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







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$endgroup$

















    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 1 hour ago









      Discrete lizard♦

      4,44011537




      4,44011537










      asked 2 hours ago









      Manoharsinh RanaManoharsinh Rana

      917




      917






















          3 Answers
          3






          active

          oldest

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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









          $endgroup$













          • $begingroup$
            I did not see @D.W's answer, possibly because I was interrupted while writing my answer.
            $endgroup$
            – Apass.Jack
            30 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
            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.♦
            33 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

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            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

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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









            $endgroup$













            • $begingroup$
              I did not see @D.W's answer, possibly because I was interrupted while writing my answer.
              $endgroup$
              – Apass.Jack
              30 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
              30 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 34 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
              30 mins ago




















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


















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






            $begingroup$
            I did not see @D.W's answer, possibly because I was interrupted while writing my answer.
            $endgroup$
            – Apass.Jack
            30 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
              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.♦
              33 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
              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.♦
              33 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 1 hour ago









            Discrete lizard♦Discrete lizard

            4,44011537




            4,44011537












            • $begingroup$
              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.♦
              33 mins ago


















            • $begingroup$
              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.♦
              33 mins ago
















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






            $begingroup$
            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




            $begingroup$
            @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.♦
            33 mins ago




            $begingroup$
            Longest common substring is much easier than longest common subsequence. See my answer.
            $endgroup$
            – D.W.♦
            33 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 52 mins ago









                D.W.♦D.W.

                102k12127291




                102k12127291






























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                    濃尾地震

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                    濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸
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