Given a string str, find the longest repeating non-overlapping substring in it. In other words find 2 identical substrings of maximum length which do not overlap. If there exists more than one such substring return any of them.
Examples:
Input : str = "neveropen"
Output : neveropen
Input : str = "aab"
Output : a
Input : str = "aabaabaaba"
Output : aaba
Input : str = "aaaaaaaaaaa"
Output : aaaaa
Input : str = "banana"
Output : an
or na
Naive Solution : The problem can be solved easily by taking all the possible substrings and for all the substrings check it for the remaining(non-overlapping) string if there exists an identical substring. There are O(n2) total substrings and checking them against the remaining string will take O(n) time. So overall time complexity of above solution is O(n3).
Dynamic Programming : This problem can be solved in O(n2) time using Dynamic Programming. The basic idea is to find the longest repeating suffix for all prefixes in the string str.
Length of longest non-repeating substring can be recursively
defined as below.
LCSRe(i, j) stores length of the matching and
non-overlapping substrings ending
with i'th and j'th characters.
If str[i-1] == str[j-1] && (j-i) > LCSRe(i-1, j-1)
LCSRe(i, j) = LCSRe(i-1, j-1) + 1,
Else
LCSRe(i, j) = 0
Where i varies from 1 to n and
j varies from i+1 to n
To avoid overlapping we have to ensure that the length of suffix is less than (j-i) at any instant.
The maximum value of LCSRe(i, j) provides the length of the longest repeating substring and the substring itself can be found using the length and the ending index of the common suffix.
Below is the implementation of the recurrence.
C++
// C++ program to find the longest repeated// non-overlapping substring#include<bits/stdc++.h>using namespace std;// Returns the longest repeating non-overlapping// substring in strstring longestRepeatedSubstring(string str){ int n = str.length(); int LCSRe[n+1][n+1]; // Setting all to 0 memset(LCSRe, 0, sizeof(LCSRe)); string res; // To store result int res_length = 0; // To store length of result // building table in bottom-up manner int i, index = 0; for (i=1; i<=n; i++) { for (int j=i+1; j<=n; j++) { // (j-i) > LCSRe[i-1][j-1] to remove // overlapping if (str[i-1] == str[j-1] && LCSRe[i-1][j-1] < (j - i)) { LCSRe[i][j] = LCSRe[i-1][j-1] + 1; // updating maximum length of the // substring and updating the finishing // index of the suffix if (LCSRe[i][j] > res_length) { res_length = LCSRe[i][j]; index = max(i, index); } } else LCSRe[i][j] = 0; } } // If we have non-empty result, then insert all // characters from first character to last // character of string if (res_length > 0) for (i = index - res_length + 1; i <= index; i++) res.push_back(str[i-1]); return res;}// Driver program to test the above functionint main(){ string str = "neveropen"; cout << longestRepeatedSubstring(str); return 0;} |
Java
// Java program to find the longest repeated // non-overlapping substring class GFG {// Returns the longest repeating non-overlapping // substring in str static String longestRepeatedSubstring(String str) { int n = str.length(); int LCSRe[][] = new int[n + 1][n + 1]; String res = ""; // To store result int res_length = 0; // To store length of result // building table in bottom-up manner int i, index = 0; for (i = 1; i <= n; i++) { for (int j = i + 1; j <= n; j++) { // (j-i) > LCSRe[i-1][j-1] to remove // overlapping if (str.charAt(i - 1) == str.charAt(j - 1) && LCSRe[i - 1][j - 1] < (j - i)) { LCSRe[i][j] = LCSRe[i - 1][j - 1] + 1; // updating maximum length of the // substring and updating the finishing // index of the suffix if (LCSRe[i][j] > res_length) { res_length = LCSRe[i][j]; index = Math.max(i, index); } } else { LCSRe[i][j] = 0; } } } // If we have non-empty result, then insert all // characters from first character to last // character of String if (res_length > 0) { for (i = index - res_length + 1; i <= index; i++) { res += str.charAt(i - 1); } } return res; }// Driver program to test the above function public static void main(String[] args) { String str = "neveropen"; System.out.println(longestRepeatedSubstring(str)); }}// This code is contributed by Rajput-JI |
