)

[Sanjay Rajpal]

Since we are given pointer to root node, we can easily find the minimum
element in the tree.

This will be the first node in the inorder traversal, now use method to find
the inorder successor of a each node. Do it iteratively.

Complexity will be O(n log n) and O(n) if tree is skewed.

Correct me if m wrong.
Sanju
:)



On Tue, Sep 27, 2011 at 8:49 AM, Nitin Garg wrote:

> Do inorder traversal, to find out the total no. of nodes.
>
> Next time, do the inorder traversal but keeping the count of nodes visited
> and stop when you visit n/2 nodes.
>
> Non recursive In-order Traversal -
>
> *inorder*(node)
>   *while* hasleftchild(node) *do*
> node = node.left
>   *do*
> visit(node)
> *if* (hasrightchild(node)) *then*
>   node = node.right
>   *while* hasleftchild(node) *do*
> node = node.left
> *else*
>   *while* node.parent ≠ *null* *and* node == node.parent.right *do*
> node = node.parent
>   node = node.parent
>   *while* node ≠ *null*
>
> Source: Wikipedia
>
>   On Tue, Sep 27, 2011 at 9:13 PM, Sanjay Rajpal  wrote:
>
>>   Recursion also requires space, so the problem is how to traverse
>> without extra space.
>>
>> Once this is done, nothing is left in the problem.
>> Sanju
>> :)
>>
>>
>>
>> On Tue, Sep 27, 2011 at 8:35 AM, Dheeraj Sharma <
>> dheerajsharma1...@gmail.com> wrote:
>>
>>> @anshu
>>> can middle element can be found if the no. of nodes are not given...
>>>
>>>
>>> On Tue, Sep 27, 2011 at 8:34 PM, vikas wrote:
>>>
 a simple one is rabit-tortoise method, and using stackless traversal,
 facing a lot of corner cases in coding this, can someone check this as
 well?

 On Sep 27, 6:41 pm, anshu mishra  wrote:
 > its not o(n) it is O(max height of tree) :P
 > i have not seen the constraint.

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>>>
>>>
>>> --
>>> *Dheeraj Sharma*
>>>
>>>
>>>
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>
>
>
> --
> Nitin Garg
>
> "Personality can open doors, but only Character can keep them open"
>
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Re: [algogeeks] Re: MS Question -> Median of a BST without using extra space and in O(n)

[Nitin Garg]

Do inorder traversal, to find out the total no. of nodes.

Next time, do the inorder traversal but keeping the count of nodes visited
and stop when you visit n/2 nodes.

Non recursive In-order Traversal -

*inorder*(node)
  *while* hasleftchild(node) *do*
node = node.left
  *do*
visit(node)
*if* (hasrightchild(node)) *then*
  node = node.right
  *while* hasleftchild(node) *do*
node = node.left
*else*
  *while* node.parent ≠ *null* *and* node == node.parent.right *do*
node = node.parent
  node = node.parent
  *while* node ≠ *null*

Source: Wikipedia

On Tue, Sep 27, 2011 at 9:13 PM, Sanjay Rajpal  wrote:

> Recursion also requires space, so the problem is how to traverse without
> extra space.
>
> Once this is done, nothing is left in the problem.
> Sanju
> :)
>
>
>
> On Tue, Sep 27, 2011 at 8:35 AM, Dheeraj Sharma <
> dheerajsharma1...@gmail.com> wrote:
>
>> @anshu
>> can middle element can be found if the no. of nodes are not given...
>>
>>
>> On Tue, Sep 27, 2011 at 8:34 PM, vikas wrote:
>>
>>> a simple one is rabit-tortoise method, and using stackless traversal,
>>> facing a lot of corner cases in coding this, can someone check this as
>>> well?
>>>
>>> On Sep 27, 6:41 pm, anshu mishra  wrote:
>>> > its not o(n) it is O(max height of tree) :P
>>> > i have not seen the constraint.
>>>
>>> --
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>>
>>
>> --
>> *Dheeraj Sharma*
>>
>>
>>
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"Personality can open doors, but only Character can keep them open"

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Re: [algogeeks] Re: MS Question -> Median of a BST without using extra space and in O(n)

[Sanjay Rajpal]

Recursion also requires space, so the problem is how to traverse without
extra space.

