Nth Even length Palindrome

Given a number n as string, find the nth even-length positive palindrome number .

Examples:

Input : n = "1"
Output : 11
1st even-length palindrome is 11 .

Input : n = "10"
Output : 1001
The first 10 even-length palindrome numbers are 11, 22, 
33, 44, 55, 66, 77, 88, 99 and 1001.

As, it is a even-length palindrome so its first half should be equal to second half and length will be 2, 4, 6, 8 …. To evaluate nth palindrome let’s just see 1st 10 even-length palindrome numbers 11, 22, 33, 44, 55, 66, 77, 88, 99 and 1001 . Here, nth palindrome is nn’ where n’ is reverse of n . Thus we just have to write n and n’ in a consecutive manner where n’ is reverse of n .

Below is implementation of this approach .



C/C++

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// C++ program to find n=th even length string.
#include <bits/stdc++.h>
using namespace std;
  
// Function to find nth even length Palindrome
string evenlength(string n)
{
    // string r to store resultant
    // palindrome. Initialize same as s
    string res = n;
  
    // In this loop string r stores
    // reverse of string s after the
    // string s in consecutive manner .
    for (int j = n.length() - 1; j >= 0; --j)
        res += n[j];
  
    return res;
}
  
// Driver code to test above function
int main()
{
    string n = "10";
    cout << evenlength(n);
    return 0;
}

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Java














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// Java program to find nth even length Palindrome 


import java.io.*; 


  


class GFG { 


    // Function to find nth even length Palindrome 


    static String evenlength(String n) 


    { 


        // string r to store resultant 


        // palindrome. Initialize same as s 


        String res = n; 


  


        // In this loop string r stores 


        // reverse of string s after the 


        // string s in consecutive manner 


        for (int j = n.length() - 1; j >= 0; --j) 


            res += n.charAt(j); 


  


        return res; 


    } 


  


    // driver program 


    public static void main(String[] args) 


    { 


        String n = "10"; 


        System.out.println(evenlength(n)); 


    } 


} 


  


// Contributed by Pramod Kumar 



















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Python3














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# Python3 program to find n=th even  


# length string. 


import math as mt 


  


# Function to find nth even length  


# Palindrome 


def evenlength(n): 


  


    # string r to store resultant 


    # palindrome. Initialize same as s 


    res = n 


  


    # In this loop string r stores 


    # reverse of string s after the 


    # string s in consecutive manner . 


    for j in range(len(n) - 1, -1, -1): 


        res += n[j] 


  


    return res 


  


# Driver code 


n = "10"


print(evenlength(n)) 


  


# This code is contributed by 


# Mohit kumar 29 



















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C#














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// C# program to find nth even 


// length Palindrome 


using System; 


  


class GFG { 


      


    // Function to find nth even 


    // length Palindrome 


    static string evenlength(string n) 


    { 


          


        // string r to store resultant 


        // palindrome. Initialize same 


        // as s 


        string res = n; 


  


        // In this loop string r stores 


        // reverse of string s after 


        // the string s in consecutive 


        // manner 


        for (int j = n.Length - 1; 


                           j >= 0; --j) 


            res += n[j]; 


  


        return res; 


    } 


  


    // driver program 


    public static void Main() 


    { 


        string n = "10"; 


          


        Console.WriteLine(evenlength(n)); 


    } 


} 


  


// This code is contributed by vt_m. 



















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PHP














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<?php 


// PHP program to find n=th even length string. 


  


// Function to find nth even length Palindrome 


function evenlength($n) 


{ 


    // string r to store resultant 


    // palindrome. Initialize same as s 


    $res = $n; 


  


    // In this loop string r stores 


    // reverse of string s after the 


    // string s in consecutive manner . 


    for ($j = strlen($n) - 1; $j >= 0; --$j) 


        $res = $res . $n[$j]; 


  


    return $res; 


} 


  


// Driver code 


$n = "10"; 


echo evenlength($n); 


  


// This code is contributed by ita_c 


?> 



















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Output:


 1001



Time Complexity : O(n)

This article is contributed by Surya Priy. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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