Python 3
# Python 3 program to find the longest repeated# non-overlapping substring# Returns the longest repeating non-overlapping# substring in strdef longestRepeatedSubstring(str): n = len(str) LCSRe = [[0 for x in range(n + 1)] for y in range(n + 1)] res = "" # To store result res_length = 0 # To store length of result # building table in bottom-up manner index = 0 for i in range(1, n + 1): for j in range(i + 1, n + 1): # (j-i) > LCSRe[i-1][j-1] to remove # overlapping if (str[i - 1] == str[j - 1] and LCSRe[i - 1][j - 1] < (j - i)): LCSRe[i][j] = LCSRe[i - 1][j - 1] + 1 # updating maximum length of the # substring and updating the finishing # index of the suffix if (LCSRe[i][j] > res_length): res_length = LCSRe[i][j] index = max(i, index) else: LCSRe[i][j] = 0 # If we have non-empty result, then insert # all characters from first character to # last character of string if (res_length > 0): for i in range(index - res_length + 1, index + 1): res = res + str[i - 1] return res# Driver Codeif __name__ == "__main__": str = "neveropen" print(longestRepeatedSubstring(str))# This code is contributed by ita_c |
C#
// C# program to find the longest repeated // non-overlapping substring using System; public class GFG { // Returns the longest repeating non-overlapping // substring in str static String longestRepeatedSubstring(String str) { int n = str.Length; int [,]LCSRe = new int[n + 1,n + 1]; String res = ""; // To store result int res_length = 0; // To store length of result // building table in bottom-up manner int i, index = 0; for (i = 1; i <= n; i++) { for (int j = i + 1; j <= n; j++) { // (j-i) > LCSRe[i-1][j-1] to remove // overlapping if (str[i - 1] == str[j - 1] && LCSRe[i - 1,j - 1] < (j - i)) { LCSRe[i,j] = LCSRe[i - 1,j - 1] + 1; // updating maximum length of the // substring and updating the finishing // index of the suffix if (LCSRe[i,j] > res_length) { res_length = LCSRe[i,j]; index = Math.Max(i, index); } } else { LCSRe[i,j] = 0; } } } // If we have non-empty result, then insert all // characters from first character to last // character of String if (res_length > 0) { for (i = index - res_length + 1; i <= index; i++) { res += str[i - 1]; } } return res; } // Driver program to test the above function public static void Main() { String str = "neveropen"; Console.WriteLine(longestRepeatedSubstring(str)); }}// This code is contributed by Rajput-JI |
Javascript
<script>// Javascript program to find the longest repeated // non-overlapping substring // Returns the longest repeating non-overlapping // substring in str function longestRepeatedSubstring(str) { let n = str.length; let LCSRe = new Array(n+1); for(let i = 0; i < n + 1; i++) { LCSRe[i] = new Array(n+1); } for(let i = 0; i < n + 1; i++) { for(let j = 0; j < n + 1; j++) { LCSRe[i][j] = 0; } } let res = ""; // To store result let res_length = 0; // To store length of result // building table in bottom-up manner let i, index = 0; for (i = 1; i <= n; i++) { for (let j = i + 1; j <= n; j++) { // (j-i) > LCSRe[i-1][j-1] to remove // overlapping if (str[i-1] == str[j-1] && LCSRe[i - 1][j - 1] < (j - i)) { LCSRe[i][j] = LCSRe[i - 1][j - 1] + 1; // updating maximum length of the // substring and updating the finishing // index of the suffix if (LCSRe[i][j] > res_length) { res_length = LCSRe[i][j]; index = Math.max(i, index); } } else { LCSRe[i][j] = 0; } } } // If we have non-empty result, then insert all // characters from first character to last // character of String if (res_length > 0) { for (i = index - res_length + 1; i <= index; i++) { res += str.charAt(i - 1); } } return res; } // Driver program to test the above function let str= "neveropen"; document.write(longestRepeatedSubstring(str)); // This code is contributed by rag2127</script> |
Output
neveropen
Time Complexity: O(n2)
Auxiliary Space: O(n2)
References:
https://www.geeksforgeeks.org/longest-common-substring/
This article is contributed by Ayush Khanduri. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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