[algogeeks] Re: linked lists

2010-10-21 Thread ligerdave
@asquare

basically i just added a flag to enable the "window slide". good catch
btw!

On Oct 20, 7:55 am, Asquare  wrote:
> @ligerdave -
> your algo will fail in the case the two arrays are:
>
> hellostl
> eeelexander
>
> ans : hellostlexander
> but according to ur method the answer would end up being
> hellostleeelexander

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[algogeeks] Re: linked lists

2010-10-21 Thread ligerdave
@Asquare

you were right. what about this?


public static char[] concat(char[] str1, char[] str2) {


boolean repeat = false;// indicates whether two neighbor chars
repeat in
// str1 array

int pointer = -1; // pointer for str2 array

for (int i = 0; i < str1.length; i++) {

if (pointer + 1 >= str2.length) {
pointer = -1;
break;
}

if (str1[i] == str2[pointer + 1]) {

pointer++;
repeat = (i > 0 && (str1[i] == str1[i - 1]));

} else {

if (!repeat)
pointer = -1;

}

}

char[] result = null;

if (pointer != -1) {
result = new char[str1.length + str2.length - (pointer 
+ 1)];

System.arraycopy(str1, 0, result, 0, str1.length);
System.arraycopy(str2, pointer + 1, result, str1.length,
str2.length
- pointer - 1);
}

return result;

}
On Oct 20, 7:55 am, Asquare  wrote:
> @ligerdave -
> your algo will fail in the case the two arrays are:
>
> hellostl
> eeelexander
>
> ans : hellostlexander
> but according to ur method the answer would end up being
> hellostleeelexander

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[algogeeks] Re: linked lists

2010-10-20 Thread Asquare
@ligerdave -
your algo will fail in the case the two arrays are:

hellostl
eeelexander

ans : hellostlexander
but according to ur method the answer would end up being
hellostleeelexander

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[algogeeks] Re: linked lists

2010-10-20 Thread Sudheendra
Is there any additional condition saying if last 'n' characters of
first list should match with first 'n' characters of 2nd list ?

On Oct 7, 12:52 pm, snehal jain  wrote:
> There are two linked list, both containing a character in each node.
>
> If one linked list contain characters  o x e n c and second contain
> characters e n c a r t a then the final linked list should contain o x
> e n c a r t a    i.e. if the end of one list is same as the start of
> second then those characters should come only once.
>
> can we do it in O(n+m) where n and m are the length of list. both are
> singly link list.

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Re: [algogeeks] Re: linked lists

2010-10-16 Thread nishaanth
@Snehal...wat ligerdave says is have ptr1 for list1
and ptr2 for list2.
if(ptr1->data==ptr2->data)increment both ptr1 and ptr2
else reset ptr2 to the head of list2 , increment ptr1

ptr2 position gives from where we need to concatenate.


On Sat, Oct 9, 2010 at 12:21 AM, ashita dadlani  wrote:

> I think we can use KMP string matching algorithm.
>
>
> On Fri, Oct 8, 2010 at 6:40 PM, shashank jain wrote:
>
>> there are 2 unsorted array we have to want a third array in sorted
>> form in mininmum time complexicity
>>
>> answer can like be this merge the two array in 3rd array then sort the
>> array
>>
>>  or
>> sort the individual array and then merge
>>
>> if feel first one is better
>>
>> can any one can help
>>
>>
>> On 10/8/10, snehal jain  wrote:
>> > @ligerdave
>> > m nt getting ur algo..can u explain with an example
>> >
>> >
>> > On 10/8/10, snehal jain  wrote:
>> >>
>> >> @neeraj
>> >> ur worst case complexity will be O(mn)
>> >>
>> >>
>> >>  On 10/8/10, snehal jain  wrote:
>> >>>
>> >>> @tech
>> >>> the ouput will be abhgrtsghgrthswert as no suffix of 1st matches with
>> >>> prefix of 2nd
>> >>>
>> >>>
>> >>>  On 10/7/10, ligerdave  wrote:
>> 
>>  use pointer. shift to left if one more leading char has been found.
>>  any unmatched char resets the pointer to first char
>> 
>>  once you went through the entire list(first one), the pointer on the
>>  second list tells you where to concatenate
>> 
>>  that gives you O(n) where n is the length of first list
>> 
>>  On Oct 7, 3:52 am, snehal jain  wrote:
>>  > There are two linked list, both containing a character in each
>> node.
>>  >
>>  > If one linked list contain characters  o x e n c and second contain
>>  > characters e n c a r t a then the final linked list should contain
>> o x
>>  > e n c a r t ai.e. if the end of one list is same as the start
>> of
>>  > second then those characters should come only once.
>>  >
>>  > can we do it in O(n+m) where n and m are the length of list. both
>> are
>>  > singly link list.
>> 
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>> 
>> >>>
>> >>
>> >
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-- 
S.Nishaanth,
Computer Science and engineering,
IIT Madras.

