@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
@asquare
basically i just added a flag to enable the window slide. good catch
btw!
On Oct 20, 7:55 am, Asquare anshika.sp...@gmail.com wrote:
@ligerdave -
your algo will fail in the case the two arrays are:
hellostl
eeelexander
ans : hellostlexander
but according to ur method the
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 learner@gmail.com wrote:
There are two linked list, both containing a character in each node.
If one linked list contain
@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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Algorithm Geeks
@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
@tech
the ouput will be abhgrtsghgrthswert as no suffix of 1st matches with prefix
of 2nd
On 10/7/10, ligerdave david.c...@gmail.com 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
@neeraj
ur worst case complexity will be O(mn)
On 10/8/10, snehal jain learner@gmail.com wrote:
@tech
the ouput will be abhgrtsghgrthswert as no suffix of 1st matches with
prefix of 2nd
On 10/7/10, ligerdave david.c...@gmail.com wrote:
use pointer. shift to left if one more leading
@ligerdave
m nt getting ur algo..can u explain with an example
On 10/8/10, snehal jain learner@gmail.com wrote:
@neeraj
ur worst case complexity will be O(mn)
On 10/8/10, snehal jain learner@gmail.com wrote:
@tech
the ouput will be abhgrtsghgrthswert as no suffix of 1st matches
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
