Related Articles
Longest Repeating Subsequence
• Difficulty Level : Medium
• Last Updated : 29 Sep, 2020

Given a string, find length of the longest repeating subseequence such that the two subsequence don’t have same string character at same position, i.e., any i’th 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 subssequence 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

```

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.
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`

## Python 3

 `# 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``]``*``(n``+``1``)]``*``(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

 ``

Output:

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

```

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`

## Python 3

 `# 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

 ``

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:

Step1: Take the input string.

Step2: Reverse the input strings.

Step3: Perform Longest common subsequence for s1 and s2 where s1[i]==s2[j] and i!=j.

Step4: Return the length.

## 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, s2, i, j, dp):` `  `  `    ``# return if we have reached the ` `    ``#end of either string` `    ``if` `i >``=` `len``(s1) ``or` `j >``=` `len``(s2):` `        ``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] ``=``=` `s2[j] ``and` `i !``=` `j: ` `            ``dp[i][j] ``=` `1``+``lrs(s1, s2, i``+``1``, j``+``1``, dp)` `        `  `        ``# else if characters at index m and n don't match` `        ``else``:  ` `            ``dp[i][j] ``=` `max``(lrs(s1, s2, i, j``+``1``, dp), ` `                                ``lrs(s1, s2, i``+``1``, j, dp))` `    `  `    ``# return answer` `    ``return` `dp[i][j]`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``s1 ``=` `"abacabaaac"` `    `  `    ``# Reversing the same string` `    ``s2 ``=` `s1[::``-``1``]  ` `    ``dp ``=` `[[``-``1` `for` `_ ``in` `range``(``len``(s1)``+``1``)]` `                          ``for` `_ ``in` `range``(``len``(s2)``+``1``)]` `    ``print``(``"LENGTH OF LONGEST REPEATING SUBSEQUENCE IS :"``, ` `                                    ``lrs(s1, s2, ``0``, ``0``, dp))` `    `  `# this code is contributed by saikumar kudikala`

Output

```LENGTH OF LONGEST REPEATING SUBSEQUENCE IS : 7

```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :