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Longest common substring in linear time

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Longest common substring in linear time

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2

$begingroup$

We know that the longest common substring of two strings can be found in O(N^2) time complexity.
Can a solution be found in only linear time?

algorithms time-complexity strings longest-common-substring

share|cite|improve this question

edited 2 hours ago

Discrete lizard♦

4,44011537

asked 2 hours ago

Manoharsinh RanaManoharsinh Rana

917

$endgroup$

add a comment | 

2

$begingroup$

We know that the longest common substring of two strings can be found in O(N^2) time complexity.
Can a solution be found in only linear time?

algorithms time-complexity strings longest-common-substring

share|cite|improve this question

edited 2 hours ago

Discrete lizard♦

4,44011537

asked 2 hours ago

Manoharsinh RanaManoharsinh Rana

917

$endgroup$

add a comment | 

$begingroup$

We know that the longest common substring of two strings can be found in O(N^2) time complexity.
Can a solution be found in only linear time?

algorithms time-complexity strings longest-common-substring

share|cite|improve this question

edited 2 hours ago

Discrete lizard♦

4,44011537

asked 2 hours ago

Manoharsinh RanaManoharsinh Rana

917

$endgroup$

share|cite|improve this question

edited 2 hours ago

Discrete lizard♦

4,44011537

asked 2 hours ago

Manoharsinh RanaManoharsinh Rana

917

share|cite|improve this question

edited 2 hours ago

Discrete lizard♦

4,44011537

asked 2 hours ago

Manoharsinh RanaManoharsinh Rana

917

edited 2 hours ago

Discrete lizard♦

4,44011537

edited 2 hours ago

Discrete lizard♦

4,44011537

asked 2 hours ago

Manoharsinh RanaManoharsinh Rana

917

asked 2 hours ago

Manoharsinh RanaManoharsinh Rana

917

3 Answers
3

active

oldest

votes

2

$begingroup$

Yes, the longest common substring of two strings can be found in $O(m+n)$ time, where $m$ and $n$ are the lengths of the two strings, assuming the size of the alphabet is constant.

Here is an excerpt from https://en.wikipedia.org/wiki/Longest_common_substring_problem.

The longest common substrings of a set of strings can be found by building a generalized suffix tree for the strings, and then finding the deepest internal nodes which have leaf nodes from all the strings in the subtree below it.

Building a generalized suffix tree for two given strings takes $Theta(m+n)$ time using the famous ingenious Ukkonen's algorithm. Finding the deepest internal nodes that come from both strings takes $Theta(m+n)$ time. Hence we can find the longest common substring in $Theta(m+n)$ time.

share|cite|improve this answer

answered 36 mins ago

Apass.JackApass.Jack

13.3k1939

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did not see @D.W's answer, possibly because I was interrupted while writing my answer.
  $endgroup$
  – Apass.Jack
  32 mins ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

It is unlikely that that a better algorithm than quadratic exists, let alone linear. For the related problem of finding subsequences, this is a known result: In the paper "Tight hardness results for LCS and other sequence similarity measures." by Abboud et al. , they show that the existence of an algorithm with a running time of $O(n^{2-varepsilon})$, for some $varepsilon>0$ refutes the Strong Exponential Time Hypothesis (SETH).

SETH is considered to be very likely true (although not universally accepted), so it is unlikely any $O(n^{2-varepsilon})$ time algorithm exists.

---

While finding a substring is a slightly different problem, it seems likely to be equally hard.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Discrete lizard♦Discrete lizard

4,44011537

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  are you talking about subsequence? I am talking about substring.
  $endgroup$
  – Manoharsinh Rana
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ManoharsinhRana Ah, I see. The problems are similar, and it is hard to find results for the string variant. I think there are similar results for the substring problem, but they are not easy to find. You could try looking at papers that cite "Quadratic conditional lower bounds for string problems and dynamic time warping" by Bringmann and Künnemann, as their program lead to a lot of results related to this problem.
  $endgroup$
  – Discrete lizard♦
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Longest common substring is much easier than longest common subsequence. See my answer.
  $endgroup$
  – D.W.♦
  34 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Yes. There's even a Wikipedia article about it! https://en.wikipedia.org/wiki/Longest_common_substring_problem

In particular, as Wikipedia explains, there is a linear-time algorithm, using suffix trees (or suffix arrays).

