Longest common substring in linear time
$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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 4 hours ago
Discrete lizard♦
4,44011537
4,44011537
asked 4 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.
$endgroup$
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.
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$begingroup$
are you talking about subsequence? I am talking about substring.
$endgroup$
– Manoharsinh Rana
4 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♦
3 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
2 hours 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.)
$endgroup$
add a comment |
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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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 2 hours ago
Apass.JackApass.Jack
13.3k1939
13.3k1939
add a comment |
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.
$endgroup$
$begingroup$
are you talking about subsequence? I am talking about substring.
$endgroup$
– Manoharsinh Rana
4 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♦
3 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
2 hours 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.
$endgroup$
$begingroup$
are you talking about subsequence? I am talking about substring.
$endgroup$
– Manoharsinh Rana
4 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♦
3 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
2 hours 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.
$endgroup$
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.
edited 3 hours ago
answered 4 hours ago
Discrete lizard♦Discrete lizard
4,44011537
4,44011537
$begingroup$
are you talking about subsequence? I am talking about substring.
$endgroup$
– Manoharsinh Rana
4 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♦
3 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
2 hours ago
add a comment |
$begingroup$
are you talking about subsequence? I am talking about substring.
$endgroup$
– Manoharsinh Rana
4 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♦
3 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
2 hours ago
$begingroup$
are you talking about subsequence? I am talking about substring.
$endgroup$
– Manoharsinh Rana
4 hours ago
$begingroup$
are you talking about subsequence? I am talking about substring.
$endgroup$
– Manoharsinh Rana
4 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♦
3 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♦
3 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
2 hours ago
$begingroup$
Longest common substring is much easier than longest common subsequence. See my answer.
$endgroup$
– D.W.♦
2 hours 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.)
$endgroup$
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.)
$endgroup$
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.)
$endgroup$
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 3 hours ago
D.W.♦D.W.
102k12127291
102k12127291
add a comment |
add a comment |
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