Once this is done, nothing is left in the problem.
Sanju
:)



On Tue, Sep 27, 2011 at 8:35 AM, Dheeraj Sharma  wrote:

> @anshu
> can middle element can be found if the no. of nodes are not given...
>
>
> On Tue, Sep 27, 2011 at 8:34 PM, vikas wrote:
>
>> a simple one is rabit-tortoise method, and using stackless traversal,
>> facing a lot of corner cases in coding this, can someone check this as
>> well?
>>
>> On Sep 27, 6:41 pm, anshu mishra  wrote:
>> > its not o(n) it is O(max height of tree) :P
>> > i have not seen the constraint.
>>
>> --
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>>
>
>
> --
> *Dheeraj Sharma*
>
>
>
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Re: [algogeeks] Re: MS Question -> Median of a BST without using extra space and in O(n)

[Dheeraj Sharma]

@anshu
can middle element can be found if the no. of nodes are not given...

On Tue, Sep 27, 2011 at 8:34 PM, vikas  wrote:

> a simple one is rabit-tortoise method, and using stackless traversal,
> facing a lot of corner cases in coding this, can someone check this as
> well?
>
> On Sep 27, 6:41 pm, anshu mishra  wrote:
> > its not o(n) it is O(max height of tree) :P
> > i have not seen the constraint.
>
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Re: [algogeeks] Re: MS Question -> Median of a BST without using extra space and in O(n)

[anshu mishra]

its not o(n) it is O(max height of tree) :P
i have not seen the constraint.

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Re: [algogeeks] Re: MS question

[Yogesh Yadav]

Suppose matrix is

1 0 0 1
1 0 1 0
0 0 0 0

then we traverse the matrix for each 1 we found at a[i][j] , we will check
for i=i to i wrote:

> If you're given that it's a sparse matrix, then you must assume
> storage is in a sparse matrix data structure to get time less than
> O(mn).
>
> In fact, if you assume the right data structure, then the operation
> can take O(1) time.
>
> For example if you say the structure is an array of sets of indices of
> the 1's in each row (so that L(i) is a list that contains j if and
> only if A(i,j) is a 1), then all you have to do is flip a bit saying
> the representation has changed, i.e. lookups will work differently.
> The old lookup is
>
> A(i,j) = if L(i) contains j then 1 else 0.
>
> The new lookup will be
>
> A(i,j) = if L(i) is nonempty or L(j) contains i then 1 else 0
>
> You'd probably want to store the sets in hash tables so that lookups
> will remain O(1). Other choices might make more sense if A has special
> structure.
>
> On Sep 26, 6:41 pm, Ankur Garg  wrote:
> > Guys an Update ,
> >
> > This has been asked in MS by me.. I suggested O(m*n) but they were
> looking
> > for a solution in nlogn ( n*n Sparse Matrix ) ..Any idea ...
> >
> > This post was discussed earlier but every1 came with O(m*n) solution so
> > instead of cluttering it ..opened a new One ...
> >
> >
> >
> > On Tue, Sep 27, 2011 at 3:06 AM, Gene  wrote:
> > > I assume we don't want to use extra storage.
> >
> > > So one way is this: Go over the matrix and mark the first row with a 1
> > > and the first column with a 1 for each 1 you find.  Because row and
> > > column 1 are used for temporary storage in this manner, you must first
> > > remember whether they contained a 1, then go ahead. With row and
> > > column 1 holding the necessary marks, you can fill in all the rows and
> > > columns except them. Finally you can fill in row and column 1 by
> > > checking the saved values.  It will look something like this.
> >
> > > row0has1 = 0;
> > > for (j = 0; j < n; j++) if (M(0,j)) { row0has1 = 1; break; }
> > > col0has1 = 0;
> > > for (i = 0; i < n; i++) if (M(i,0)) { col0has1 = 1; break; }
> > > for (i = 1; i < m; i++)
> > >  for (j = 1; j < n; j++)
> > >if (M(i,j)) M(i,0) = M(0,j) = 1;
> > > for (i = 1; i < m; i++)
> > >  for (j = 1; j < n; j++)
> > >if (M(i,0) || M(0,j)) M(i, j) = 1;
> > > if (row0has1)
> > >  for (j = 0; j < n; j++) M(0,j) = 1;
> > > if (col0has1)
> > >  for (i = 0; i < n; i++) M(i,0) = 1;
> >
> > > Maybe there's a slicker way, but this is O(mn)
> >
> > > On Sep 26, 9:46 pm, Ankur Garg  wrote:
> > > > Saw this question in one of the algo communities.
> >
> > > > Amazon telephonic interview question on Matrix
> > > > Input is a matrix of size n x m of 0's and 1's. eg:
> > > > 1 0 0 1
> > > > 0 0 1 0
> > > > 0 0 0 0
> >
> > > > If a location has 1; make all the elements of that row and column =
> 1. eg
> > > > 1 1 1 1
> > > > 1 1 1 1
> > > > 1 0 1 1
> >
> > > > Solution should be with Time complexity = O(n*m) and space complexity
> =
> > > O(1)
> >
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Re: [algogeeks] Re: MS question

[Ankur Garg]

Guys an Update ,

This has been asked in MS by me.. I suggested O(m*n) but they were looking
for a solution in nlogn ( n*n Sparse Matrix ) ..Any idea ...

