Longest Repeating Subsequence

Difficulty Level : Medium
Last Updated : 17 Feb, 2023

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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 i^th character in the two subsequences shouldn’t have the same index in the original string.

longest-repeating-subsequence

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

Method 1: 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.

Initialize the input string, which is to be checked.
Initialize the length of string to the variable.
create a DP table using 2D matrix and set each element to 0.
Fill the table if the characters are same and indexes are different.
Return the values inside the table
Print the String.

Below is the implementation of the idea.

C++

// C++ program to find the longest repeating
// subsequence
#include <iostream>
#include <string>
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

<?php
// PHP 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. 
function findLongestRepeatingSubSeq($str)
{
    $n = strlen($str);
 
    // Create and initialize
    // DP table
    $dp = array(array());
    for ($i = 0; $i <= $n; $i++)
        for ($j = 0; $j <= $n; $j++)
            $dp[$i][$j] = 0;
 
    // Fill dp table 
    // (similar to LCS loops)
    for ($i = 1; $i <= $n; $i++)
    {
        for ($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 Code
$str = "aabb";
echo "The length of the largest ". 
     "subsequence that repeats itself is : ",
            findLongestRepeatingSubSeq($str);
 
// This code is contributed
// by shiv_bhakt.
?>

Javascript

<script>
    // Javascript 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. 
    function findLongestRepeatingSubSeq(str)
    {
        var n = str.length;
      
        // Create and initialize DP table
        var dp = new Array(n + 1);
         
        for (var i=0; i<=n; i++)
        {
            dp[i] = new Array(n + 1);
            for (var j=0; j<=n; j++)
            {
                dp[i][j] = 0;
            }
        }
             
        // Fill dp table (similar to LCS loops)
        for (var i=1; i<=n; i++)
        {
            for (var 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 Code
     
    var str = "aabb";
    document.write("The length of the largest subsequence that repeats itself is : " + findLongestRepeatingSubSeq(str));
 
</script>

Output

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

Time Complexity: O(n²)
Auxiliary Space: O(n²)

Method 2: (Using recursion)

C++

// C++ program to find the longest repeating
// subsequence using recursion
#include <bits/stdc++.h>
using namespace std;
 
int dp[1000][1000];
 
// 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

<?php
// PHP program to find the longest repeating
// subsequence using recursion
 
$dp = array_fill(0, 1000, array_fill(0, 1000, -1));
 
// This function mainly returns LCS(str, str) 
// with a condition that same characters at 
// same index are not considered. 
 
function findLongestRepeatingSubSeq($X, $m, $n)
{
    global $dp;
     
    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));
}
 
// Driver code
 
    $str = "aabb";
    $m = strlen($str);
 
    echo "The length of the largest subsequence". 
    "that repeats itself is : ".findLongestRepeatingSubSeq($str,$m,$m); 
 
 
// this code is contributed by mits
?>

Javascript

<script>
 
let dp=new Array(1000);
 
for(let i=0;i<1000;i++)
{
    dp[i]=new Array(1000);
    for(let j=0;j<1000;j++)
    {
        dp[i][j]=-1;
    }
     
}
 
function 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 || 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));
}
 
let str = "aabb";
let m = str.length;
 
document.write("The length of the largest subsequence that"
                   + " repeats itself is : "
                   + findLongestRepeatingSubSeq(str.split(""), m, m));
 
 
// This code is contributed by ab2127
</script>

Output

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

Time Complexity: O(m*n)
Auxiliary Space: O(m*n)

Method 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 <bits/stdc++.h>
using namespace std;
 
int lrs(string s1,int i,int j, vector<vector<int>>&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<vector<int>>dp(1000,vector<int>(1000,-1));
cout<<"LENGTH OF LONGEST REPEATING SUBSEQUENCE IS : "<<lrs(s1, 0, 0, dp);
 
}
 
// This code is contributed by shinjanpatra

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

<script>
 
function lrs(s1, i, j, 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
let  s1 = "aabb";
 
// Append a string into StringBuilder input1
let input1 = s1.split("");
 
// Reverse StringBuilder input1
input1.reverse();
let dp = new Array(1000);
for(let i = 0; i < 1000; i++)
{
    dp[i] = new Array(1000);
    for(let j = 0; j < 1000; j++)
    {
        dp[i][j] = -1;
    }
}
 
document.write("LENGTH OF LONGEST REPEATING " + 
               "SUBSEQUENCE IS :" +
               lrs(input1, 0, 0, dp));
                
// This code is contributed by unknown2108
 
</script>

Output

LENGTH OF LONGEST REPEATING SUBSEQUENCE IS : 2

Time Complexity: O(n²)
Auxiliary Space: O(n²)

Method 4: If we look closely at solution 1, we can analyze that we are using only the previous column and element just above the current element.

