Check if there exist a string Y such that (X+Y) and (Y+X) are palindromes

Given a string X of length N and an integer K, the task is to determine if there exists a string (say Y) of length K such that its concatenation on either side of X makes the resultant string palindrome i.e. X+Y and Y+X both are palindromes.

Examples:

Input: N = 6, K = 2, X = “abbaab”
Output: Yes
Explanation: Let string Y = “ba”. Then, 
X + Y = “abbaabba” (which is a palindrome)
Y + X = “baabbaab” (which is also a palindrome).

Input: N = 12, K = 400378271514996652, X = “njvhhvjnnjvh”
Output: Yes

Approach: To solve the problem follow the below observations:

Observations:

Case 1: K < 2*N

Let rev(X) represent the reverse of string X. If K<2*N, the leftmost min(N, K) characters of Y should be equal to rev(X), and so the rightmost min(N, K) characters. So, we know that there exist a string that can satisfy the conditions. We just have to find it and check if it satisfies the conditions.

Case 2: K >= 2*N

K can be replaced with K%(2*N) so that Case-2 becomes equivalent to Case-1, however the problem won’t be affected.

Proof:

There is string Y (|Y| = K) such that X+Y, Y+X are palindromes. Let string Y (|Y| = K – 2N) be such that X + rev(X) + Y + rev(X),  
rev(X) + Y + rev(X) + X are palindromes

• rev(X)+Y, Y+rev(X) are palindromes
• rev(rev(X)+Y), rev(Y+rev(X)) are palindromes
• rev(Y)+X, X+rev(Y) are palindromes
• Y+X, X+Y are palindromes.

Based on the above observations, the following steps can be used to solve the problem :

• Reverse the given string X
• Replace K with K%(2N)
• Initialize a string (say temp) of size K.
• Iterate through the initial min(N, K) characters of X and make the i^th character of temp equal to the i^th character from the end of the string X.
• Again, iterate through min(N, K). If an i^th character from the end of temp is not filled yet, make it equal to i^th character of X. Else, if it is filled and not equal to X[i], return “No”.
• At the end of the iteration, check if both X+temp and temp+X are palindromes. If so, return “Yes”, else return “No”.

Following is the code based on the above approach:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
#define int long long

// Function to check if given string is
// a palindrome or not
bool isPalindrome(string S)
{
    for (int i = 0; i < S.size(); i++) {
        if (S[i] != S.end()[-1 - i])
            return false;
    }
    return true;
}

// Function to Check if there exist a
// string Y such that (X+Y) and (Y+X)
// are palindromes
void isPossible(string X, int N, int K)
{

    // Reverse string X
    reverse(X.begin(), X.end());

    // Updating K
    K %= (2 * N);

    // Declaring and Initializing a
    // string temp
    string temp(K, '.');

    // Iterating from i=0 to min(K-1, N-1)
    for (int i = 0; i < K && i < N; i++) {

        // Making ith character of temp
        // equal to ith character from
        // the end of X
        temp[i] = X[N - 1 - i];
    }

    // Iterating from i=0 to min(K-1, N-1)
    for (int i = 0; i < K && i < N; i++) {
        // If ith character from end of
        // temp is not filled yet, make it
        // equal to ith character of X
        if (temp[K - 1 - i] == '.')
            temp[K - 1 - i] = X[i];

        // Else if it is filled and not
        // equal to X[i], print "No"
        // and return
        else {
            if (X[i] != temp[K - 1 - i]) {
                cout << "No" << endl;
                return;
            }
        }
    }

    // If both (X+temp) and (temp+X) are
    // palindrome, print yes
    if (isPalindrome(X + temp) && isPalindrome(temp + X))
        cout << "Yes" << endl;

    // Else print No
    else
        cout << "No" << endl;
}

// Driver code
int32_t main()
{
    int N = 6, K = 2;
    string S = "abbaab";

    // Function Call
    isPossible(S, N, K);
    return 0;
}

Java

import java.util.*;

class Main {
    // Function to check if given string is
    // a palindrome or not
    public static boolean isPalindrome(String S)
    {
        for (int i = 0; i < S.length(); i++) {
            if (S.charAt(i)
                != S.charAt(S.length() - 1 - i)) {
                return false;
            }
        }
        return true;
    }

    // Function to Check if there exist a
    // string Y such that (X+Y) and (Y+X)
    // are palindromes
    public static void isPossible(String X, int N, int K)
    {
        // Reverse string X
        StringBuilder reversed = new StringBuilder(X);
        reversed = reversed.reverse();
        X = reversed.toString();

        // Updating K
        K %= (2 * N);

        // Declaring and Initializing a
        // string temp
        char[] temp = new char[K];
        Arrays.fill(temp, '.');

        // Iterating from i=0 to min(K-1, N-1)
        for (int i = 0; i < K && i < N; i++) {
            // Making ith character of temp
            // equal to ith character from
            // the end of X
            temp[i] = X.charAt(N - 1 - i);
        }

        // Iterating from i=0 to min(K-1, N-1)
        for (int i = 0; i < K && i < N; i++) {
            // If ith character from end of
            // temp is not filled yet, make it
            // equal to ith character of X
            if (temp[K - 1 - i] == '.') {
                temp[K - 1 - i] = X.charAt(i);
            }
            // Else if it is filled and not
            // equal to X[i], print "No"
            // and return
            else {
                if (X.charAt(i) != temp[K - 1 - i]) {
                    System.out.println("No");
                    return;
                }
            }
        }