This post was discussed earlier but every1 came with O(m*n) solution so
instead of cluttering it ..opened a new One ...



On Tue, Sep 27, 2011 at 3:06 AM, Gene  wrote:

> I assume we don't want to use extra storage.
>
> So one way is this: Go over the matrix and mark the first row with a 1
> and the first column with a 1 for each 1 you find.  Because row and
> column 1 are used for temporary storage in this manner, you must first
> remember whether they contained a 1, then go ahead. With row and
> column 1 holding the necessary marks, you can fill in all the rows and
> columns except them. Finally you can fill in row and column 1 by
> checking the saved values.  It will look something like this.
>
> row0has1 = 0;
> for (j = 0; j < n; j++) if (M(0,j)) { row0has1 = 1; break; }
> col0has1 = 0;
> for (i = 0; i < n; i++) if (M(i,0)) { col0has1 = 1; break; }
> for (i = 1; i < m; i++)
>  for (j = 1; j < n; j++)
>if (M(i,j)) M(i,0) = M(0,j) = 1;
> for (i = 1; i < m; i++)
>  for (j = 1; j < n; j++)
>if (M(i,0) || M(0,j)) M(i, j) = 1;
> if (row0has1)
>  for (j = 0; j < n; j++) M(0,j) = 1;
> if (col0has1)
>  for (i = 0; i < n; i++) M(i,0) = 1;
>
> Maybe there's a slicker way, but this is O(mn)
>
> On Sep 26, 9:46 pm, Ankur Garg  wrote:
> > Saw this question in one of the algo communities.
> >
> > Amazon telephonic interview question on Matrix
> > Input is a matrix of size n x m of 0's and 1's. eg:
> > 1 0 0 1
> > 0 0 1 0
> > 0 0 0 0
> >
> > If a location has 1; make all the elements of that row and column = 1. eg
> > 1 1 1 1
> > 1 1 1 1
> > 1 0 1 1
> >
> > Solution should be with Time complexity = O(n*m) and space complexity =
> O(1)
>
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Re: [algogeeks] Re: MS Question

[Shrey Choudhary]

i made a boo boo

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Re: [algogeeks] Re: MS Question

[SANDEEP CHUGH]

@shrey   but its not given in question that we have to return the position
or the occurence in the document.. we hav to only the tell whether the word
is in document..
read
.design a code which efficiently  search the word  and find occurrence of it
in given document .

its nt given that return the position or something like that.

isn't so ??
On Thu, Aug 25, 2011 at 7:38 PM, SANDEEP CHUGH wrote:

> ohh sry my mistake ..  got it
>
>
> On Thu, Aug 25, 2011 at 7:36 PM, Shrey Choudhary <
> choudharyshre...@gmail.com> wrote:
>
>> "design a code which efficiently  search
>> the word  and find occurrence of it in given document"
>>
>>
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Re: [algogeeks] Re: MS Question

[SANDEEP CHUGH]

ohh sry my mistake ..  got it

On Thu, Aug 25, 2011 at 7:36 PM, Shrey Choudhary  wrote:

> "design a code which efficiently  search
> the word  and find occurrence of it in given document"
>
>
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Re: [algogeeks] Re: MS Question

[SANDEEP CHUGH]

no ..
question says that  there is a function  that gives the next word in the
document..
this means after hashing one word , we hav to go to another word ... for
this going to next word in the document , we hav already provided wid a
function..

nothing to do wid the occurence

On Thu, Aug 25, 2011 at 7:30 PM, Shrey Choudhary  wrote:

> But it says finding the "next word" and also it's position of
> occurence. If we use frequency..won't position of occurence overlap?
>
> Am not sure whether I got the question right or not!
>
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Re: [algogeeks] Re: MS Question

[SANDEEP CHUGH]

for every word in the document , apply hash funtion..  store the string n
its frequency too..
if we get the same word , then  increment the frequency.

after storing.

whenver we want to search the word , searching in O(1) time , jst applying
hash function again on the word to be searched ..

correct me if i am wrong?