C++

#include <bits/stdc++.h>
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] = {0};
 
    // Fill dp table (similar to LCS loops)
    for (int i=1; i<=n; i++)
    {
        int new_a[n+1] = {0};
        for (int j=1; j<=n; j++)
        {
            // If characters match and indexes are
            // not same
            if (str[i-1] == str[j-1] && i != j)
            {
                new_a[j] = 1 + dp[j-1];
            }
 
            // If characters do not match
            else
            {
                new_a[j] = max(dp[j], new_a[j-1]);
            }
        }
        for (int j=0; j<=n; j++)
            dp[j] = new_a[j];
    }
    return dp[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 Longest Repeating
// Subsequence
import java.util.*;
 
class GFG {
 
  // This function mainly returns LCS(str, str)
  // with a condition that same characters at
  // same index are not considered.
  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
  public static void main(String[] args)
  {
    String str = "aabb";
    System.out.println("The length of the largest subsequence that "
                       + "repeats itself is : "+
                       findLongestRepeatingSubSeq(str));
  }
}

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)]
 
    # Fill dp table (similar to LCS loops)
    for i in range(1, n + 1):
        new_a = [0]
        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:
                new_a.append(1 + dp[j - 1])
 
                # If characters do not match
            else:
                new_a.append(max(dp[j], new_a[-1]))
        dp = new_a[:]
    return dp[-1]
 
 
# 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#

using System;
 
namespace findLongestRepeatingSubSeq {
class GFG {
    static int findLongestRepeatingSubSeq(string str)
    {
        int n = str.Length;
 
        // Create and initialize DP table
        int[] dp = new int[n + 1];
 
        // Fill dp table (similar to LCS loops)
        for (int i = 1; i <= n; i++) {
            int[] new_a = new int[n + 1];
            for (int j = 1; j <= n; j++) {
                // If characters match and indexes are
                // not same
                if (str[i - 1] == str[j - 1] && i != j) {
                    new_a[j] = 1 + dp[j - 1];
                }
                // If characters do not match
                else {
                    new_a[j]
                        = Math.Max(dp[j], new_a[j - 1]);
                }
            }
            for (int j = 0; j <= n; j++)
                dp[j] = new_a[j];
        }
        return dp[n];
    }
 
    // Driver Program
    static void Main(string[] args)
    {
        string str = "aabb";
        Console.WriteLine(
            "The length of the largest subsequence that"
            + " repeats itself is : "
            + findLongestRepeatingSubSeq(str));
    }
}
}
// This code is contributed by Susobhan Akhuli

Javascript

// This function mainly returns LCS(str, str)
// with a condition that same characters at
// same index are not considered.
function findLongestRepeatingSubSeq( str)
{
    let n = str.length;
 
    // Create and initialize DP table
    let dp=new Array(n+1).fill(0);
 
    // Fill dp table (similar to LCS loops)
    for (let i=1; i<=n; i++)
    {
        let new_a=new Array(n+1).fill(0);
        for (let j=1; j<=n; j++)
        {
            // If characters match and indexes are
            // not same
            if (str[i-1] == str[j-1] && i != j)
            {
                new_a[j] = 1 + dp[j-1];
            }
 
            // If characters do not match
            else
            {
                new_a[j] = Math.max(dp[j], new_a[j-1]);
            }
        }
        for (let j=0; j<=n; j++)
            dp[j] = new_a[j];
    }
    return dp[n];
}
 
// Driver Program
let str = "aabb";
console.log("The length of the largest subsequence that"
     + " repeats itself is : " + findLongestRepeatingSubSeq(str));

Output

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

Time Complexity: O(n²)
Auxiliary Space: O(n)

This article is contributed by Ekta Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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