        // If both (X+temp) and (temp+X) are
        // palindrome, print yes
        if (isPalindrome(X + new String(temp))
            && isPalindrome(new String(temp) + X)) {
            System.out.println("Yes");
        }
        // Else print No
        else {
            System.out.println("No");
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        int N = 6, K = 2;
        String S = "abbaab";

        // Function Call
        isPossible(S, N, K);
    }
}

Python3

# Python code for the above approach

# Function to check if given string is
# a palindrome or not
def isPalindrome(S):
    for i in range(len(S)):
        if S[i] != S[-1 - i]:
            return False
    return True

# Function to check if there exist a
# string Y such that (X+Y) and (Y+X)
# are palindromes
def isPossible(X, N, K):

    # Reverse string X
    X = X[::-1]

    # Updating K
    K %= (2 * N)

    # Declaring and Initializing a
    # string temp
    temp = K * '.'

    # Iterating from i=0 to min(K-1, N-1)
    for i in range(min(K, N)):

        # Making ith character of temp
        # equal to ith character from
        # the end of X
        temp = temp[:i] + X[N - 1 - i] + temp[i+1:]

    # Iterating from i=0 to min(K-1, N-1)
    for i in range(min(K, N)):

        # If ith character from end of
        # temp is not filled yet, make it
        # equal to ith character of X
        if temp[K - 1 - i] == '.':
            temp = temp[:K - 1 - i] + X[i] + temp[K - i:]

        # Else if it is filled and not
        # equal to X[i], print "No"
        # and return
        else:
            if X[i] != temp[K - 1 - i]:
                print("No")
                return

    # If both (X+temp) and (temp+X) are
    # palindrome, print "Yes"
    if isPalindrome(X + temp) and isPalindrome(temp + X):
        print("Yes")

    # Else print "No"
    else:
        print("No")


# Driver code
N, K = 6, 2
S = "abbaab"

# Function Call
isPossible(S, N, K)

# This code is contributed by prasad264

C#

using System;
using System.Text;

public class GFG{
    
    // Function to check if given string is
    // a palindrome or not
    public static bool isPalindrome(string S)
    {
        for (int i = 0; i < S.Length; i++) {
            if (S[i]
                != S[S.Length - 1 - i]) {
                return false;
            }
        }
        return true;
    }

    // Function to Check if there exist a
    // string Y such that (X+Y) and (Y+X)
    // are palindromes
    public static void isPossible(string X, int N, int K)
    {
        // Reverse string X
        StringBuilder sb = new StringBuilder();
 
        for (int i = X.Length - 1; i >=0 ; i--) {
            sb.Append(X[i]);
        }
 
        X = sb.ToString();

        // Updating K
        K %= (2 * N);

        // Declaring and Initializing a
        // string temp
        char[] temp = new char[K];
        Array.Fill(temp, '.');

        // Iterating from i=0 to min(K-1, N-1)
        for (int i = 0; i < K && i < N; i++) {
            // Making ith character of temp
            // equal to ith character from
            // the end of X
            temp[i] = X[N - 1 - i];
        }

        // Iterating from i=0 to min(K-1, N-1)
        for (int i = 0; i < K && i < N; i++) {
            // If ith character from end of
            // temp is not filled yet, make it
            // equal to ith character of X
            if (temp[K - 1 - i] == '.') {
                temp[K - 1 - i] = X[i];
            }
            // Else if it is filled and not
            // equal to X[i], print "No"
            // and return
            else {
                if (X[i] != temp[K - 1 - i]) {
                    Console.WriteLine("No");
                    return;
                }
            }
        }

        // If both (X+temp) and (temp+X) are
        // palindrome, print yes
        if (isPalindrome(X + new string(temp))
            && isPalindrome(new string(temp) + X)) {
           Console.WriteLine("Yes");
        }
        // Else print No
        else {
            Console.WriteLine("No");
        }
    }

    // Driver code
    static public void Main (){

        
        int N = 6, K = 2;
        String S = "abbaab";

        // Function Call
        isPossible(S, N, K);
    }
}

Javascript

// JavaScript code for the above approach
function isPalindrome(S) {
    for (let i = 0; i < S.length; i++) {
        if (S[i] !== S[S.length - 1 - i]) {
            return false;
        }
    }
    return true;
}

function isPossible(X, N, K) {
    // Reverse string X
    X = X.split("").reverse().join("");
    
    // Updating K
    K %= 2 * N;
    
    // Declaring and Initializing a
    // string temp
    let temp = ".".repeat(K);
    
    // Iterating from i=0 to min(K-1, N-1)
    for (let i = 0; i < Math.min(K, N); i++) {
        // Making ith character of temp
        // equal to ith character from
        // the end of X
        temp = temp.substr(0, i) + X[N - 1 - i] + temp.substr(i + 1);
    }
    
    // Iterating from i=0 to min(K-1, N-1)
    for (let i = 0; i < Math.min(K, N); i++) {
        // If ith character from end of
        // temp is not filled yet, make it
        // equal to ith character of X
        if (temp[K - 1 - i] === ".") {
            temp = temp.substr(0, K - 1 - i) + X[i] + temp.substr(K - i);
        }
        // Else if it is filled and not
        // equal to X[i], print "No"
        // and return
        else if (X[i] !== temp[K - 1 - i]) {
            console.log("No");
            return;
        }
    }

    // If both (X+temp) and (temp+X) are
    // palindrome, print yes
    if (isPalindrome(X + temp) && isPalindrome(temp + X)) {
        console.log("Yes");
    }
    // Else print No
    else {
        console.log("No");
    }
}

// Driver code
let N = 6,
K = 2;
let S = "abbaab";

// Function Call
isPossible(S, N, K);

Output

Yes

Time Complexity : O(min(N, K))
Auxiliary Space : O(K)