On Thu, Aug 25, 2011 at 7:14 PM, Shrey Choudhary  wrote:

> how will we exactly implement hashtables in this? What will be
> appropriate keys? ??
>
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Re: [algogeeks] Re: MS Question

[nishaanth]

ya why not hashing ?

On Thu, Aug 25, 2011 at 3:31 PM, SANDEEP CHUGH wrote:

> wat abt doing wid hashing?
>
>
> On Thu, Aug 25, 2011 at 3:55 PM, vikas wrote:
>
>> yep, trie needs to be built
>>
>> On Aug 24, 10:49 pm, Ankur Garg  wrote:
>> > It means when u call that func u get the next word in the document
>> >
>> > Regards
>> > Ankur
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> > On Wed, Aug 24, 2011 at 6:59 PM, vikas 
>> wrote:
>> > > what do you mean by "a function for finding the next word is given" ?
>> >
>> > > On Aug 22, 1:56 am, Ankur Garg  wrote:
>> > > > Question-->  Given a document containing some words ...and a
>> function
>> > > > for finding the next word is given .design a code which
>> efficiently
>> > >  search
>> > > > the word  and find occurrence of it in given document .
>> >
>> > > >  -which data structure will be used?
>> >
>> > > >  -write  algorithm for implementing
>> complexity?
>> >
>> > > > Guys any Ideas here .. I think tries can be used but can anyone
>> > > > explain/suggest/discuss proper implementation /technique to solve
>> this
>> > > > problem
>> >
>> > > > Regards
>> >
>> > > > Ankur
>> >
>> > > --
>> > > You received this message because you are subscribed to the Google
>> Groups
>> > > "Algorithm Geeks" group.
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-- 
S.Nishaanth,
Computer Science and engineering,
IIT Madras.

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Re: [algogeeks] Re: MS Question

[SANDEEP CHUGH]

wat abt doing wid hashing?

On Thu, Aug 25, 2011 at 3:55 PM, vikas  wrote:

> yep, trie needs to be built
>
> On Aug 24, 10:49 pm, Ankur Garg  wrote:
> > It means when u call that func u get the next word in the document
> >
> > Regards
> > Ankur
> >
> >
> >
> >
> >
> >
> >
> > On Wed, Aug 24, 2011 at 6:59 PM, vikas 
> wrote:
> > > what do you mean by "a function for finding the next word is given" ?
> >
> > > On Aug 22, 1:56 am, Ankur Garg  wrote:
> > > > Question-->  Given a document containing some words ...and a
> function
> > > > for finding the next word is given .design a code which
> efficiently
> > >  search
> > > > the word  and find occurrence of it in given document .
> >
> > > >  -which data structure will be used?
> >
> > > >  -write  algorithm for implementing
> complexity?
> >
> > > > Guys any Ideas here .. I think tries can be used but can anyone
> > > > explain/suggest/discuss proper implementation /technique to solve
> this
> > > > problem
> >
> > > > Regards
> >
> > > > Ankur
> >
> > > --
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> Groups
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Re: [algogeeks] Re: MS Question

[Ankur Garg]

It means when u call that func u get the next word in the document

Regards
Ankur

On Wed, Aug 24, 2011 at 6:59 PM, vikas  wrote:

> what do you mean by "a function for finding the next word is given" ?
>
> On Aug 22, 1:56 am, Ankur Garg  wrote:
> > Question-->  Given a document containing some words ...and a function
> > for finding the next word is given .design a code which efficiently
>  search
> > the word  and find occurrence of it in given document .
> >
> >  -which data structure will be used?
> >
> >  -write  algorithm for implementing complexity?
> >
> > Guys any Ideas here .. I think tries can be used but can anyone
> > explain/suggest/discuss proper implementation /technique to solve this
> > problem
> >
> > Regards
> >
> > Ankur
>
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Re: [algogeeks] Re: MS question

[abhijeet srivastva]

Below is a solution -

#include 
int main(int  argc, char* argv[]){
int i = 0;
char *array = argv[1];
char prev = array[0];
while(array[i]){
int count = 0;
while(prev == array[i]){
i +=1;
count++;
}
printf("%d%c,",count,prev);
prev = array[i];
}
printf("\n");
}

Regards
Abhijeet Srivastva



On Wed, Aug 10, 2011 at 9:04 AM, monish001  wrote:

> Given : ddbbccae
> O/P : 2d4a2b2c1a1e
> What is happening? What algo?
>
> Thanks and regards
> Monish
>
> On Aug 9, 5:59 pm, ankit sambyal  wrote:
> > Given an array of characters, change the array to something as shown in
> the
> > following example.
> > Given : ddbbccae
> > O/P : 2d4a2b2c1a1e
> >
> > Do it in the most efficient manner both in terms of time and space ...
>
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Re: [algogeeks] Re: MS question

[Priyanshu]

@gopi.. didnt got you...

