Longest common substring in linear time












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











$endgroup$

















    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











    $endgroup$















      2












      2








      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











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 4 hours ago









      Discrete lizard♦

      4,44011537




      4,44011537










      asked 4 hours ago









      Manoharsinh RanaManoharsinh Rana

      917




      917






















          3 Answers
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          active

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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









          $endgroup$





















            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











            $endgroup$













            • $begingroup$
              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



















            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









            $endgroup$













              Your Answer





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              3 Answers
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              3 Answers
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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









              $endgroup$


















                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









                $endgroup$
















                  2












                  2








                  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









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Apass.JackApass.Jack

                  13.3k1939




                  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











                      $endgroup$













                      • $begingroup$
                        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
















                      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











                      $endgroup$













                      • $begingroup$
                        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














                      1












                      1








                      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











                      $endgroup$



                      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.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      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


















                      • $begingroup$
                        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











                      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









                      $endgroup$


















                        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









                        $endgroup$
















                          1












                          1








                          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









                          $endgroup$



                          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.)







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          D.W.♦D.W.

                          102k12127291




                          102k12127291






























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