Longest repeating and non-overlapping substring
Last Updated :
16 Nov, 2024
Given a string s, the task is to find the longest repeating non-overlapping substring in it. In other words, find 2 identical substrings of maximum length which do not overlap. Return -1 if no such string exists.
Note: Multiple Answers are possible but we have to return the substring whose first occurrence is earlier.
Examples:
Input: s = "acdcdacdc"
Output: "acdc"
Explanation: The string "acdc" is the longest Substring of s which is repeating but not overlapping.
Input: s = "geeksforgeeks"
Output: "geeks"
Explanation: The string "geeks" is the longest subString of s which is repeating but not overlapping.
Using Brute Force Method - O(n^3) Time and O(n) Space
The idea is to generate all the possible substrings and check if the substring exists in the remaining string. If substring exists and its length is greater than answer substring, then set answer to current substring.
C++
// C++ program to find longest repeating
// and non-overlapping substring
// using recursion
#include <bits/stdc++.h>
using namespace std;
string longestSubstring(string& s) {
int n = s.length();
string ans = "";
int len = 0;
int i = 0, j = 0;
while (i < n && j < n) {
string curr = s.substr(i, j - i + 1);
// If substring exists, compare its length
// with ans
if (s.find(curr, j + 1) != string::npos
&& j - i + 1 > len) {
len = j - i + 1;
ans = curr;
}
// Otherwise increment i
else
i++;
j++;
}
return len > 0 ? ans : "-1";
}
int main() {
string s = "geeksforgeeks";
cout << longestSubstring(s) << endl;
return 0;
}
// Java program to find longest repeating
// and non-overlapping substring
// using recursion
class GfG {
static String longestSubstring(String s) {
int n = s.length();
String ans = "";
int len = 0;
int i = 0, j = 0;
while (i < n && j < n) {
String curr = s.substring(i, j + 1);
// If substring exists, compare its length
// with ans
if (s.indexOf(curr, j + 1) != -1
&& j - i + 1 > len) {
len = j - i + 1;
ans = curr;
}
// Otherwise increment i
else
i++;
j++;
}
return len > 0 ? ans : "-1";
}
public static void main(String[] args) {
String s = "geeksforgeeks";
System.out.println(longestSubstring(s));
}
}
# Python program to find longest repeating
# and non-overlapping substring
# using recursion
def longestSubstring(s):
n = len(s)
ans = ""
lenAns = 0
i, j = 0, 0
while i < n and j < n:
curr = s[i:j + 1]
# If substring exists, compare its length
# with ans
if s.find(curr, j + 1) != -1 and j - i + 1 > lenAns:
lenAns = j - i + 1
ans = curr
# Otherwise increment i
else:
i += 1
j += 1
if lenAns > 0:
return ans
return "-1"
if __name__ == "__main__":
s = "geeksforgeeks"
print(longestSubstring(s))
// C# program to find longest repeating
// and non-overlapping substring
// using recursion
using System;
class GfG {
static string longestSubstring(string s) {
int n = s.Length;
string ans = "";
int len = 0;
int i = 0, j = 0;
while (i < n && j < n) {
string curr = s.Substring(i, j - i + 1);
// If substring exists, compare its length
// with ans
if (s.IndexOf(curr, j + 1) != -1
&& j - i + 1 > len) {
len = j - i + 1;
ans = curr;
}
// Otherwise increment i
else
i++;
j++;
}
return len > 0 ? ans : "-1";
}
static void Main(string[] args) {
string s = "geeksforgeeks";
Console.WriteLine(longestSubstring(s));
}
}
// JavaScript program to find longest repeating
// and non-overlapping substring
// using recursion
function longestSubstring(s) {
const n = s.length;
let ans = "";
let len = 0;
let i = 0, j = 0;
while (i < n && j < n) {
const curr = s.substring(i, j + 1);
// If substring exists, compare its length
// with ans
if (s.indexOf(curr, j + 1) !== -1
&& j - i + 1 > len) {
len = j - i + 1;
ans = curr;
}
// Otherwise increment i
else
i++;
j++;
}
return len > 0 ? ans : "-1";
}
const s = "geeksforgeeks";
console.log(longestSubstring(s));
Using Top-Down DP (Memoization) - O(n^2) Time and O(n^2) Space
The approach is to compute the longest repeating suffix for all prefix pairs in the string s. For indices i and j, if s[i] == s[j], then recursively compute suffix(i+1, j+1) and set suffix(i, j) as min(suffix(i+1, j+1) + 1, j - i - 1) to prevent overlap. If the characters do not match, set suffix(i, j) = 0.