@adi.. ya thats what i am talking about...


Regards,
Priyanshu Gupta



On Tue, Aug 9, 2011 at 11:18 AM, Aditya Virmani wrote:

> i guess ur qn was how will u decide which frame shud b ur thumbnail...as
> after tht, the frame cud be set as a thumbnail which is pretty much an eay
> task in windows programming.i think most, dun hav much info abt it though,
> thumbnails shud contain more of data; i mean shudnt be unicolour(blank
> screen).
>
>
> On Tue, Aug 9, 2011 at 11:30 PM, *$*  wrote:
>
>> I guess , it can be done using indexing , with time stamp as key , and
>> frame pointer as data ..
>> Please correct me if I am wrong.
>>
>>
>> On Tue, Aug 9, 2011 at 11:03 PM, Priyanshu wrote:
>>
>>> Anyone??
>>>
>>> Regards,
>>> Priyanshu Gupta
>>>
>>>
>>> On Fri, Aug 5, 2011 at 6:09 PM, priyanshu wrote:
>>>
 Give an efficient algorithm  to determine which part of the video
 should be displayed as a thumbnail??
>>>
>>>
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>>
>>
>> --
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>> --Gopi
>>
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Re: [algogeeks] Re: MS question

[Aditya Virmani]

i guess ur qn was how will u decide which frame shud b ur thumbnail...as
after tht, the frame cud be set as a thumbnail which is pretty much an eay
task in windows programming.i think most, dun hav much info abt it though,
thumbnails shud contain more of data; i mean shudnt be unicolour(blank
screen).

On Tue, Aug 9, 2011 at 11:30 PM, *$*  wrote:

> I guess , it can be done using indexing , with time stamp as key , and
> frame pointer as data ..
> Please correct me if I am wrong.
>
>
> On Tue, Aug 9, 2011 at 11:03 PM, Priyanshu wrote:
>
>> Anyone??
>>
>> Regards,
>> Priyanshu Gupta
>>
>>
>> On Fri, Aug 5, 2011 at 6:09 PM, priyanshu wrote:
>>
>>> Give an efficient algorithm  to determine which part of the video
>>> should be displayed as a thumbnail??
>>
>>
>>  --
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>
>
>
> --
> Thx,
> --Gopi
>
>
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Re: [algogeeks] Re: MS question

[*$*]

I guess , it can be done using indexing , with time stamp as key , and frame
pointer as data ..
Please correct me if I am wrong.

On Tue, Aug 9, 2011 at 11:03 PM, Priyanshu  wrote:

> Anyone??
>
> Regards,
> Priyanshu Gupta
>
>
> On Fri, Aug 5, 2011 at 6:09 PM, priyanshu wrote:
>
>> Give an efficient algorithm  to determine which part of the video
>> should be displayed as a thumbnail??
>
>
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Thx,
--Gopi

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Re: [algogeeks] Re: MS question

[rajeev bharshetty]

Test cases for MSN Search Engine

 1 : Check for auto completion feature  (The auto completion based on the
prefix should provide suggestions for the most searched keyword with that
prefix)
.
 2 : The spelling correction feature which shows as to what the user
actually meant. This should be highly relevant to the input query.

3 : Speed of page searches is done using usually caching the pages , check
for that required speed with various benchmark tests.

4 : then you can test for the protocols , the maximum load the server can
bear by stress testing it.

5 :The results returned usually should be highly relevant  (test for that) ,
the count of results returned should be validated  .

6 : If the search engine supports various languages , then it should be
tested on various languages .

7 :You can test for working of various buttons on the search web page .Test
various links on the page.

8 : Test for the validity of GET or POST requests sent by the search engine
.

9 :Check for validity of links on the results page . check various links are
working or not through a bot.

10 : The Validity of Page rank algorithm used can also be tested for various
search queries.The machine learning object fitted  in search engine should
be tested with various kinds of inputs to make it learn the behavior .

