@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
>> >
>> > > --
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>> > regards Pratima :)
>>
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>
>
>
> --
> Saurabh Singh
> B.Tech (Computer Science)
> MNNIT ALLAHABAD
>
>
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Sunny Aggrawal
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