# Longest Repeating Subsequence

• Difficulty Level : Medium
• Last Updated : 22 Jun, 2022

Given a string, find the length of the longest repeating subsequence, such that the two subsequences don’t have same string character at the same position, i.e. any ith character in the two subsequences shouldn’t have the same index in the original string. Examples:

```Input: str = "abc"
Output: 0
There is no repeating subsequence

Input: str = "aab"
Output: 1
The two subsequence are 'a'(first) and 'a'(second).
Note that 'b' cannot be considered as part of subsequence
as it would be at same index in both.

Input: str = "aabb"
Output: 2

Input: str = "axxxy"
Output: 2```
Recommended Practice

This problem is just the modification of Longest Common Subsequence problem. The idea is to find the LCS(str, str) where, str is the input string with the restriction that when both the characters are same, they shouldn’t be on the same index in the two strings.

Algorithm:

Step 1: Initialize the input string, which is to be checked.

Step 2: Initialize the length of string to the variable.

Step 3: Create a DP table using 2D matrix and set each element to 0.

Step 4: Fill the table if  characters are same and indexes are different.

Step 5: Return the values inside the table

Step 6: Print the String.

Below is the implementation of the idea.

## C++

 `// C++ program to find the longest repeating` `// subsequence` `#include ` `#include ` `using` `namespace` `std;`   `// This function mainly returns LCS(str, str)` `// with a condition that same characters at` `// same index are not considered. ` `int` `findLongestRepeatingSubSeq(string str)` `{` `    ``int` `n = str.length();`   `    ``// Create and initialize DP table` `    ``int` `dp[n+1][n+1];` `    ``for` `(``int` `i=0; i<=n; i++)` `        ``for` `(``int` `j=0; j<=n; j++)` `            ``dp[i][j] = 0;`   `    ``// Fill dp table (similar to LCS loops)` `    ``for` `(``int` `i=1; i<=n; i++)` `    ``{` `        ``for` `(``int` `j=1; j<=n; j++)` `        ``{` `            ``// If characters match and indexes are ` `            ``// not same` `            ``if` `(str[i-1] == str[j-1] && i != j)` `                ``dp[i][j] =  1 + dp[i-1][j-1];          ` `                     `  `            ``// If characters do not match` `            ``else` `                ``dp[i][j] = max(dp[i][j-1], dp[i-1][j]);` `        ``}` `    ``}` `    ``return` `dp[n][n];` `}`   `// Driver Program` `int` `main()` `{` `    ``string str = ``"aabb"``;` `    ``cout << ``"The length of the largest subsequence that"` `            ``" repeats itself is : "` `        ``<< findLongestRepeatingSubSeq(str);` `    ``return` `0;` `}`

## Java

 `// Java program to find the longest ` `// repeating subsequence` `import` `java.io.*;` `import` `java.util.*;`   `class` `LRS ` `{` `    ``// Function to find the longest repeating subsequence` `    ``static` `int` `findLongestRepeatingSubSeq(String str)` `    ``{` `        ``int` `n = str.length();` ` `  `        ``// Create and initialize DP table` `        ``int``[][] dp = ``new` `int``[n+``1``][n+``1``];` ` `  `        ``// Fill dp table (similar to LCS loops)` `        ``for` `(``int` `i=``1``; i<=n; i++)` `        ``{` `            ``for` `(``int` `j=``1``; j<=n; j++)` `            ``{` `                ``// If characters match and indexes are not same` `                ``if` `(str.charAt(i-``1``) == str.charAt(j-``1``) && i!=j)` `                    ``dp[i][j] =  ``1` `+ dp[i-``1``][j-``1``];          ` `                      `  `                ``// If characters do not match` `                ``else` `                    ``dp[i][j] = Math.max(dp[i][j-``1``], dp[i-``1``][j]);` `            ``}` `        ``}` `        ``return` `dp[n][n];` `    ``}` `    `  `    ``// driver program to check above function` `    ``public` `static` `void` `main (String[] args) ` `    ``{` `        ``String str = ``"aabb"``;` `        ``System.out.println(``"The length of the largest subsequence that"` `            ``+``" repeats itself is : "``+findLongestRepeatingSubSeq(str));` `    ``}` `}`   `// This code is contributed by Pramod Kumar`

