Longest common substring in linear timeComputing the longest common substring of two strings using suffix arraysFind longest common substring using a rolling hashWhich algorithm to use to find all common substring (LCS case) with really big stringsFinding the longest repeating subsequenceHow to find longest recurring pattern from lage string data set?Longest substring with consecutive repetitionsDoes the Longest Common Subsequence problem reduce to its binary version?Substring problems in suffix treesNumber of optimal solutions for Longest Common Subsequence (Substring) problemLongest common sequence matrix giving wrong answer
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Longest common substring in linear time
Computing the longest common substring of two strings using suffix arraysFind longest common substring using a rolling hashWhich algorithm to use to find all common substring (LCS case) with really big stringsFinding the longest repeating subsequenceHow to find longest recurring pattern from lage string data set?Longest substring with consecutive repetitionsDoes the Longest Common Subsequence problem reduce to its binary version?Substring problems in suffix treesNumber of optimal solutions for Longest Common Subsequence (Substring) problemLongest common sequence matrix giving wrong answer
$begingroup$
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
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
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
edited 2 hours ago
Discrete lizard♦
4,44011537
asked 2 hours ago
Manoharsinh RanaManoharsinh Rana
917
$endgroup$
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
algorithms time-complexity strings longest-common-substring
edited 2 hours ago
Discrete lizard♦
4,44011537
asked 2 hours ago
Manoharsinh RanaManoharsinh Rana
917
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
edited 2 hours ago
Discrete lizard♦
4,44011537
4,44011537
asked 2 hours ago
Manoharsinh RanaManoharsinh Rana
917
asked 2 hours ago
Manoharsinh RanaManoharsinh Rana
917
asked 2 hours ago
Manoharsinh RanaManoharsinh Rana
917
917
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
answered 56 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
52 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.
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
2 hours 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♦
2 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
54 mins ago
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.)
answered 1 hour ago
D.W.♦D.W.
102k12127291
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 56 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
52 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.
answered 56 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
52 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.
answered 56 mins ago
Apass.JackApass.Jack
13.3k1939
$endgroup$
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.
answered 56 mins ago
Apass.JackApass.Jack
13.3k1939
answered 56 mins ago
Apass.JackApass.Jack
13.3k1939
answered 56 mins ago
Apass.JackApass.Jack
13.3k1939
answered 56 mins ago
Apass.JackApass.Jack
13.3k1939
13.3k1939
$begingroup$
I did not see @D.W's answer, possibly because I was interrupted while writing my answer.
$endgroup$
– Apass.Jack
52 mins ago
add a comment |
$begingroup$
I did not see @D.W's answer, possibly because I was interrupted while writing my answer.
$endgroup$
– Apass.Jack
52 mins ago
$begingroup$
I did not see @D.W's answer, possibly because I was interrupted while writing my answer.
$endgroup$
– Apass.Jack
52 mins ago
$begingroup$
I did not see @D.W's answer, possibly because I was interrupted while writing my answer.
$endgroup$
– Apass.Jack
52 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.
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
2 hours 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♦
2 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
54 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.
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
2 hours 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♦
2 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
54 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.
answered 2 hours ago
Discrete lizard♦Discrete lizard
4,44011537
$endgroup$
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.
answered 2 hours ago
Discrete lizard♦Discrete lizard
4,44011537
edited 2 hours 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
4,44011537
$begingroup$
are you talking about subsequence? I am talking about substring.
$endgroup$
– Manoharsinh Rana
2 hours 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♦
2 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
54 mins ago
add a comment |
$begingroup$
are you talking about subsequence? I am talking about substring.
$endgroup$
– Manoharsinh Rana
2 hours 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♦
2 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
54 mins ago
$begingroup$
are you talking about subsequence? I am talking about substring.
$endgroup$
– Manoharsinh Rana
2 hours ago
$begingroup$
are you talking about subsequence? I am talking about substring.
$endgroup$
– Manoharsinh Rana
2 hours 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♦
2 hours 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♦
2 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
54 mins ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
54 mins ago
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.)
answered 1 hour 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.)
answered 1 hour 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.)
answered 1 hour ago
D.W.♦D.W.
102k12127291
$endgroup$
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.)
answered 1 hour ago
D.W.♦D.W.
102k12127291
answered 1 hour ago
D.W.♦D.W.
102k12127291
answered 1 hour ago
D.W.♦D.W.
102k12127291
answered 1 hour ago
D.W.♦D.W.
102k12127291
102k12127291
add a comment |
add a comment |
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