On Fri, Jul 29, 2011 at 11:13 PM, vaibhav agarwal <
vibhu.bitspil...@gmail.com> wrote:

> what has palindrome to do with this?
>
>
> On Mon, Jul 18, 2011 at 1:11 AM, swetha rahul wrote:
>
>> Thanks everyone!!
>>
>>
>> On Sun, Jul 17, 2011 at 10:56 PM, ankit sambyal 
>> wrote:
>>
>>> Check for case senstivity also:
>>> eg:   "Madam" and "madam" both are palindromes.
>>>
>>>
>>> Also for clarify with the interviewer whether whitespace and
>>> punctuation can be ignored.
>>> eg: is the following string a palindrome or not:
>>> "A man, a plan, a canal, Panama"
>>> And check for this condition accordingly
>>>
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Rajeev N B 

"*Winners Don't do Different things , they do things Differently"*

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Re: [algogeeks] Re: MS question

[vaibhav agarwal]

what has palindrome to do with this?

On Mon, Jul 18, 2011 at 1:11 AM, swetha rahul wrote:

> Thanks everyone!!
>
>
> On Sun, Jul 17, 2011 at 10:56 PM, ankit sambyal wrote:
>
>> Check for case senstivity also:
>> eg:   "Madam" and "madam" both are palindromes.
>>
>>
>> Also for clarify with the interviewer whether whitespace and
>> punctuation can be ignored.
>> eg: is the following string a palindrome or not:
>> "A man, a plan, a canal, Panama"
>> And check for this condition accordingly
>>
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Re: [algogeeks] Re: MS question

[swetha rahul]

Thanks everyone!!

On Sun, Jul 17, 2011 at 10:56 PM, ankit sambyal wrote:

> Check for case senstivity also:
> eg:   "Madam" and "madam" both are palindromes.
>
>
> Also for clarify with the interviewer whether whitespace and
> punctuation can be ignored.
> eg: is the following string a palindrome or not:
> "A man, a plan, a canal, Panama"
> And check for this condition accordingly
>
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Re: [algogeeks] Re: MS question

[ankit sambyal]

Check for case senstivity also:
eg:   "Madam" and "madam" both are palindromes.


Also for clarify with the interviewer whether whitespace and
punctuation can be ignored.
eg: is the following string a palindrome or not:
"A man, a plan, a canal, Panama"
And check for this condition accordingly

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Re: [algogeeks] Re: MS question

[vaibhav agarwal]

check for query injection,string like (spaces,nothing) entered.

On Sun, Jul 17, 2011 at 9:04 PM, Dumanshu  wrote:

> Well, for the third case, you can elaborate on the implementation. We
> need to compare the first half with the later half. So, in case of
> even no. of characters it splits perfectly and in case of odd number
> of characters, just ignore the middle character.
>
> On Jul 17, 8:28 pm, swetha rahul  wrote:
> > is this ans sufficient..? any other possible test cases for palindrome ?
> >
> >
> >
> > On Sun, Jul 17, 2011 at 8:53 PM, Nishant 
> wrote:
> > > 1.If string is NULL then it should return 1 i.e. string is palindrome
> > > 2.If there is only one character in string then it is palindrome
> > > 3.If reverse of given string is same as string
> >
> > > On Jul 17, 8:18 pm, swetha rahul  wrote:
> > > > ya got it..thanks...
> >
> > > > how abt test cases for program to check whether a given string is
> > > palindrome
> > > > or not..?
> >
> > > > On Sun, Jul 17, 2011 at 8:35 PM, ankit sambyal <
> ankitsamb...@gmail.com
> > > >wrote:
> >
> > > > > 1. If u entered nothing and just pressed search, it should display
> > > nothing.
> > > > > 2. If u just entered a space and just pressed search, it should
> display
> > > > > nothing.
> > > > > 3.Verify the results are really related to give word or not
> > > > > 4.Check if proper Result is displayed for key word.
> > > > > 5. Check for the Order of the Result set, whether most relevant
> > > > > results are displayed first.
> >
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Re: [algogeeks] Re: MS question

[swetha rahul]

is this ans sufficient..? any other possible test cases for palindrome ?

On Sun, Jul 17, 2011 at 8:53 PM, Nishant  wrote:

> 1.If string is NULL then it should return 1 i.e. string is palindrome
> 2.If there is only one character in string then it is palindrome
> 3.If reverse of given string is same as string
>
> On Jul 17, 8:18 pm, swetha rahul  wrote:
> > ya got it..thanks...
> >
> > how abt test cases for program to check whether a given string is
> palindrome
> > or not..?
> >
> > On Sun, Jul 17, 2011 at 8:35 PM, ankit sambyal  >wrote:
> >
> > > 1. If u entered nothing and just pressed search, it should display
> nothing.
> > > 2. If u just entered a space and just pressed search, it should display
> > > nothing.
> > > 3.Verify the results are really related to give word or not
> > > 4.Check if proper Result is displayed for key word.
> > > 5. Check for the Order of the Result set, whether most relevant
> > > results are displayed first.
> >
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Re: [algogeeks] Re: MS question

[rizwan hudda]

Sorry dave, your solution works.
Thanks for the answer, What I was assuming to be a Counting sort was a
variant of it
for integers.