## Python3

 `# Python 3 program to find the longest repeating ` `# subsequence `     `# This function mainly returns LCS(str, str) ` `# with a condition that same characters at ` `# same index are not considered. ` `def` `findLongestRepeatingSubSeq( ``str``): `   `    ``n ``=` `len``(``str``) `   `    ``# Create and initialize DP table ` `    ``dp``=``[[``0` `for` `i ``in` `range``(n``+``1``)] ``for` `j ``in` `range``(n``+``1``)]`   `    ``# Fill dp table (similar to LCS loops) ` `    ``for` `i ``in` `range``(``1``,n``+``1``):` `        ``for` `j ``in` `range``(``1``,n``+``1``):` `            ``# If characters match and indexes are ` `            ``# not same ` `            ``if` `(``str``[i``-``1``] ``=``=` `str``[j``-``1``] ``and` `i !``=` `j): ` `                ``dp[i][j] ``=` `1` `+` `dp[i``-``1``][j``-``1``]         ` `                        `  `            ``# If characters do not match ` `            ``else``:` `                ``dp[i][j] ``=` `max``(dp[i][j``-``1``], dp[i``-``1``][j]) ` `        `  `    `  `    ``return` `dp[n][n] `     `# Driver Program ` `if` `__name__``=``=``'__main__'``:` `    ``str` `=` `"aabb"` `    ``print``(``"The length of the largest subsequence that repeats itself is : "` `          ``,findLongestRepeatingSubSeq(``str``))`   `# this code is contributed by ash264`

## C#

 `// C# program to find the longest repeating ` `// subsequence` `using` `System;`   `class` `GFG {` `    `  `    ``// Function to find the longest repeating` `    ``// subsequence` `    ``static` `int` `findLongestRepeatingSubSeq(``string` `str)` `    ``{` `        ``int` `n = str.Length;`   `        ``// Create and initialize DP table` `        ``int` `[,]dp = ``new` `int``[n+1,n+1];`   `        ``// Fill dp table (similar to LCS loops)` `        ``for` `(``int` `i = 1; i <= n; i++)` `        ``{` `            ``for` `(``int` `j = 1; j <= n; j++)` `            ``{` `                `  `                ``// If characters match and indexes` `                ``// are not same` `                ``if` `(str[i-1] == str[j-1] && i != j)` `                    ``dp[i,j] = 1 + dp[i-1,j-1];         ` `                        `  `                ``// If characters do not match` `                ``else` `                    ``dp[i,j] = Math.Max(dp[i,j-1], ` `                                       ``dp[i-1,j]);` `            ``}` `        ``}` `        ``return` `dp[n,n];` `    ``}` `    `  `    ``// driver program to check above function` `    ``public` `static` `void` `Main () ` `    ``{` `        ``string` `str = ``"aabb"``;` `        ``Console.Write(``"The length of the largest "` `         ``+ ``"subsequence that repeats itself is : "` `               ``+ findLongestRepeatingSubSeq(str));` `    ``}` `}`   `// This code is contributed by nitin mittal.`

## PHP

 ``

## Javascript

 ``

Output:

`The length of the largest subsequence that repeats itself is : 2`

Time Complexity: O(n2)

Auxiliary Space: O(n2)

Another approach: (Using recursion)

## C++

 `// C++ program to find the longest repeating` `// subsequence using recursion` `#include ` `using` `namespace` `std;`   `int` `dp;`   `// This function mainly returns LCS(str, str) ` `// with a condition that same characters at ` `// same index are not considered. `   `int` `findLongestRepeatingSubSeq(string X, ``int` `m, ``int` `n)` `{` `    `  `    ``if``(dp[m][n]!