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
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 Programint 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 subsequenceimport 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 // subsequenceusing 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 Problemint 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 recursionpublic 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 recursiondp = [[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 Problemif __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 Codeint main(){string s1 = "aabb"; // Reversing the same stringstring 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 Codeif __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 codelet s1 = "aabb";// Append a string into StringBuilder input1let input1 = s1.split("");// Reverse StringBuilder input1input1.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 Programint 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// Subsequenceimport 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 Programif __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 Programlet 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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