Searching on "longest common substring" turns up that Wikipedia article as the first hit (for me). In the future, please research the problem before asking here. (See, e.g., https://meta.stackoverflow.com/q/261592/781723.)

share|cite|improve this answer

answered 54 mins ago

D.W.♦D.W.

102k12127291

$endgroup$

add a comment | 

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2

$begingroup$

Yes, the longest common substring of two strings can be found in $O(m+n)$ time, where $m$ and $n$ are the lengths of the two strings, assuming the size of the alphabet is constant.

Here is an excerpt from https://en.wikipedia.org/wiki/Longest_common_substring_problem.

The longest common substrings of a set of strings can be found by building a generalized suffix tree for the strings, and then finding the deepest internal nodes which have leaf nodes from all the strings in the subtree below it.

Building a generalized suffix tree for two given strings takes $Theta(m+n)$ time using the famous ingenious Ukkonen's algorithm. Finding the deepest internal nodes that come from both strings takes $Theta(m+n)$ time. Hence we can find the longest common substring in $Theta(m+n)$ time.

share|cite|improve this answer

answered 36 mins ago

Apass.JackApass.Jack

13.3k1939

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did not see @D.W's answer, possibly because I was interrupted while writing my answer.
  $endgroup$
  – Apass.Jack
  32 mins ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Yes, the longest common substring of two strings can be found in $O(m+n)$ time, where $m$ and $n$ are the lengths of the two strings, assuming the size of the alphabet is constant.

Here is an excerpt from https://en.wikipedia.org/wiki/Longest_common_substring_problem.

The longest common substrings of a set of strings can be found by building a generalized suffix tree for the strings, and then finding the deepest internal nodes which have leaf nodes from all the strings in the subtree below it.

Building a generalized suffix tree for two given strings takes $Theta(m+n)$ time using the famous ingenious Ukkonen's algorithm. Finding the deepest internal nodes that come from both strings takes $Theta(m+n)$ time. Hence we can find the longest common substring in $Theta(m+n)$ time.

share|cite|improve this answer

answered 36 mins ago

Apass.JackApass.Jack

13.3k1939

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did not see @D.W's answer, possibly because I was interrupted while writing my answer.
  $endgroup$
  – Apass.Jack
  32 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Yes, the longest common substring of two strings can be found in $O(m+n)$ time, where $m$ and $n$ are the lengths of the two strings, assuming the size of the alphabet is constant.

Here is an excerpt from https://en.wikipedia.org/wiki/Longest_common_substring_problem.

The longest common substrings of a set of strings can be found by building a generalized suffix tree for the strings, and then finding the deepest internal nodes which have leaf nodes from all the strings in the subtree below it.

Building a generalized suffix tree for two given strings takes $Theta(m+n)$ time using the famous ingenious Ukkonen's algorithm. Finding the deepest internal nodes that come from both strings takes $Theta(m+n)$ time. Hence we can find the longest common substring in $Theta(m+n)$ time.

share|cite|improve this answer

answered 36 mins ago

Apass.JackApass.Jack

13.3k1939

$endgroup$

share|cite|improve this answer

answered 36 mins ago

Apass.JackApass.Jack

13.3k1939

answered 36 mins ago

Apass.JackApass.Jack

13.3k1939

answered 36 mins ago

Apass.JackApass.Jack

13.3k1939

1

$begingroup$

It is unlikely that that a better algorithm than quadratic exists, let alone linear. For the related problem of finding subsequences, this is a known result: In the paper "Tight hardness results for LCS and other sequence similarity measures." by Abboud et al. , they show that the existence of an algorithm with a running time of $O(n^{2-varepsilon})$, for some $varepsilon>0$ refutes the Strong Exponential Time Hypothesis (SETH).

SETH is considered to be very likely true (although not universally accepted), so it is unlikely any $O(n^{2-varepsilon})$ time algorithm exists.

---

While finding a substring is a slightly different problem, it seems likely to be equally hard.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Discrete lizard♦Discrete lizard

4,44011537

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  are you talking about subsequence? I am talking about substring.
  $endgroup$
  – Manoharsinh Rana
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ManoharsinhRana Ah, I see. The problems are similar, and it is hard to find results for the string variant. I think there are similar results for the substring problem, but they are not easy to find. You could try looking at papers that cite "Quadratic conditional lower bounds for string problems and dynamic time warping" by Bringmann and Künnemann, as their program lead to a lot of results related to this problem.
  $endgroup$
  – Discrete lizard♦
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Longest common substring is much easier than longest common subsequence. See my answer.
  $endgroup$
  – D.W.♦
  34 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

It is unlikely that that a better algorithm than quadratic exists, let alone linear. For the related problem of finding subsequences, this is a known result: In the paper "Tight hardness results for LCS and other sequence similarity measures." by Abboud et al. , they show that the existence of an algorithm with a running time of $O(n^{2-varepsilon})$, for some $varepsilon>0$ refutes the Strong Exponential Time Hypothesis (SETH).

SETH is considered to be very likely true (although not universally accepted), so it is unlikely any $O(n^{2-varepsilon})$ time algorithm exists.

---

While finding a substring is a slightly different problem, it seems likely to be equally hard.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Discrete lizard♦Discrete lizard

4,44011537

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  are you talking about subsequence? I am talking about substring.
  $endgroup$
  – Manoharsinh Rana
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ManoharsinhRana Ah, I see. The problems are similar, and it is hard to find results for the string variant. I think there are similar results for the substring problem, but they are not easy to find. You could try looking at papers that cite "Quadratic conditional lower bounds for string problems and dynamic time warping" by Bringmann and Künnemann, as their program lead to a lot of results related to this problem.
  $endgroup$
  – Discrete lizard♦
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Longest common substring is much easier than longest common subsequence. See my answer.
  $endgroup$
  – D.W.♦
  34 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

It is unlikely that that a better algorithm than quadratic exists, let alone linear. For the related problem of finding subsequences, this is a known result: In the paper "Tight hardness results for LCS and other sequence similarity measures." by Abboud et al. , they show that the existence of an algorithm with a running time of $O(n^{2-varepsilon})$, for some $varepsilon>0$ refutes the Strong Exponential Time Hypothesis (SETH).

SETH is considered to be very likely true (although not universally accepted), so it is unlikely any $O(n^{2-varepsilon})$ time algorithm exists.

---

While finding a substring is a slightly different problem, it seems likely to be equally hard.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Discrete lizard♦Discrete lizard

4,44011537

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Discrete lizard♦Discrete lizard

4,44011537

answered 2 hours ago

Discrete lizard♦Discrete lizard

4,44011537

answered 2 hours ago

Discrete lizard♦Discrete lizard

4,44011537

1

$begingroup$

Yes. There's even a Wikipedia article about it! https://en.wikipedia.org/wiki/Longest_common_substring_problem

In particular, as Wikipedia explains, there is a linear-time algorithm, using suffix trees (or suffix arrays).

Searching on "longest common substring" turns up that Wikipedia article as the first hit (for me). In the future, please research the problem before asking here. (See, e.g., https://meta.stackoverflow.com/q/261592/781723.)

share|cite|improve this answer

answered 54 mins ago

D.W.♦D.W.

102k12127291

$endgroup$

add a comment | 

1

$begingroup$

Yes. There's even a Wikipedia article about it! https://en.wikipedia.org/wiki/Longest_common_substring_problem

In particular, as Wikipedia explains, there is a linear-time algorithm, using suffix trees (or suffix arrays).

Searching on "longest common substring" turns up that Wikipedia article as the first hit (for me). In the future, please research the problem before asking here. (See, e.g., https://meta.stackoverflow.com/q/261592/781723.)

share|cite|improve this answer

answered 54 mins ago

D.W.♦D.W.

102k12127291

$endgroup$

add a comment | 

$begingroup$

Yes. There's even a Wikipedia article about it! https://en.wikipedia.org/wiki/Longest_common_substring_problem

In particular, as Wikipedia explains, there is a linear-time algorithm, using suffix trees (or suffix arrays).

Searching on "longest common substring" turns up that Wikipedia article as the first hit (for me). In the future, please research the problem before asking here. (See, e.g., https://meta.stackoverflow.com/q/261592/781723.)

share|cite|improve this answer

answered 54 mins ago

D.W.♦D.W.

102k12127291

$endgroup$

share|cite|improve this answer

answered 54 mins ago

D.W.♦D.W.

102k12127291

answered 54 mins ago

D.W.♦D.W.

102k12127291

answered 54 mins ago

D.W.♦D.W.

102k12127291