=-1)` `    ``return` `dp[m][n];` `    `  `    ``// return if we have reached the end of either string` `    ``if` `(m == 0 || n == 0)` `        ``return` `dp[m][n] = 0;`   `    ``// if characters at index m and n matches ` `    ``// and index is different` `    ``if` `(X[m - 1] == X[n - 1] && m != n)` `        ``return` `dp[m][n] = findLongestRepeatingSubSeq(X, ` `                            ``m - 1, n - 1) + 1;`   `    ``// else if characters at index m and n don't match` `    ``return` `dp[m][n] = max (findLongestRepeatingSubSeq(X, m, n - 1), ` `                           ``findLongestRepeatingSubSeq(X, m - 1, n));` `}`   `// Longest Repeated Subsequence Problem` `int` `main()` `{` `    ``string str = ``"aabb"``;` `    ``int` `m = str.length();`   `memset``(dp,-1,``sizeof``(dp));` `cout << ``"The length of the largest subsequence that"` `            ``" repeats itself is : "` `        ``<< findLongestRepeatingSubSeq(str,m,m); `   `    ``return` `0;` `// this code is contributed by Kushdeep Mittal` `}`

## Java

 `import` `java.util.Arrays;`   `// Java program to find the longest repeating` `// subsequence using recursion` `public` `class` `GFG {`   `    ``static` `int` `dp[][] = ``new` `int``[``1000``][``1000``];`   `// This function mainly returns LCS(str, str) ` `// with a condition that same characters at ` `// same index are not considered. ` `    ``static` `int` `findLongestRepeatingSubSeq(``char` `X[], ``int` `m, ``int` `n) {`   `        ``if` `(dp[m][n] != -``1``) {` `            ``return` `dp[m][n];` `        ``}`   `        ``// return if we have reached the end of either string` `        ``if` `(m == ``0` `|| n == ``0``) {` `            ``return` `dp[m][n] = ``0``;` `        ``}`   `        ``// if characters at index m and n matches ` `        ``// and index is different` `        ``if` `(X[m - ``1``] == X[n - ``1``] && m != n) {` `            ``return` `dp[m][n] = findLongestRepeatingSubSeq(X,` `                    ``m - ``1``, n - ``1``) + ``1``;` `        ``}`   `        ``// else if characters at index m and n don't match` `        ``return` `dp[m][n] = Math.max(findLongestRepeatingSubSeq(X, m, n - ``1``),` `                ``findLongestRepeatingSubSeq(X, m - ``1``, n));` `    ``}`   `// Longest Repeated Subsequence Problem` `    ``static` `public` `void` `main(String[] args) {` `        ``String str = ``"aabb"``;` `        ``int` `m = str.length();` `        ``for` `(``int``[] row : dp) {` `            ``Arrays.fill(row, -``1``);` `        ``}` `        ``System.out.println(``"The length of the largest subsequence that"` `                ``+ ``" repeats itself is : "` `                ``+ findLongestRepeatingSubSeq(str.toCharArray(), m, m));`   `    ``}` `}`   `// This code is contributed by 29AjayKumar`

## Python3

 `# Python 3 program to find the longest repeating` `# subsequence using recursion`   `dp ``=` `[[``0` `for` `i ``in` `range``(``1000``)] ``for` `j ``in` `range``(``1000``)]`   `# This function mainly returns LCS(str, str) ` `# with a condition that same characters at ` `# same index are not considered. `   `def` `findLongestRepeatingSubSeq( X, m, n):` `    `  `    ``if``(dp[m][n]!``=``-``1``):` `        ``return` `dp[m][n]` `    `  `    ``# return if we have reached the end of either string` `    ``if` `(m ``=``=` `0` `or` `n ``=``=` `0``):` `        ``dp[m][n] ``=` `0` `        ``return` `dp[m][n]`   `    ``# if characters at index m and n matches ` `    ``# and index is different` `    ``if` `(X[m ``-` `1``] ``=``=` `X[n ``-` `1``] ``and` `m !``=` `n):` `        ``dp[m][n] ``=` `findLongestRepeatingSubSeq(X, ` `                            ``m ``-` `1``, n ``-` `1``) ``+` `1` `        `  `        ``return` `dp[m][n]`   `    ``# else if characters at index m and n don't match` `    ``dp[m][n] ``=` `max` `(findLongestRepeatingSubSeq(X, m, n ``-` `1``), ` `                        ``findLongestRepeatingSubSeq(X, m ``-` `1``, n))` `    ``return` `dp[m][n]`   `# Longest Repeated Subsequence Problem` `if` `__name__ ``=``=` `"__main__"``:` `    ``str` `=` `"aabb"` `    ``m ``=` `len``(``str``)`   `dp ``=``[[``-``1` `for` `i ``in` `range``(``1000``)] ``for` `j ``in` `range``(``1000``)]` `print``( ``"The length of the largest subsequence that"` `            ``" repeats itself is : "` `        ``, findLongestRepeatingSubSeq(``str``,m,m))` `        `  `# this code is contributed by` `# ChitraNayal`