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Re: [algogeeks] Re: linked lists

2010-10-08 Thread ashita dadlani
I think we can use KMP string matching algorithm.

On Fri, Oct 8, 2010 at 6:40 PM, shashank jain wrote:

> there are 2 unsorted array we have to want a third array in sorted
> form in mininmum time complexicity
>
> answer can like be this merge the two array in 3rd array then sort the
> array
>
>  or
> sort the individual array and then merge
>
> if feel first one is better
>
> can any one can help
>
>
> On 10/8/10, snehal jain  wrote:
> > @ligerdave
> > m nt getting ur algo..can u explain with an example
> >
> >
> > On 10/8/10, snehal jain  wrote:
> >>
> >> @neeraj
> >> ur worst case complexity will be O(mn)
> >>
> >>
> >>  On 10/8/10, snehal jain  wrote:
> >>>
> >>> @tech
> >>> the ouput will be abhgrtsghgrthswert as no suffix of 1st matches with
> >>> prefix of 2nd
> >>>
> >>>
> >>>  On 10/7/10, ligerdave  wrote:
> 
>  use pointer. shift to left if one more leading char has been found.
>  any unmatched char resets the pointer to first char
> 
>  once you went through the entire list(first one), the pointer on the
>  second list tells you where to concatenate
> 
>  that gives you O(n) where n is the length of first list
> 
>  On Oct 7, 3:52 am, snehal jain  wrote:
>  > There are two linked list, both containing a character in each node.
>  >
>  > If one linked list contain characters  o x e n c and second contain
>  > characters e n c a r t a then the final linked list should contain o
> x
>  > e n c a r t ai.e. if the end of one list is same as the start of
>  > second then those characters should come only once.
>  >
>  > can we do it in O(n+m) where n and m are the length of list. both
> are
>  > singly link list.
> 
>  --
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> 
> 
> >>>
> >>
> >
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Re: [algogeeks] Re: linked lists

2010-10-08 Thread shashank jain
there are 2 unsorted array we have to want a third array in sorted
form in mininmum time complexicity

answer can like be this merge the two array in 3rd array then sort the array

  or
sort the individual array and then merge

if feel first one is better

can any one can help


On 10/8/10, snehal jain  wrote:
> @ligerdave
> m nt getting ur algo..can u explain with an example
>
>
> On 10/8/10, snehal jain  wrote:
>>
>> @neeraj
>> ur worst case complexity will be O(mn)
>>
>>
>>  On 10/8/10, snehal jain  wrote:
>>>
>>> @tech
>>> the ouput will be abhgrtsghgrthswert as no suffix of 1st matches with
>>> prefix of 2nd
>>>
>>>
>>>  On 10/7/10, ligerdave  wrote:

 use pointer. shift to left if one more leading char has been found.
 any unmatched char resets the pointer to first char

 once you went through the entire list(first one), the pointer on the
 second list tells you where to concatenate

 that gives you O(n) where n is the length of first list

 On Oct 7, 3:52 am, snehal jain  wrote:
 > There are two linked list, both containing a character in each node.
 >
 > If one linked list contain characters  o x e n c and second contain
 > characters e n c a r t a then the final linked list should contain o x
 > e n c a r t ai.e. if the end of one list is same as the start of
 > second then those characters should come only once.
 >
 > can we do it in O(n+m) where n and m are the length of list. both are
 > singly link list.