On Thu, Jun 30, 2011 at 7:56 AM, Dave  wrote:

> @Rizwan: I don't see the problem. Initialize an array of counters, one
> for each possible last character, to zero. If there are no
> restrictions on the last character, use an array of 256 counters.
> Then, for each string, increment the counter corresponding to the last
> character of the string by the length of the string (or 1 + the length
> of the string if you want to store the null byte). After all of the
> strings are processed, compute the running sum of the counters to get
> the position in the output array of the first entry for each last
> character. Then copy the strings into the output array at the position
> indicated by the appropriate counter, and increment that counter by
> the number of characters copied.
>
> Why doesn't this work? And isn't it a counting sort?
>
> Dave
>
> On Jun 29, 7:05 pm, rizwan hudda  wrote:
> > @Dave: Think of it again counting sort won't work, If you are just
> > considering the last character as
> > the even though key is just last character, the value is the whole
> string..
> >
> >
> >
> >
> >
> > On Tue, Jun 28, 2011 at 1:53 PM, juver++  wrote:
> > > List[letter] - linked list of all words with the last character as
> letter.
> > > Then iterate over all letters and concatenate lists.
> >
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Re: [algogeeks] Re: MS question

[rizwan hudda]

@Dave: Think of it again counting sort won't work, If you are just
considering the last character as
the even though key is just last character, the value is the whole string..


On Tue, Jun 28, 2011 at 1:53 PM, juver++  wrote:

> List[letter] - linked list of all words with the last character as letter.
> Then iterate over all letters and concatenate lists.
>
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Re: [algogeeks] Re: MS question

[juver++]

List[letter] - linked list of all words with the last character as letter.
Then iterate over all letters and concatenate lists.

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Re: [algogeeks] Re: MS question

[Kamakshii Aggarwal]

@Dave:can u please explain the second method.

On Mon, Jun 27, 2011 at 11:24 PM, Dave  wrote:

> @Nishant: Here are a couple of O(n) alternatives, given that there are
> a limited number of "last characters" and no requirement to break ties
> in any particular way:
>
> 1. A counting sort.
>
> 2. Form a linked list of entries corresponding to each last character,
> and then merge these lists in collating sequence.
>
> Dave
>
> On Jun 27, 3:58 am, Nishant Mittal  wrote:
> > WAP to sort an array of character strings on the basis of last
> > character of each string
> > eg:- {xxxc , yyya, zzzb} => {yyya , zzzb, xxxc}
>
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Re: [algogeeks] Re: MS Question

[sunny agrawal]

Nice Confusion... :)

Consider the following case
A[M] = {1,1,3,12};
B[N] = {1,2,12}

here again i think answer should be {1,1,12} , why are u binding one
occurrence of 1 in array A with one in B.  Question is which elements of
first array is present in second array. so for this case A[0], A[1], A[3]
are present so answer should be still {1,1,12} whether or not it occurs
multiple times in B.

now the question remains is whether to report duplicate entries or not. both
have been handled above with same time and space complexity.

Any Questions are Welcome :)

On Wed, Jun 15, 2011 at 1:08 AM, Arpit Sood  wrote:

> i meant if N = { 1, 1, 1, 2, 12}
> and M = { 1, 1, 3, 12}
> then answer should be = {1, 1, 12}
>
>
> On Mon, Jun 13, 2011 at 8:06 PM, sunny agrawal wrote:
>
>> no we can take care of duplicates without any extra memory
>> modify 2nd step of my previous solution as follows
>>
>> if T[a[i]] is set then this element is there in second array so report
>> this element and Reset T[a[i]].
>>
>> now no duplicates will be reported. and only 1025 bits will be required.
>> any failures ??
>>
>>
>> --
>> Sunny Aggrawal
>> B-Tech IV year,CSI
>> Indian Institute Of Technology,Roorkee
>>
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Re: [algogeeks] Re: MS Question

[Arpit Sood]

i meant if N = { 1, 1, 1, 2, 12}
and M = { 1, 1, 3, 12}
then answer should be = {1, 1, 12}


On Mon, Jun 13, 2011 at 8:06 PM, sunny agrawal wrote:

> no we can take care of duplicates without any extra memory
> modify 2nd step of my previous solution as follows
>
> if T[a[i]] is set then this element is there in second array so report this
> element and Reset T[a[i]].
>
> now no duplicates will be reported. and only 1025 bits will be required.
> any failures ??
>
>
> --
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> B-Tech IV year,CSI
> Indian Institute Of Technology,Roorkee
>
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Re: [algogeeks] Re: MS Question

[Divye Kapoor]

Good one! That halved the memory requirement. :)

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Re: [algogeeks] Re: MS Question

[sunny agrawal]

no we can take care of duplicates without any extra memory
modify 2nd step of my previous solution as follows

if T[a[i]] is set then this element is there in second array so report this
element and Reset T[a[i]].

now no duplicates will be reported. and only 1025 bits will be required.
any failures ??

