Longest Repeated Subsequence
Last Updated :
15 Nov, 2024
Given a string s, the task is to find the longest repeating subsequence, such that the two subsequences don’t have the 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: s = "aabb"
Output: "ab"
Explanation: The longest repeated subsequence is "ab", formed by matching 'a' from positions 1 and 2, and 'b' from positions 3 and 4.
Input: s = "aab"
Output: "a"
Explanation: The two subsequences are 'a'(first) and 'a' (second). Note that 'b' cannot be considered as part of a subsequence as it would be at the same index in both.
This problem is just the modification of the Longest Common Subsequence problem. The idea is to find the LCS(s, s) where s is the input string with the restriction that when both the characters are the same, they shouldn't be on the same index in the two strings. We have discussed a solution to find the length of the longest repeated subsequence.
C++
// C++ program to find the longest repeating
// subsequence
#include <iostream>
#include <string>
using namespace std;
int longestRepeatingSubsequence(string s) {
int n = s.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 (s[i - 1] == s[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];
}
int main() {
string s = "aabb";
int res = longestRepeatingSubsequence(s);
cout << res << endl;
return 0;
}
// Java program to find the longest
// repeating subsequence
import java.io.*;
import java.util.*;
class GfG {
// Function to find the longest repeating subsequence
static int longestRepeatingSubsequence(String s) {
int n = s.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 (s.charAt(i - 1) == s.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];
}
public static void main(String[] args) {
String s = "aabb";
int res = longestRepeatingSubsequence(s);
System.out.print(res);
}
}
# Python 3 program to find the longest repeating
# subsequence
def longestRepeatingSubsequence(s):
n = len(s)
# 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 (s[i - 1] == s[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]
if __name__ == '__main__':
s = "aabb"
print(longestRepeatingSubsequence(s))
// C# program to find the longest repeating
// subsequence
using System;
class GfG {
// Function to find the longest repeating
// subsequence
static int longestRepeatingSubsequence(string s) {
int n = s.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 (s[i - 1] == s[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];
}
static void Main() {
string s = "aabb";
int res = longestRepeatingSubsequence(s);
Console.Write(res);
}
}
// Javascript program to find the longest repeating
// subsequence
function longestRepeatingSubsequence(s) {
var n = s.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 ((s[i - 1] == s[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];
}
var s = "aabb";
let res = longestRepeatingSubsequence(s);
console.log(res);
Time Complexity: O(n*n), where n is the length of string s.
Auxilairy Space: O(n*n)
How to print the subsequence?
The above solution only finds length of subsequence. We can print the subsequence using dp[n+1][n+1] table The idea is similar to printing LCS.
C++
// C++ program to print the longest repeated
// subsequence
#include <bits/stdc++.h>
using namespace std;
string longestRepeatedSubSeq(string s) {
// THIS PART OF CODE IS
// FILLS dp[][]
int n = s.length();
int dp[n + 1][n + 1];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= n; j++) {
dp[i][j] = 0;
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (s[i - 1] == s[j - 1] && i != j) {
dp[i][j] = 1 + dp[i - 1][j - 1];
}
else {
dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);
}
}
}
// THIS PART OF CODE FINDS THE RESULT STRING USING DP[][]
// Initialize result
string res = "";
// Traverse dp[][] from bottom right
int i = n, j = n;
while (i > 0 && j > 0) {
// If this cell is same as diagonally
// adjacent cell just above it, then
// same characters are present at
// str[i-1] and str[j-1]. Append any
// of them to result.
if (dp[i][j] == dp[i - 1][j - 1] + 1) {
res = res + s[i - 1];
i--;
j--;
}
// Otherwise we move to the side
// that gave us maximum result
else if (dp[i][j] == dp[i - 1][j]) {
i--;
}
else {
j--;
}
}
// Since we traverse dp[][] from bottom,
// we get result in reverse order.
reverse(res.begin(), res.end());
return res;
}
int main() {
string s = "AABEBCDD";
cout << longestRepeatedSubSeq(s);
return 0;
}
// Java program to print the longest repeated
// subsequence
import java.util.*;
class GfG {
static String longestRepeatedSubSeq(String s) {
// THIS PART OF CODE
// FILLS dp[][]
int n = s.length();
int[][] dp = new int[n + 1][n + 1];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= n; j++) {
dp[i][j] = 0;
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (s.charAt(i - 1) == s.charAt(j - 1)
&& i != j) {
dp[i][j] = 1 + dp[i - 1][j - 1];
}
else {
dp[i][j] = Math.max(dp[i][j - 1],
dp[i - 1][j]);
}
}
}
// THIS PART OF CODE FINDS
// THE RESULT STRING USING DP[][]
// Initialize result
String res = "";
// Traverse dp[][] from bottom right
int i = n, j = n;
while (i > 0 && j > 0) {
// If this cell is same as diagonally
// adjacent cell just above it, then
// same characters are present at
// str[i-1] and str[j-1]. Append any
// of them to result.
if (dp[i][j] == dp[i - 1][j - 1] + 1) {
res = res + s.charAt(i - 1);
i--;
j--;
}
// Otherwise we move to the side
// that gave us maximum result
else if (dp[i][j] == dp[i - 1][j]) {
i--;
}
else {
j--;
}
}
// Since we traverse dp[][] from bottom,
// we get result in reverse order.
