Make a string non-palindromic by inserting a given character

Given a string S and a character X, the task is to generate a non-palindromic string by inserting the character X in the string S. If it is not possible to obtain a non-palindromic string, then print “-1”.

Examples:

Input: S = “ababab”, X = ‘a’
Output: “aababab”
Explanation: Inserting the character ‘a’ at the beginning of the string S modifies the string to “aababab”, which is non-palindromic.

Input: S = “rrr”, X = ‘r’
Output: -1

Approach: The given problem can be solved based on the following observation that if the string contains only character X, then it is impossible to make the string non-palindromic. Otherwise, the string can be made non-palindromic. Follow the steps below to solve the problem:

1. If the count of character X in the string S is the same as the length of the string S, then print “-1”.
2. Otherwise, check if concatenations of string S and character X is palindromic or not. If found to be true, then print the string S + X. Otherwise print (X + S).

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if a
// string is palindromic or not
bool Palindrome(string str)
{
    // Traverse the string str
    for (int i = 0, j = str.length() - 1;
         i < j; i++, j--) {

        // Check if i-th character from
        // both ends are the same or not
        if (str[i] != str[j])
            return false;
    }

    // Return true, as str is palindrome
    return true;
}

// Function to make the non-palindromic
// string by inserting the character X
void NonPalindrome(string str, char X)
{

    // If all the characters
    // in the string are X
    if (count(str.begin(), str.end(), X)
        == str.length()) {

        cout << "-1";
        return;
    }

    // Check if X + str is
    // palindromic or not
    if (Palindrome(X + str))
        cout << str + X << endl;
    else
        cout << X + str << endl;
}

// Driver Code
int main()
{
    string S = "geek";
    char X = 's';
    NonPalindrome(S, X);

    return 0;
}

Java

// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

public class GFG {

    // Function to check if a
    // string is palindromic or not
    static boolean Palindrome(String str)
    {
        // Traverse the string str
        for (int i = 0, j = str.length() - 1; i < j;
             i++, j--) {

            // Check if i-th character from
            // both ends are the same or not
            if (str.charAt(i) != str.charAt(j))
                return false;
        }

        // Return true, as str is palindrome
        return true;
    }

    // Function to make the non-palindromic
    // string by inserting the character X
    static void NonPalindrome(String str, char X)
    {

        // stores the count of char X in str
        int count = 0;
        for (int i = 0; i < str.length(); i++)
            if (str.charAt(i) == X)
                count++;

        // If all the characters
        // in the string are X
        if (count == str.length()) {

            System.out.println("-1");
            return;
        }

        // Check if X + str is
        // palindromic or not
        if (Palindrome(X + str))
            System.out.println(str + X);
        else
            System.out.println(X + str);
    }

    // Driver Code
    public static void main(String[] args)
    {

        String S = "geek";
        char X = 's';
        NonPalindrome(S, X);
    }
}

// This code is contributed by Kingash.

Python3

# Python3 program for the above approach

# Function to check if a
# string is palindromic or not
def Palindrome(str):

    if str == str[::-1]:
        return True

    # Return true, as str is palindrome
    return False

# Function to make the non-palindromic
# string by inserting the character X
def NonPalindrome(str, X):

    # If all the characters
    # in the string are X
    if (str.count(X) == len(str)):
        print("-1")
        return

    # Check if X + str is
    # palindromic or not
    if (Palindrome(X + str)):
        print(str + X)
    else:
        print(X + str)

# Driver Code
if __name__ == '__main__':

    S = "geek"
    X = 's'

    NonPalindrome(S, X)

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach

using System;
using System.Linq;
class GFG {
    // Function to check if a
    // string is palindromic or not
    static bool Palindrome(string str)
    {
        // Traverse the string str
        for (int i = 0, j = str.Length - 1; i < j;
             i++, j--) {

            // Check if i-th character from
            // both ends are the same or not
            if (str[i] != str[j])
                return false;
        }

        // Return true, as str is palindrome
        return true;
    }

    // Function to make the non-palindromic
    // string by inserting the character X
    static void NonPalindrome(string str, char X)
    {

        // If all the characters
        // in the string are X
        if (str.Count(p => p == X) == str.Length) {

            Console.Write("-1");
            return;
        }

        // Check if X + str is
        // palindromic or not
        if (Palindrome(X + str))
            Console.WriteLine(str + X);
        else
            Console.WriteLine(X + str);
    }

    // Driver Code
    public static void Main()
    {
        string S = "geek";
        char X = 's';
        NonPalindrome(S, X);
    }
}

// This code is contributed by ukasp.

Javascript

<script>

// javascript program for the above approach

// Function to check if a
// string is palindromic or not
function Palindrome( str)
{
    // Traverse the string str
    for (let i = 0, j = str.length - 1; i < j;
        i++, j--) {

        // Check if i-th character from
        // both ends are the same or not
        if (str.charAt(i) != str.charAt(j))
            return false;
    }

    // Return true, as str is palindrome
    return true;
}

// Function to make the non-palindromic
// string by inserting the character X
function NonPalindrome( str, X)
{

    // stores the count of char X in str
    var count = 0;
    for (let i = 0; i < str.length; i++)
        if (str.charAt(i) == X)
            count++;

    // If all the characters
    // in the string are X
    if (count == str.length) {

        document.write("-1");
        return;
    }

    // Check if X + str is
    // palindromic or not
    if (Palindrome(X + str))
        document.write(str + X);
    else
        document.write(X + str);
}


// Driver Code

var S = "geek";
var X = 's';
NonPalindrome(S, X);

</script>

Output:

sgeek

Time Complexity: O(N)
Auxiliary Space: O(1),  since no extra space has been taken.