Repeated subsequence of length 2 or more

Given a string, find if there is any subsequence of length 2 or more that repeats itself such that the two subsequences don’t have the same character at the same position, i.e., any 0’th or 1st character in the two subsequences shouldn’t have the same index in the original string.

Example:

Input: ABCABD
Output: Repeated Subsequence Exists (A B is repeated) 

Input: ABBB
Output: Repeated Subsequence Exists (B B is repeated)

Input: AAB
Output: Repeated Subsequence Doesn't Exist (Note that 
A B cannot be considered as repeating because B is at
same position in two subsequences).

Input: AABBC
Output: Repeated Subsequence Exists (A B is repeated)

Input: ABCDACB
Output: Repeated Subsequence Exists (A B is repeated)

Input: ABCD
Output: Repeated Subsequence Doesn't Exist

The problem is a classic variation of longest common subsequence problem. We have discussed Dynamic programming solution here. Dynamic programming solution takes O(n²) time and space. 

In this post, O(n) time and space approach is discussed.

The idea is to remove all the non-repeated characters from the string and check if the resultant string is palindrome or not. If the remaining string is palindrome then it is not repeated, else there is a repetition. One special case we need to handle for inputs like “AAA”, which are palindrome but their repeated subsequence exists. Repeated subsequence exists for a palindrome string if it is of odd length and its middle letter is the same as the left(or right) character.

Algorithm: 

• Step 1: Initialize the input string.
• Step 2: Find the length of input string
• Step 3: Create an array and store all characters and their frequencies in it.
• Step 4: Traverse the input string and store the frequency of all characters in another array.
• Step 5: increment the count if the letters are repeating, if count is more than two return true.
• Step 6: In place of non repeating characters put ‘\0’ .
• Step 7: check if the string is palindrome, if it’s palindrome return false else true.
• Step 7: Print the output.

Below is the implementation of the above idea. 

C++

// C++ program to check if any repeated
// subsequence exists in the string
#include <bits/stdc++.h>
#define MAX_CHAR 256
using namespace std;

// A function to check if a string str is palindrome
bool isPalindrome(char str[], int l, int h)
{
    // l and h are leftmost and rightmost corners of str
    // Keep comparing characters while they are same
    while (h > l)
        if (str[l++] != str[h--])
            return false;

    return true;
}

// The main function that checks if repeated
// subsequence exists in the string
int check(char str[])
{
    // Find length of input string
    int n = strlen(str);

    // Create an array to store all characters and their
    // frequencies in str[]
    int freq[MAX_CHAR] = { 0 };

    // Traverse the input string and store frequencies
    // of all characters in freq[] array.
    for (int i = 0; i < n; i++)
    {
        freq[str[i]]++;

        // If the character count is more than 2
        // we found a repetition
        if (freq[str[i]] > 2)
            return true;
    }

    // In-place remove non-repeating characters
    // from the string
    int k = 0;
    for (int i = 0; i < n; i++)
        if (freq[str[i]] > 1)
            str[k++] = str[i];
    str[k] = '\0';

    // check if the resultant string is palindrome
    if (isPalindrome(str, 0, k-1))
    {
        // special case - if length is odd
        // return true if the middle character is
        // same as previous one
        if (k & 1)
            return str[k/2] == str[k/2 - 1];

        // return false if string is a palindrome
        return false;
    }

    // return true if string is not a palindrome
    return true;
}

// Driver code
int main()
{
    char str[] = "ABCABD";

    if (check(str))
        cout << "Repeated Subsequence Exists";
    else
        cout << "Repeated Subsequence Doesn't Exists";

    return 0;
}

Java

// Java program to check if any repeated
// subsequence exists in the string
import java.io.*;
class GFG {
    static int MAX_CHAR = 256;

    // A function to check
    // if a string str is palindrome
    public static boolean isPalindrome(String str, int l,
                                       int h)
    {