## C#

 `//C# program to find the longest repeating ` `// subsequence using recursion ` `using` `System;` `public` `class` `GFG { `   `    ``static` `int` `[,]dp = ``new` `int``[1000,1000]; `   `// This function mainly returns LCS(str, str) ` `// with a condition that same characters at ` `// same index are not considered. ` `    ``static` `int` `findLongestRepeatingSubSeq(``char` `[]X, ``int` `m, ``int` `n) { `   `        ``if` `(dp[m,n] != -1) { ` `            ``return` `dp[m,n]; ` `        ``} `   `        ``// return if we have reached the end of either string ` `        ``if` `(m == 0 || n == 0) { ` `            ``return` `dp[m,n] = 0; ` `        ``} `   `        ``// if characters at index m and n matches ` `        ``// and index is different ` `        ``if` `(X[m - 1] == X[n - 1] && m != n) { ` `            ``return` `dp[m,n] = findLongestRepeatingSubSeq(X, ` `                    ``m - 1, n - 1) + 1; ` `        ``} `   `        ``// else if characters at index m and n don't match ` `        ``return` `dp[m,n] = Math.Max(findLongestRepeatingSubSeq(X, m, n - 1), ` `                ``findLongestRepeatingSubSeq(X, m - 1, n)); ` `    ``} `   `// Longest Repeated Subsequence Problem ` `    ``static` `public` `void` `Main() { ` `        ``String str = ``"aabb"``; ` `        ``int` `m = str.Length; ` `        ``for` `(``int` `i = 0; i < dp.GetLength(0); i++)` `            ``for` `(``int` `j = 0; j < dp.GetLength(1); j++)` `                ``dp[i, j] = -1;` `        ``Console.WriteLine(``"The length of the largest subsequence that"` `                ``+ ``" repeats itself is : "` `                ``+ findLongestRepeatingSubSeq(str.ToCharArray(), m, m)); `   `    ``}` `} `   `// This code is contributed by 29AjayKumar `

## PHP

 ``

## Javascript

 ``

Output:

`The length of the largest subsequence that repeats itself is : 2`

Approach 3:

To find the length of the Longest Repeating Subsequence  dynamic  programming Top-down Approach:

• Take the input string.
• Perform the Longest common subsequence where s1[i]==s1[j] and i!=j.
• Return the length.

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `int` `lrs(string s1,``int` `i,``int` `j, vector>&dp){`   `    ``// return if we have reached the` `    ``//end of either string` `    ``if``(i >= s1.length() || j >= s1.length())` `        ``return` `0;`   `    ``if``(dp[i][j] != -1)` `        ``return` `dp[i][j];` `    `  `    ``// while dp[i][j] is not computed earlier` `    ``if``(dp[i][j] == -1){` `    `  `        ``// if characters at index m and n matches` `        ``// and index is different` `        ``// Index should not match` `        ``if``(s1[i] == s1[j] && i != j)` `            ``dp[i][j] = 1+lrs(s1, i+1, j+1, dp);` `        `  `        ``// else if characters at index m and n don't match` `        ``else` `            ``dp[i][j] = max(lrs(s1, i, j+1, dp),` `                                ``lrs(s1, i+1, j, dp));` `    ``}` `    `  `    ``// return answer` `    ``return` `dp[i][j];` `}`   `// Driver Code` `int` `main(){`   `string s1 = ``"aabb"``;` `    `  `// Reversing the same string` `string s2 = s1;` `reverse(s2.begin(),s2.end());` `vector>dp(1000,vector<``int``>(1000,-1));` `cout<<``"LENGTH OF LONGEST REPEATING SUBSEQUENCE IS : "``<