Note:
- To avoid overlapping we have to ensure that the length of suffix is less than (j-i) at any instant.
- The maximum value of suffix(i, j) provides the length of the longest repeating substring and the substring itself can be found using the length and the starting index of the common suffix.
- suffix(i, j) stores the length of the longest common suffix between indices i and j, ensuring it doesn’t exceed j - i - 1 to avoid overlap.
C++
// C++ program to find longest repeating
// and non-overlapping substring
// using memoization
#include <bits/stdc++.h>
using namespace std;
int findSuffix(int i, int j, string &s,
vector<vector<int>> &memo) {
// base case
if (j == s.length())
return 0;
// return memoized value
if (memo[i][j] != -1)
return memo[i][j];
// if characters match
if (s[i] == s[j]) {
memo[i][j] = 1 + min(findSuffix(i + 1, j + 1, s, memo),
j - i - 1);
}
else {
memo[i][j] = 0;
}
return memo[i][j];
}
string longestSubstring(string s) {
int n = s.length();
vector<vector<int>> memo(n, vector<int>(n, -1));
// find length of non-overlapping
// substrings for all pairs (i,j)
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
findSuffix(i, j, s, memo);
}
}
string ans = "";
int ansLen = 0;
// If length of suffix is greater
// than ansLen, update ans and ansLen
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (memo[i][j] > ansLen) {
ansLen = memo[i][j];
ans = s.substr(i, ansLen);
}
}
}
return ansLen > 0 ? ans : "-1";
}
int main() {
string s = "geeksforgeeks";
cout << longestSubstring(s) << endl;
return 0;
}
// Java program to find longest repeating
// and non-overlapping substring
// using memoization
import java.util.Arrays;
class GfG {
static int findSuffix(int i, int j, String s,
int[][] memo) {
// base case
if (j == s.length())
return 0;
// return memoized value
if (memo[i][j] != -1)
return memo[i][j];
// if characters match
if (s.charAt(i) == s.charAt(j)) {
memo[i][j] = 1
+ Math.min(findSuffix(i + 1, j + 1,
s, memo),
j - i - 1);
}
else {
memo[i][j] = 0;
}
return memo[i][j];
}
static String longestSubstring(String s) {
int n = s.length();
int[][] memo = new int[n][n];
for (int[] row : memo) {
Arrays.fill(row, -1);
}
// find length of non-overlapping
// substrings for all pairs (i, j)
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
findSuffix(i, j, s, memo);
}
}
String ans = "";
int ansLen = 0;
// If length of suffix is greater
// than ansLen, update ans and ansLen
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (memo[i][j] > ansLen) {
ansLen = memo[i][j];
ans = s.substring(i, i + ansLen);
}
}
}
return ansLen > 0 ? ans : "-1";
}
public static void main(String[] args) {
String s = "geeksforgeeks";
System.out.println(longestSubstring(s));
}
}
# Python program to find longest repeating
# and non-overlapping substring
# using memoization
def findSuffix(i, j, s, memo):
# base case
if j == len(s):
return 0
# return memoized value
if memo[i][j] != -1:
return memo[i][j]
# if characters match
if s[i] == s[j]:
memo[i][j] = 1 + min(findSuffix(i + 1, j + 1, s, memo), \
j - i - 1)
else:
memo[i][j] = 0
return memo[i][j]
def longestSubstring(s):
n = len(s)
memo = [[-1] * n for _ in range(n)]
# find length of non-overlapping
# substrings for all pairs (i, j)
for i in range(n):
for j in range(i + 1, n):
findSuffix(i, j, s, memo)
ans = ""
ansLen = 0