        // l and h are leftmost and rightmost corners of str
        // Keep comparing characters while they are same
        while (h > l)
            if (str.charAt(l++) != str.charAt(h--))
                return false;

        return true;
    }

    // The main function that checks if repeated
    // subsequence exists in the string
    public static boolean check(String str)
    {

        // Find length of input string
        int n = str.length();

        // Create an array to store all characters
        // and their frequencies in str[]
        int[] freq = new int[MAX_CHAR];

        // Traverse the input string and store frequencies
        // of all characters in freq[] array.
        for (int i = 0; i < n; i++) {
            freq[str.charAt(i)]++;

            // If the character count is more than 2
            // we found a repetition
            if (freq[str.charAt(i)] > 2)
                return true;
        }

        // In-place remove non-repeating characters
        // from the string
        int k = 0;
        for (int i = 0; i < n; i++)
            if (freq[str.charAt(i)] > 1)
                str.replace(str.charAt(k++), str.charAt(i));
        str.replace(str.charAt(k), '\0');

        // check if the resultant string is palindrome
        if (isPalindrome(str, 0, k - 1)) {

            // special case - if length is odd
            // return true if the middle character is
            // same as previous one
            if ((k & 1) == 1) {

                // It is checked so that
                // StringIndexOutOfBounds can be avoided
                if (k / 2 >= 1)
                    return (str.charAt(k / 2)
                            == str.charAt(k / 2 - 1));
            }

            // return false if string is a palindrome
            return false;
        }

        // return true if string is not a palindrome
        return true;
    }

    // Driver Code
    public static void main(String[] args)
    {
        String str = "ABCABD";

        if (check(str))
            System.out.println(
                "Repeated Subsequence Exists");
        else
            System.out.println("Repeated Subsequence"
                               + " Doesn't Exists");
    }
}

// This code is contributed by sanjeev2552

Python3

# Python3 program to check if any repeated
# subsequence exists in the String
MAX_CHAR = 256

# A function to check
# if a String Str is palindrome
def isPalindrome(Str, l, h):

    # l and h are leftmost and rightmost corners of Str
    # Keep comparing characters while they are same
    while (h > l):
        if (Str[l] != Str[h]):
            l += 1
            h -= 1
            return False

    return True

# The main function that checks if repeated
# subsequence exists in the String
def check(Str):

    # Find length of input String
    n = len(Str)

    # Create an array to store all characters
    # and their frequencies in Str[]
    freq = [0 for i in range(MAX_CHAR)]

    # Traverse the input String and
    # store frequencies of all characters
    # in freq[] array.
    for i in range(n):

        freq[ord(Str[i])] += 1

        # If the character count is more than 2
        # we found a repetition
        if (freq[ord(Str[i])] > 2):
            return True

    # In-place remove non-repeating characters
    # from the String
    k = 0
    for i in range(n):
        if (freq[ord(Str[i])] > 1):
            Str[k] = Str[i]
            k += 1
    Str[k] = '\0'

    # check if the resultant String is palindrome
    if (isPalindrome(Str, 0, k - 1)):

        # special case - if length is odd
        # return true if the middle character is
        # same as previous one
        if (k & 1):
            return Str[k // 2] == Str[k // 2 - 1]

        # return false if String is a palindrome
        return False

    # return true if String is not a palindrome
    return True

# Driver code
S = "ABCABD"
Str = [i for i in S]

if (check(Str)):
    print("Repeated Subsequence Exists")
else:
    print("Repeated Subsequence Doesn't Exists")

# This code is contributed by Mohit Kumar

C#

// C# program to check if any repeated
// subsequence exists in the string
using System;

class GFG
{
    static int MAX_CHAR = 256;

    // A function to check
    // if a string str is palindrome
    public static Boolean isPalindrome(String str,
                                    int l, int h)
    {