String reverse = "";
for (int k = res.length() - 1; k >= 0; k--) {
reverse = reverse + res.charAt(k);
}
return reverse;
}
public static void main(String args[]) {
String str = "AABEBCDD";
System.out.println(longestRepeatedSubSeq(str));
}
}
# Python3 program to print the
# longest repeated subsequence
def longestRepeatedSubSeq(s):
# This part of code
# fills dp[][]
n = len(s)
dp = [[0 for i in range(n + 1)] for j in range(n + 1)]
for i in range(1, n + 1):
for j in range(1, n + 1):
if (s[i - 1] == s[j - 1] and i != j):
dp[i][j] = 1 + dp[i - 1][j - 1]
else:
dp[i][j] = max(dp[i][j - 1], dp[i - 1][j])
# This part of code finds the result
# string using dp[][] Initialize result
res = ''
# Traverse dp[][] from bottom right
i = n
j = n
while (i > 0 and j > 0):
# If this cell is same as diagonally
# adjacent cell just above it, then
# same characters are present at
# str[i-1] and str[j-1]. Append any
# of them to result.
if (dp[i][j] == dp[i - 1][j - 1] + 1):
res += s[i - 1]
i -= 1
j -= 1
# Otherwise we move to the side
# that gave us maximum result.
elif (dp[i][j] == dp[i - 1][j]):
i -= 1
else:
j -= 1
# Since we traverse dp[][] from bottom,
# we get result in reverse order.
res = ''.join(reversed(res))
return res
s = 'AABEBCDD'
print(longestRepeatedSubSeq(s))
// C# program to print the longest repeated
// subsequence
using System;
using System.Collections.Generic;
class GfG {
static String longestRepeatedSubSeq(String s) {
// THIS PART OF CODE
// FILLS dp[,]
int n = s.Length, i, j;
int[, ] dp = new int[n + 1, n + 1];
for (i = 0; i <= n; i++) {
for (j = 0; j <= n; j++) {
dp[i, j] = 0;
}
}
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if (s[i - 1] == s[j - 1] && i != j) {
dp[i, j] = 1 + dp[i - 1, j - 1];
}
else {
dp[i, j] = Math.Max(dp[i, j - 1],
dp[i - 1, j]);
}
}
}
// THIS PART OF CODE FINDS
// THE RESULT STRING USING DP[,]
// Initialize result
String res = "";
// Traverse dp[,] from bottom right
i = n;
j = n;
while (i > 0 && j > 0) {
// If this cell is same as diagonally
// adjacent cell just above it, then
// same characters are present at
// str[i-1] and str[j-1]. Append any
// of them to result.
if (dp[i, j] == dp[i - 1, j - 1] + 1) {
res = res + s[i - 1];
i--;
j--;
}
// Otherwise we move to the side
// that gave us maximum result
else if (dp[i, j] == dp[i - 1, j])
i--;
else
j--;
}
// Since we traverse dp[,] from bottom,
// we get result in reverse order.
String reverse = "";
for (int k = res.Length - 1; k >= 0; k--) {
reverse = reverse + res[k];
}
return reverse;
}
static void Main(String[] args) {
String s = "AABEBCDD";
Console.WriteLine(longestRepeatedSubSeq(s));
}
}
// Javascript program to print the longest repeated
// subsequence
function longestRepeatedSubSeq(s) {
// THIS PART OF CODE
// FILLS dp[][]
let n = s.length;
let dp = new Array(n + 1);
for (let i = 0; i <= n; i++) {
dp[i] = new Array(n + 1);
for (let j = 0; j <= n; j++) {
dp[i][j] = 0;
}
}
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= n; j++) {
if (s[i - 1] == s[j - 1] && i != j) {
dp[i][j] = 1 + dp[i - 1][j - 1];
} else {
dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
}
}
}
// THIS PART OF CODE FINDS
// THE RESULT STRING USING DP[][]
// Initialize result
let res = "";
// Traverse dp[][] from bottom right
let i = n, j = n;
while (i > 0 && j > 0) {
// If this cell is same as diagonally
// adjacent cell just above it, then
// same characters are present at
// str[i-1] and str[j-1]. Append any
// of them to result.
if (dp[i][j] == dp[i - 1][j - 1] + 1) {
res = res + s[i - 1];
i--;
j--;
}
// Otherwise we move to the side
// that gave us maximum result
else if (dp[i][j] == dp[i - 1][j]) {
i--;
} else {
j--;
}
}
// Since we traverse dp[][] from bottom,
// we get result in reverse order.
let reverse = "";
for (let k = res.length - 1; k >= 0; k--) {
reverse = reverse + res[k];
}
return reverse;
}
let s = "AABEBCDD";
console.log(longestRepeatedSubSeq(s));
Time Complexity: O(n*n), where n is the length of string s.
Auxilairy Space: O(n*n)
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