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Re: [algogeeks] Re: MS Question

[Arpit Sood]

we can take care of the duplicate entries, but then that would cost more
space(int), as of now we are working with bool

On Mon, Jun 13, 2011 at 5:51 PM, sunny agrawal wrote:

> that will report duplicate entries multiple times :(
>
>
> On Mon, Jun 13, 2011 at 5:38 PM, sunny agrawal wrote:
>
>> why do we need 2 bits at all ??
>> i think single bit per table entry will do.
>> say table is T[1025], and array is A[M]
>>
>>
>> 1. Go through the N sized array and set bit 0 of the hash table entry to 1
>> if it is present in the first array.
>> 2. Go through the M sized array and if T[a[i]] is set then this element is
>> there in second array else not.
>>
>> any cases this can fails??
>>
>> On Mon, Jun 13, 2011 at 5:28 PM, Divye Kapoor wrote:
>>
>>> Use a hash table of size 1025 with bits per table entry = 2.
>>> 1. Go through the N sized array and set bit 0 of the hash table entry to
>>> 1 if it is present in the first array.
>>> 2. Go through the M sized array and set bit 1 of the hash table entry to
>>> 1 if the element belongs to 0 to 1024.
>>> 3. Go through the hash table and print entries with bit values 11.
>>>
>>> Space usage = O(1), 2050 bits (rounded off to the corresponding nearest
>>> byte). Time complexity = O(N + M)
>>>
>>> --
>>> DK
>>>
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>>
>>
>>
>> --
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>> B-Tech IV year,CSI
>> Indian Institute Of Technology,Roorkee
>>
>>
>
>
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Re: [algogeeks] Re: MS Question

[sunny agrawal]

that will report duplicate entries multiple times :(

On Mon, Jun 13, 2011 at 5:38 PM, sunny agrawal wrote:

> why do we need 2 bits at all ??
> i think single bit per table entry will do.
> say table is T[1025], and array is A[M]
>
>
> 1. Go through the N sized array and set bit 0 of the hash table entry to 1
> if it is present in the first array.
> 2. Go through the M sized array and if T[a[i]] is set then this element is
> there in second array else not.
>
> any cases this can fails??
>
> On Mon, Jun 13, 2011 at 5:28 PM, Divye Kapoor wrote:
>
>> Use a hash table of size 1025 with bits per table entry = 2.
>> 1. Go through the N sized array and set bit 0 of the hash table entry to 1
>> if it is present in the first array.
>> 2. Go through the M sized array and set bit 1 of the hash table entry to 1
>> if the element belongs to 0 to 1024.
>> 3. Go through the hash table and print entries with bit values 11.
>>
>> Space usage = O(1), 2050 bits (rounded off to the corresponding nearest
>> byte). Time complexity = O(N + M)
>>
>> --
>> DK
>>
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>
>
>
> --
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> B-Tech IV year,CSI
> Indian Institute Of Technology,Roorkee
>
>


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Re: [algogeeks] Re: MS Question

[sunny agrawal]

why do we need 2 bits at all ??
i think single bit per table entry will do.
say table is T[1025], and array is A[M]

1. Go through the N sized array and set bit 0 of the hash table entry to 1
if it is present in the first array.
2. Go through the M sized array and if T[a[i]] is set then this element is
there in second array else not.

any cases this can fails??

On Mon, Jun 13, 2011 at 5:28 PM, Divye Kapoor  wrote:

> Use a hash table of size 1025 with bits per table entry = 2.
> 1. Go through the N sized array and set bit 0 of the hash table entry to 1
> if it is present in the first array.
> 2. Go through the M sized array and set bit 1 of the hash table entry to 1
> if the element belongs to 0 to 1024.
> 3. Go through the hash table and print entries with bit values 11.
>
> Space usage = O(1), 2050 bits (rounded off to the corresponding nearest
> byte). Time complexity = O(N + M)
>
> --
> DK
>
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