# If length of suffix is greater
# than ansLen, update ans and ansLen
for i in range(n):
for j in range(i + 1, n):
if memo[i][j] > ansLen:
ansLen = memo[i][j]
ans = s[i:i + ansLen]
if ansLen > 0:
return ans
return "-1"
if __name__ == "__main__":
s = "geeksforgeeks"
print(longestSubstring(s))
// C# program to find longest repeating
// and non-overlapping substring
// using memoization
using System;
class GfG {
static int findSuffix(int i, int j, string s,
int[, ] memo) {
// base case
if (j == s.Length)
return 0;
// return memoized value
if (memo[i, j] != -1)
return memo[i, j];
// if characters match
if (s[i] == s[j]) {
memo[i, j] = 1
+ Math.Min(findSuffix(i + 1, j + 1,
s, memo),
j - i - 1);
}
else {
memo[i, j] = 0;
}
return memo[i, j];
}
static string longestSubstring(string s) {
int n = s.Length;
int[, ] memo = new int[n, n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
memo[i, j] = -1;
}
}
// find length of non-overlapping
// substrings for all pairs (i, j)
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
findSuffix(i, j, s, memo);
}
}
string ans = "";
int ansLen = 0;
// If length of suffix is greater
// than ansLen, update ans and ansLen
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (memo[i, j] > ansLen) {
ansLen = memo[i, j];
ans = s.Substring(i, ansLen);
}
}
}
return ansLen > 0 ? ans : "-1";
}
static void Main(string[] args) {
string s = "geeksforgeeks";
Console.WriteLine(longestSubstring(s));
}
}
// JavaScript program to find longest repeating
// and non-overlapping substring
// using memoization
function findSuffix(i, j, s, memo) {
// base case
if (j === s.length)
return 0;
// return memoized value
if (memo[i][j] !== -1)
return memo[i][j];
// if characters match
if (s[i] === s[j]) {
memo[i][j]
= 1
+ Math.min(findSuffix(i + 1, j + 1, s, memo),
j - i - 1);
}
else {
memo[i][j] = 0;
}
return memo[i][j];
}
function longestSubstring(s) {
const n = s.length;
const memo
= Array.from({length : n}, () => Array(n).fill(-1));
// find length of non-overlapping
// substrings for all pairs (i, j)
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
findSuffix(i, j, s, memo);
}
}
let ans = "";
let ansLen = 0;
// If length of suffix is greater
// than ansLen, update ans and ansLen
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
if (memo[i][j] > ansLen) {
ansLen = memo[i][j];
ans = s.substring(i, i + ansLen);
}
}
}
return ansLen > 0 ? ans : "-1";
}
const s = "geeksforgeeks";
console.log(longestSubstring(s));
Using Bottom-Up DP (Tabulation) - O(n^2) Time and O(n^2) Space
The idea is to create a 2D matrix of size (n+1)*(n+1) and calculate the longest repeating suffixes for all index pairs (i, j) iteratively. We start from the end of the string and work backwards to fill the table. For each (i, j), if s[i] == s[j], we set suffix[i][j] to min(suffix[i+1][j+1]+1, j-i-1) to avoid overlap; otherwise, suffix[i][j] = 0.
C++
// C++ program to find longest repeating
// and non-overlapping substring
// using tabulation
#include <bits/stdc++.h>
using namespace std;
string longestSubstring(string s) {
int n = s.length();
vector<vector<int>> dp(n+1, vector<int>(n+1, 0));
string ans = "";
int ansLen = 0;
// find length of non-overlapping
// substrings for all pairs (i,j)
for (int i=n-1; i>=0; i--) {
for (int j=n-1; j>i; j--) {
// if characters match, set value
// and compare with ansLen.
