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













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 2 hours ago









      Discrete lizard♦

      4,44011537




      4,44011537










      asked 2 hours ago









      Manoharsinh RanaManoharsinh Rana

      917




      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









          $endgroup$












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










          • $begingroup$
            Longest common substring is much easier than longest common subsequence. See my answer.
            $endgroup$
            – D.W.♦
            48 mins 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
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            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









            $endgroup$












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
















            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









            $endgroup$












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














            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 49 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
              45 mins ago

















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
















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





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










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










            • $begingroup$
              Longest common substring is much easier than longest common subsequence. See my answer.
              $endgroup$
              – D.W.♦
              48 mins 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 1 hour ago

























            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♦
              1 hour ago










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










            • $begingroup$
              Longest common substring is much easier than longest common subsequence. See my answer.
              $endgroup$
              – D.W.♦
              48 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♦
            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.♦
            48 mins ago




            $begingroup$
            Longest common substring is much easier than longest common subsequence. See my answer.
            $endgroup$
            – D.W.♦
            48 mins 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 1 hour ago









                D.W.♦D.W.

                102k12127291




                102k12127291



























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