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>>>
>>
>
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Re: [algogeeks] Re: linked lists

2010-10-08 Thread snehal jain
@ligerdave
m nt getting ur algo..can u explain with an example


On 10/8/10, snehal jain  wrote:
>
> @neeraj
> ur worst case complexity will be O(mn)
>
>
>  On 10/8/10, snehal jain  wrote:
>>
>> @tech
>> the ouput will be abhgrtsghgrthswert as no suffix of 1st matches with
>> prefix of 2nd
>>
>>
>>  On 10/7/10, ligerdave  wrote:
>>>
>>> use pointer. shift to left if one more leading char has been found.
>>> any unmatched char resets the pointer to first char
>>>
>>> once you went through the entire list(first one), the pointer on the
>>> second list tells you where to concatenate
>>>
>>> that gives you O(n) where n is the length of first list
>>>
>>> On Oct 7, 3:52 am, snehal jain  wrote:
>>> > There are two linked list, both containing a character in each node.
>>> >
>>> > If one linked list contain characters  o x e n c and second contain
>>> > characters e n c a r t a then the final linked list should contain o x
>>> > e n c a r t ai.e. if the end of one list is same as the start of
>>> > second then those characters should come only once.
>>> >
>>> > can we do it in O(n+m) where n and m are the length of list. both are
>>> > singly link list.
>>>
>>> --
>>> You received this message because you are subscribed to the Google Groups
>>> "Algorithm Geeks" group.
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>>> .
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>>> http://groups.google.com/group/algogeeks?hl=en.
>>>
>>>
>>
>

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Re: [algogeeks] Re: linked lists

2010-10-08 Thread snehal jain
@neeraj
ur worst case complexity will be O(mn)


On 10/8/10, snehal jain  wrote:
>
> @tech
> the ouput will be abhgrtsghgrthswert as no suffix of 1st matches with
> prefix of 2nd
>
>
>  On 10/7/10, ligerdave  wrote:
>>
>> use pointer. shift to left if one more leading char has been found.
>> any unmatched char resets the pointer to first char
>>
>> once you went through the entire list(first one), the pointer on the
>> second list tells you where to concatenate
>>
>> that gives you O(n) where n is the length of first list
>>
>> On Oct 7, 3:52 am, snehal jain  wrote:
>> > There are two linked list, both containing a character in each node.
>> >
>> > If one linked list contain characters  o x e n c and second contain
>> > characters e n c a r t a then the final linked list should contain o x
>> > e n c a r t ai.e. if the end of one list is same as the start of
>> > second then those characters should come only once.
>> >
>> > can we do it in O(n+m) where n and m are the length of list. both are
>> > singly link list.
>>
>> --
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>>
>

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Re: [algogeeks] Re: linked lists

2010-10-08 Thread snehal jain
@tech
the ouput will be abhgrtsghgrthswert as no suffix of 1st matches with prefix
of 2nd


On 10/7/10, ligerdave  wrote:
>
> use pointer. shift to left if one more leading char has been found.
> any unmatched char resets the pointer to first char
>
> once you went through the entire list(first one), the pointer on the
> second list tells you where to concatenate
>
> that gives you O(n) where n is the length of first list
>
> On Oct 7, 3:52 am, snehal jain  wrote:
> > There are two linked list, both containing a character in each node.
> >
> > If one linked list contain characters  o x e n c and second contain
> > characters e n c a r t a then the final linked list should contain o x
> > e n c a r t ai.e. if the end of one list is same as the start of
> > second then those characters should come only once.
> >
> > can we do it in O(n+m) where n and m are the length of list. both are
> > singly link list.
>
> --
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>

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[algogeeks] Re: linked lists

2010-10-07 Thread ligerdave
use pointer. shift to left if one more leading char has been found.
any unmatched char resets the pointer to first char

once you went through the entire list(first one), the pointer on the
second list tells you where to concatenate

that gives you O(n) where n is the length of first list

On Oct 7, 3:52 am, snehal jain  wrote:
> There are two linked list, both containing a character in each node.
>
> If one linked list contain characters  o x e n c and second contain
> characters e n c a r t a then the final linked list should contain o x
> e n c a r t a    i.e. if the end of one list is same as the start of
> second then those characters should come only once.
>
> can we do it in O(n+m) where n and m are the length of list. both are
> singly link list.

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