## Java

 `import` `java.lang.*;` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG ` `{    ` `  ``static` `int` `lrs(StringBuilder s1, ``int` `i, ``int` `j, ``int``[][] dp)` `  ``{` `    ``if``(i >= s1.length() || j >= s1.length())` `    ``{` `      ``return` `0``;` `    ``}`   `    ``if``(dp[i][j] != -``1``)` `    ``{` `      ``return` `dp[i][j];` `    ``}`   `    ``if``(dp[i][j] == -``1``)` `    ``{` `      ``if``(s1.charAt(i) == s1.charAt(j) && i != j)` `      ``{` `        ``dp[i][j] = ``1` `+ lrs(s1, i + ``1``, j + ``1``, dp);` `      ``}` `      ``else` `      ``{` `        ``dp[i][j] = Math.max(lrs(s1, i, j + ``1``, dp), lrs(s1, i + ``1``, j, dp));` `      ``}` `    ``}` `    ``return` `dp[i][j];`   `  ``}`   `  ``// Driver code` `  ``public` `static` `void` `main (String[] args) ` `  ``{    ` `    ``String s1 = ``"aabb"``;   ` `    ``StringBuilder input1 = ``new` `StringBuilder();`   `    ``// append a string into StringBuilder input1` `    ``input1.append(s1);`   `    ``// reverse StringBuilder input1` `    ``input1.reverse();` `    ``int``[][] dp = ``new` `int``[``1000``][``1000``];` `    ``for``(``int``[] row : dp)` `    ``{` `      ``Arrays.fill(row, -``1``);` `    ``}` `    ``System.out.println(``"LENGTH OF LONGEST REPEATING SUBSEQUENCE IS :"` `+` `                       ``lrs(input1, ``0``, ``0``, dp));` `  ``}` `}`   `// This code is contributed by rag2127.`

## Python3

 `# Python 3 program to find the longest repeating` `# subsequence Length`   `# This function mainly returns LRS(str, str,i,j,dp)` `# with a condition that same characters at` `# same index are not considered.` `def` `lrs(s1, i, j, dp):` `  `  `    ``# return if we have reached the ` `    ``#end of either string` `    ``if` `i >``=` `len``(s1) ``or` `j >``=` `len``(s1):` `        ``return` `0` `  `  `    ``if` `dp[i][j] !``=` `-``1``:` `        ``return` `dp[i][j]` `      `  `    ``# while dp[i][j] is not computed earlier` `    ``if` `dp[i][j] ``=``=` `-``1``: ` `      `  `        ``# if characters at index m and n matches` `        ``# and index is different` `        ``# Index should not match` `        ``if` `s1[i] ``=``=` `s1[j] ``and` `i !``=` `j: ` `            ``dp[i][j] ``=` `1``+``lrs(s1, i``+``1``, j``+``1``, dp)` `        `  `        ``# else if characters at index m and n don't match` `        ``else``:  ` `            ``dp[i][j] ``=` `max``(lrs(s1, i, j``+``1``, dp), ` `                                ``lrs(s1, i``+``1``, j, dp))` `    `  `    ``# return answer` `    ``return` `dp[i][j]`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``s1 ``=` `"aabb"` `    `  `    ``# Reversing the same string` `    ``s2 ``=` `s1[::``-``1``]  ` `    ``dp ``=``[[``-``1` `for` `i ``in` `range``(``1000``)] ``for` `j ``in` `range``(``1000``)]` `    ``print``(``"LENGTH OF LONGEST REPEATING SUBSEQUENCE IS :"``, ` `                                    ``lrs(s1, ``0``, ``0``, dp))` `    `  `# this code is contributed by saikumar kudikala`

## C#

 `using` `System;`   `public` `class` `GFG{`   `  ``static` `int` `lrs(``string` `s1, ``int` `i, ``int` `j, ``int``[,] dp)` `  ``{` `    ``if``(i >= s1.Length || j >= s1.Length)` `    ``{` `      ``return` `0;` `    ``}`   `    ``if``(dp[i, j] != -1)` `    ``{` `      ``return` `dp[i, j];` `    ``}`   `    ``if``(dp[i, j] == -1)` `    ``{` `      ``if``(s1[i] == s1[j] && i != j)` `      ``{` `        ``dp[i, j] = 1 + lrs(s1, i + 1, j + 1, dp);` `      ``}` `      ``else` `      ``{` `        ``dp[i, j] = Math.Max(lrs(s1, i, j + 1, dp), lrs(s1, i + 1, j, dp));` `      ``}` `    ``}` `    ``return` `dp[i, j];`   `  ``}`   `  ``// Driver code` `  ``static` `public` `void` `Main (){` `    ``string` `s1 = ``"aabb"``;` `    ``char``[] chars = s1.ToCharArray();` `    ``Array.Reverse(chars);` `    ``s1= ``new` `String(chars);`   `    ``int``[,] dp = ``new` `int``[1000,1000];` `    ``for``(``int` `i = 0; i < 1000; i++)` `    ``{` `      ``for``(``int` `j = 0; j < 1000; j++)` `      ``{` `        ``dp[i, j] = -1;` `      ``}` `    ``}` `    ``Console.WriteLine(``"LENGTH OF LONGEST REPEATING SUBSEQUENCE IS :"` `+` `                      ``lrs(s1, 0, 0, dp));` `  ``}` `}`   `// This code is contributed by avanitrachhadiya2155`

## Javascript

 ``

Output

`LENGTH OF LONGEST REPEATING SUBSEQUENCE IS : 2`

Time Complexity: O(n2)

Auxiliary Space: O(n2)

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