hmm o(nlogn) was for constructing the tree.Btw sorry for being unclear.
The best solution is the obvious one in this case.

On Wed, Jul 27, 2011 at 2:10 PM, surender sanke <[email protected]> wrote:

> @ sunny , ur right!!
>
> surender
>
> On Wed, Jul 27, 2011 at 1:58 PM, sunny agrawal <[email protected]>wrote:
>
>> i don't find difference between your suffix tree approach and my simple
>> O(N^2) method
>> in both cases printf statement will be executed O(N^2) times
>> and in Suffix Tree approach will take some extra time of construction of
>> tree and extra space too !!!!!
>>
>> On Wed, Jul 27, 2011 at 1:45 PM, surender sanke <[email protected]>wrote:
>>
>>>         *
>>>      /  \    \
>>>    a     b    c
>>>   /        \
>>> b          c
>>> /
>>> c
>>>
>>> prints *a*
>>> comes to b, appends a with b    prints *ab*
>>> comes to c ,appends ab with c   prints *abc*
>>> starts with new child
>>> prints *b*
>>> prints *bc*
>>> prints *c*
>>>
>>> surender
>>>
>>>
>>> On Wed, Jul 27, 2011 at 12:46 PM, sunny agrawal <[email protected]
>>> > wrote:
>>>
>>>> But still Printing O(N^2) substrings will take O(N^2) time isn't it ?
>>>>
>>>> On Wed, Jul 27, 2011 at 12:39 PM, surender sanke 
>>>> <[email protected]>wrote:
>>>>
>>>>>
>>>>>
>>>>> @sunny
>>>>> consider *uncompressed* suffix tree, even with distinct elements
>>>>> maximum number of nodes with string length n formed will be 2n.
>>>>>  once suffix tree is constructed, needs to traverse in dfs order
>>>>> appending the node found on the way.
>>>>> total complexity would be O(construction of suffix tree ) + O(traverse
>>>>> time).
>>>>>
>>>>> surender
>>>>>
>>>>>
>>>>> On Wed, Jul 27, 2011 at 12:00 PM, sunny agrawal <
>>>>> [email protected]> wrote:
>>>>>
>>>>>> @shiva viknesh
>>>>>> this is a different Question.......
>>>>>>
>>>>>> @saurabh
>>>>>> how is nlgn possible, total no of possible substrings are n^2
>>>>>>
>>>>>>
>>>>>> this is the O(n^2) solutionOn Wed, Jul 27, 2011 at 8:09 AM, saurabh
>>>>>> singh
>>>>>>
>>>>>> for(int l = 0; l < len; l++){
>>>>>>                 for(int i = 0; i < len-l; i++){
>>>>>>                         int j = i+l;
>>>>>>                         char temp = s[j+1];
>>>>>>                         s[j+1] = 0;
>>>>>>                         printf("%s\n", s+i);
>>>>>>                         s[j+1] = temp;
>>>>>>                 }
>>>>>>         }
>>>>>>
>>>>>> <[email protected]> wrote:
>>>>>> >
>>>>>> > using suffix tree this can be done in o(nlogn) though will take
>>>>>> extra space.
>>>>>> >
>>>>>> > On Wed, Jul 27, 2011 at 12:47 AM, siva viknesh <
>>>>>> [email protected]> wrote:
>>>>>> >>
>>>>>> >> http://geeksforgeeks.org/?p=767
>>>>>> >>
>>>>>> >> On Jul 26, 11:49 pm, Pratz mary <[email protected]> wrote:
>>>>>> >> > how?
>>>>>> >> >
>>>>>> >> > On 26 July 2011 23:18, ankit sambyal <[email protected]>
>>>>>> wrote:
>>>>>> >> >
>>>>>> >> > > @vivin : Suffix trees are memory intensive..
>>>>>> >> >
>>>>>> >> > > This problem can be solved just by running 2 nested loops in
>>>>>> O(1)
>>>>>> >> > > space and O(n^2) time
>>>>>> >> >
>>>>>> >> > > --
>>>>>> >> > > You received this message because you are subscribed to the
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>>>>>> >> > > "Algorithm Geeks" group.
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>>>>>> >> >
>>>>>> >> > --
>>>>>> >> > regards Pratima :)
>>>>>> >>
>>>>>> >> --
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>>>>>> >
>>>>>> >
>>>>>> >
>>>>>> > --
>>>>>> > Saurabh Singh
>>>>>> > B.Tech (Computer Science)
>>>>>> > MNNIT ALLAHABAD
>>>>>> >
>>>>>> >
>>>>>> > --
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>>>>>>
>>>>>>
>>>>>> --
>>>>>> Sunny Aggrawal
>>>>>> B-Tech IV year,CSI
>>>>>> Indian Institute Of Technology,Roorkee
>>>>>>
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>>>>>>
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>>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> Sunny Aggrawal
>>>> B-Tech IV year,CSI
>>>> Indian Institute Of Technology,Roorkee
>>>>
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>>>
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>>
>>
>>
>> --
>> Sunny Aggrawal
>> B-Tech IV year,CSI
>> Indian Institute Of Technology,Roorkee
>>
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>
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-- 
Saurabh Singh
B.Tech (Computer Science)
MNNIT ALLAHABAD

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