if (s[i]==s[j]) {
dp[i][j] = 1 + min(dp[i+1][j+1], j-i-1);
if (dp[i][j]>=ansLen) {
ansLen = dp[i][j];
ans = s.substr(i, ansLen);
}
}
}
}
return ansLen>0?ans:"-1";
}
int main() {
string s = "geeksforgeeks";
cout << longestSubstring(s) << endl;
return 0;
}
// Java program to find longest repeating
// and non-overlapping substring
// using tabulation
class GfG {
static String longestSubstring(String s) {
int n = s.length();
int[][] dp = new int[n + 1][n + 1];
String ans = "";
int ansLen = 0;
// find length of non-overlapping
// substrings for all pairs (i, j)
for (int i = n - 1; i >= 0; i--) {
for (int j = n - 1; j > i; j--) {
// if characters match, set value
// and compare with ansLen.
if (s.charAt(i) == s.charAt(j)) {
dp[i][j] = 1 + Math.min(dp[i + 1][j + 1], j - i - 1);
if (dp[i][j] >= ansLen) {
ansLen = dp[i][j];
ans = s.substring(i, i + ansLen);
}
}
}
}
return ansLen > 0 ? ans : "-1";
}
public static void main(String[] args) {
String s = "geeksforgeeks";
System.out.println(longestSubstring(s));
}
}
# Python program to find longest repeating
# and non-overlapping substring
# using tabulation
def longestSubstring(s):
n = len(s)
dp = [[0] * (n + 1) for _ in range(n + 1)]
ans = ""
ansLen = 0
# find length of non-overlapping
# substrings for all pairs (i, j)
for i in range(n - 1, -1, -1):
for j in range(n - 1, i, -1):
# if characters match, set value
# and compare with ansLen.
if s[i] == s[j]:
dp[i][j] = 1 + min(dp[i + 1][j + 1], j - i - 1)
if dp[i][j] >= ansLen:
ansLen = dp[i][j]
ans = s[i:i + ansLen]
return ans if ansLen > 0 else "-1"
if __name__ == "__main__":
s = "geeksforgeeks"
print(longestSubstring(s))
// C# program to find longest repeating
// and non-overlapping substring
// using tabulation
using System;
class GfG {
static string longestSubstring(string s) {
int n = s.Length;
int[,] dp = new int[n + 1, n + 1];
string ans = "";
int ansLen = 0;
// find length of non-overlapping
// substrings for all pairs (i, j)
for (int i = n - 1; i >= 0; i--) {
for (int j = n - 1; j > i; j--) {
// if characters match, set value
// and compare with ansLen.
if (s[i] == s[j]) {
dp[i, j] = 1 + Math.Min(dp[i + 1, j + 1], j - i - 1);
if (dp[i, j] >= ansLen) {
ansLen = dp[i, j];
ans = s.Substring(i, ansLen);
}
}
}
}
return ansLen > 0 ? ans : "-1";
}
static void Main(string[] args) {
string s = "geeksforgeeks";
Console.WriteLine(longestSubstring(s));
}
}
// JavaScript program to find longest repeating
// and non-overlapping substring
// using tabulation
function longestSubstring(s) {
const n = s.length;
const dp = Array.from({ length: n + 1 }, () => Array(n + 1).fill(0));
let ans = "";
let ansLen = 0;
// find length of non-overlapping
// substrings for all pairs (i, j)
for (let i = n - 1; i >= 0; i--) {
for (let j = n - 1; j > i; j--) {
// if characters match, set value
// and compare with ansLen.
if (s[i] === s[j]) {
dp[i][j] = 1 + Math.min(dp[i + 1][j + 1], j - i - 1);
if (dp[i][j] >= ansLen) {
ansLen = dp[i][j];
ans = s.substring(i, i + ansLen);
}
}
}
}
return ansLen > 0 ? ans : "-1";
}
const s = "geeksforgeeks";
console.log(longestSubstring(s));
Using Space Optimized DP – O(n^2) Time and O(n) Space
The idea is to use a single 1D array instead of a 2D matrix by keeping track of only the "next row" values required to compute suffix[i][j]. Since each value suffix[i][j] depends only on suffix[i+1][j+1] in the row below, we can maintain the previous row's values in a 1D array and update them iteratively for each row.
