10829 - L-Gap Substrings

[filigabriel]

Any hint for solving this problem ??

In the worst case (a string with 50,000 same character), the solution is greater than 500,000,000 L-GAP. O.K. We can manage that case.

I think, even using KMP's algrithm is not possible count all L-GAPs in time

Please, help.

[ibrahim]

hi filigabriel,
I can't understand the porblem. Can you help me?

[filigabriel]

A g-Gap is defined as a string in the form UVU such that |V| = g.

For example: AALLAA is a 2-Gap taking U = AA and V = LL.
furthermore is 4-GAP because of U = A and V = ALLA and |V| = 4,
finally is not a 6-GAP, because U can not be empty.

The problem consists in given a string S, and a positive integer G, count all substrings S' in S such that S' is a G-Gap.

[ibrahim]

Thanks filigabriel, i got the problem. You got any hints????


Regards,
Ibrahim

[ardiankp]

Hmm.. still no hint? I also stuck in this problems..

any help will be appreciated

Thx

[Gigi's Baby]

Any hints ?
I could only get an O(n^2) method, but it surely got TLE.

[polone]

hmm...

suffix tree is not bad

we can count L-Gap strings during making the tree

I think it should work

I also wonder how the guys on ranklist deal this problem..

[Adrian Kuegel]

I only solved this problem because for an bioinformatics course I had read a paper which described an algorithm how to find branching tandem repeats. I modified this algorithm for this problem. I know that the problemsetter had a different solution, he told me it is something with divide and conquer and extended kmp (knuth morris pratt algorithm) (but I don't know what he meant with extended).

[polch]

!!!!!!!!!!!!!!! SPOILER - ATTENTION !!!!!!!!!!!!!!!!!!!!!!!!!!!







Hi,I solved this task using KMR + binary search in O(n*(logn)^2).

For each k being length of U I counted all subwords UVU of text in O(n/k*logk). (|V|=g)

Adrian, how fast (asympt.) algo have you got?

[serur]

to Adrian: did you store all the (i,D) pairs -- L-gap tandem repeats -- explicitly? I'm going to use an O(nlogn) algo for branching tandem repeats

[serur]

Never mind, the question above is silly: we are asked to find the number of occurences, not the number of distinct substrings.

[serur]

I find all branching tandem repeats with gap 0 <= g <= L and length > L-g in O(L*n*log(n)).
Specifically, for

Code: Select all

bbaabaaaaa$

my code finds pairs

Code: Select all

(6,10), (3,7), (8,10), (4,6), (7,9) and (2,6)

but fails to find

Code: Select all

(6,8),

because the corresponding tandem repeat (g = 0) is not branching.
The question is: how to handle non-branching repeats in O(nlogn)?

[serur]

for the testcase

Code: Select all

s = a^n

the output size is

Code: Select all

(n-L-1) + (n-L-3) + (n-L-5) + ...,

which amounts to O(n^2). Now, given we can find all branching tandem repeats in O(nlogn), how
to handle non-branching tandem repeats?

[serur]

Well, with Adrian's help I managed to get AC with an O(|GAP|n(logn)^2) algo, and I can't see as yet the way of shortening this to O(nlogn).
My code is 580 lines long, pure C code.
Many thanks to Adrian and to Rujia!

[luyuncheng]

can anyone give me some hint for this problem?i don't know how to solve this? as i think ,may be suffix array,but i don't know how to deal with the gap?can anyone help me? thanks very much! my email:luyuncheng@sina.com