        // l and h are leftmost and rightmost corners of str
        // Keep comparing characters while they are same
        while(h > l)
            if (str[l++] != str[h--])
                return false;

        return true;
    }

    // The main function that checks if repeated
    // subsequence exists in the string
    public static Boolean check(String str)
    {

        // Find length of input string
        int n = str.Length;

        // Create an array to store all characters
        // and their frequencies in str[]
        int[] freq = new int[MAX_CHAR];

        // Traverse the input string and store frequencies
        // of all characters in freq[] array.
        for (int i = 0; i < n; i++)
        {
            freq[str[i]]++;

            // If the character count is more than 2
            // we found a repetition
            if (freq[str[i]] > 2)
                return true;
        }

        // In-place remove non-repeating characters
        // from the string
        int k = 0;
        for (int i = 0; i < n; i++)
            if (freq[str[i]] > 1)
                str.Replace(str[k++],
                            str[i]);
        str.Replace(str[k], '\0');

        // check if the resultant string is palindrome
        if (isPalindrome(str, 0, k - 1))
        {

            // special case - if length is odd
            // return true if the middle character is
            // same as previous one
            if ((k & 1) == 1)
            {

                // It is checked so that
                // StringIndexOutOfBounds can be avoided
                if (k / 2 >= 1)
                    return (str[k / 2] ==
                            str[k / 2 - 1]);
            }

            // return false if string is a palindrome
            return false;
        }

        // return true if string is not a palindrome
        return true;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        String str = "ABCABD";

        if (check(str))
            Console.WriteLine("Repeated Subsequence Exists");
        else
            Console.WriteLine("Repeated Subsequence" +
                            " Doesn't Exists");
    }
}

// This code is contributed by Princi Singh

Javascript

<script>
// Javascript program to check if any repeated
// subsequence exists in the string

    let MAX_CHAR = 256;

    // A function to check
    // if a string str is palindrome
    function isPalindrome(str,l,h)
    {
        // l and h are leftmost and rightmost corners of str
        // Keep comparing characters while they are same
        while(h > l)
            if (str[l++] != str[h--])
                return false;

        return true;
    }

    // The main function that checks if repeated
    // subsequence exists in the string
    function check(str)
    {
        // Find length of input string
        let n = str.length;

        // Create an array to store all characters
        // and their frequencies in str[]
        let freq = new Array(MAX_CHAR);
        for(let i=0;i<freq.length;i++)
        {
            freq[i]=0;
        }

        // Traverse the input string and store frequencies
        // of all characters in freq[] array.
        for (let i = 0; i < n; i++)
        {
            freq[str[i].charCodeAt(0)]++;

            // If the character count is more than 2
            // we found a repetition
            if (freq[str[i].charCodeAt(0)] > 2)
                return true;
        }

        // In-place remove non-repeating characters
        // from the string
        let k = 0;
        for (let i = 0; i < n; i++)
            if (freq[str[i].charCodeAt(0)] > 1)
                str.replace(str[k++],
                            str[i]);
        str.replace(str[k], '\0');

        // check if the resultant string is palindrome
        if (isPalindrome(str, 0, k - 1))
        {

            // special case - if length is odd
            // return true if the middle character is
            // same as previous one
            if ((k & 1) == 1)
            {

                // It is checked so that
                // StringIndexOutOfBounds can be avoided
                if (k / 2 >= 1)
                    return (str[(Math.floor(k/2))] ==
                            str[(Math.floor(k/2) -1)]);
            }

            // return false if string is a palindrome
            return false;
        }

        // return true if string is not a palindrome
        return true;
    }

    // Driver Code
    let str = "ABCABD";
    if (check(str))
        document.write("Repeated Subsequence Exists");
    else
        document.write("Repeated Subsequence" +
                               " Doesn't Exists");

    // This code is contributed by unknown2108
</script>

Output

Repeated Subsequence Exists

Time Complexity: O(n).
 Auxiliary Space: O(256), because we are using frequency array size of MAX_CHAR.

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.