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++

// 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; }

## Java

// 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

Output:

1001

**Time Complexity : O(n)**

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