C++
// C++ program to find longest repeating
// and non-overlapping substring
// using space optimised
#include <bits/stdc++.h>
using namespace std;
string longestSubstring(string s) {
int n = s.length();
vector<int> dp(n+1,0);
string ans = "";
int ansLen = 0;
// find length of non-overlapping
// substrings for all pairs (i,j)
for (int i=n-1; i>=0; i--) {
for (int j=i; j<n; j++) {
// if characters match, set value
// and compare with ansLen.
if (s[i]==s[j]) {
dp[j] = 1 + min(dp[j+1], j-i-1);
if (dp[j]>=ansLen) {
ansLen = dp[j];
ans = s.substr(i, ansLen);
}
}
else dp[j] = 0;
}
}
return ansLen>0?ans:"-1";
}
int main() {
string s = "geeksforgeeks";
cout << longestSubstring(s) << endl;
return 0;
}
// Java program to find longest repeating
// and non-overlapping substring
// using space optimised
class GfG {
static String longestSubstring(String s) {
int n = s.length();
int[] dp = new int[n + 1];
String ans = "";
int ansLen = 0;
// find length of non-overlapping
// substrings for all pairs (i, j)
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
// if characters match, set value
// and compare with ansLen.
if (s.charAt(i) == s.charAt(j)) {
dp[j] = 1 + Math.min(dp[j + 1], j - i - 1);
if (dp[j] >= ansLen) {
ansLen = dp[j];
ans = s.substring(i, i + ansLen);
}
} else {
dp[j] = 0;
}
}
}
return ansLen > 0 ? ans : "-1";
}
public static void main(String[] args) {
String s = "geeksforgeeks";
System.out.println(longestSubstring(s));
}
}
# Python program to find longest repeating
# and non-overlapping substring
# using space optimised
def longestSubstring(s):
n = len(s)
dp = [0] * (n + 1)
ans = ""
ansLen = 0
# find length of non-overlapping
# substrings for all pairs (i, j)
for i in range(n - 1, -1, -1):
for j in range(i, n):
# if characters match, set value
# and compare with ansLen.
if s[i] == s[j]:
dp[j] = 1 + min(dp[j + 1], j - i - 1)
if dp[j] >= ansLen:
ansLen = dp[j]
ans = s[i:i + ansLen]
else:
dp[j] = 0
return ans if ansLen > 0 else "-1"
if __name__ == "__main__":
s = "geeksforgeeks"
print(longestSubstring(s))
// C# program to find longest repeating
// and non-overlapping substring
// using space optimised
using System;
class GfG {
static string longestSubstring(string s) {
int n = s.Length;
int[] dp = new int[n + 1];
string ans = "";
int ansLen = 0;
// find length of non-overlapping
// substrings for all pairs (i, j)
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
// if characters match, set value
// and compare with ansLen.
if (s[i] == s[j]) {
dp[j] = 1 + Math.Min(dp[j + 1], j - i - 1);
if (dp[j] >= ansLen) {
ansLen = dp[j];
ans = s.Substring(i, ansLen);
}
} else {
dp[j] = 0;
}
}
}
return ansLen > 0 ? ans : "-1";
}
static void Main(string[] args) {
string s = "geeksforgeeks";
Console.WriteLine(longestSubstring(s));
}
}
// JavaScript program to find longest repeating
// and non-overlapping substring
// using space optimised
function longestSubstring(s) {
const n = s.length;
const dp = new Array(n + 1).fill(0);
let ans = "";
let ansLen = 0;
// find length of non-overlapping
// substrings for all pairs (i, j)
for (let i = n - 1; i >= 0; i--) {
for (let j = i; j < n; j++) {
// if characters match, set value
// and compare with ansLen.
if (s[i] === s[j]) {
dp[j] = 1 + Math.min(dp[j + 1], j - i - 1);
if (dp[j] >= ansLen) {
ansLen = dp[j];
ans = s.substring(i, i + ansLen);
}
} else {
dp[j] = 0;
}
}
}
return ansLen > 0 ? ans : "-1";
}
const s = "geeksforgeeks";
console.log(longestSubstring(s));
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Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
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Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
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Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
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Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
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Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
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Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
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Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
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Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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