Longest double string from a Palindrome

Given a palindrome String, the task is to find the maximum length of the double string and its length that can be obtained from the given palindromic string. A double string is a string that has two clear repetition of a substring one after the other.
Examples:

Input: 
abba
Output:
abab 
4
Explanation: 
abab is double string 
which can be obtained 
by changing the order of letters

Input:
abcba
Output:
abab 4
Explanation: 
abab is double string 
which can be obtained 
by changing the order of letters
and deleting letter c

Approach: The double string can be considered in two cases:

• Case 1: If the length of the string is even then the length of double string will always be the length of string. 
  
• Case 2: If the length of the string is odd then the length of double string will always be the length of string – 1. 
  

Below is the implementation of the above approach

C++

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

// Function to return the required position
int lenDoubleString(string s)
{

    int l = s.length();
    string first_half = s.substr(0, l / 2);
    string second_half = "";

    if (l % 2 == 0)
        second_half = s.substr(l / 2);
    else
        second_half = s.substr(l / 2 + 1);

    reverse(second_half.begin(), second_half.end());

    // Print the double string
    cout << first_half << second_half << endl;

    // Print the length of the double string
    if (l % 2 == 0)
        cout << l << endl;
    else
        cout << l - 1 << endl;
}

int main()
{
    string n = "abba";
    lenDoubleString(n);

    n = "abcdedcba";
    lenDoubleString(n);
    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the required position
static int lenDoubleString(String s)
{

    int l = s.length();
    String first_half = s.substring(0, l / 2);
    String second_half = "";

    if (l % 2 == 0)
        second_half = s.substring(l / 2);
    else
        second_half = s.substring(l / 2 + 1);

    second_half = reverse(second_half);

    // Print the double String
    System.out.println(first_half + second_half);

    // Print the length of the double String
    if (l % 2 == 0)
        System.out.println(l);
    else
        System.out.println(l - 1);
        return Integer.MIN_VALUE;
}
static String reverse(String input)
{
    char[] temparray = input.toCharArray();
    int left, right = 0;
    right = temparray.length - 1;

    for (left = 0; left < right; left++, right--)
    {
        // Swap values of left and right
        char temp = temparray[left];
        temparray[left] = temparray[right];
        temparray[right] = temp;
    }
    return String.valueOf(temparray);
}

// Driver code
public static void main(String[] args)
{
    String n = "abba";
    lenDoubleString(n);

    n = "abcdedcba";
    lenDoubleString(n);
}
}

// This code contributed by Rajput-Ji

Python3

# Python3 implementation of the approach.

# Function to return the required position
def lenDoubleString(s):

    l = len(s)
    first_half = s[0 : l // 2]
    second_half = ""

    if l % 2 == 0:
        second_half = s[l // 2 : ]
    else:
        second_half = s[l // 2 + 1 : ]

    second_half = second_half[::-1]

    # Print the double string
    print(first_half + second_half)

    # Print the length of the double string
    if l % 2 == 0:
        print(l)
    else:
        print(l - 1)

# Driver Code
if __name__ == "__main__":

    n = "abba"
    lenDoubleString(n)

    n = "abcdedcba"
    lenDoubleString(n)

# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the required position
static int lenDoubleString(String s)
{

    int l = s.Length;
    String first_half = s.Substring(0, l / 2);
    String second_half = "";

    if (l % 2 == 0)
        second_half = s.Substring(l / 2);
    else
        second_half = s.Substring(l / 2 + 1);

    second_half = reverse(second_half);

    // Print the double String
    Console.WriteLine(first_half + second_half);

    // Print the length of the double String
    if (l % 2 == 0)
        Console.WriteLine(l);
    else
        Console.WriteLine(l - 1);
        return int.MinValue;
}

static String reverse(String input)
{
    char[] temparray = input.ToCharArray();
    int left, right = 0;
    right = temparray.Length - 1;

    for (left = 0; left < right; left++, right--)
    {
        // Swap values of left and right
        char temp = temparray[left];
        temparray[left] = temparray[right];
        temparray[right] = temp;
    }
    return String.Join("",temparray);
}

// Driver code
public static void Main(String[] args)
{
    String n = "abba";
    lenDoubleString(n);

    n = "abcdedcba";
    lenDoubleString(n);
}
}

// This code has been contributed by 29AjayKumar

PHP

<?php
// PHP implementation of the approach

// Function to return the required position
function lenDoubleString($s)
{
    $l = strlen($s);
    $first_half = substr($s, 0, (int)($l / 2));
    $second_half = "";

    if ($l % 2 == 0)
        $second_half = substr($s, (int)($l / 2));
    else
        $second_half = substr($s, (int)($l / 2 + 1));

    $second_half = strrev($second_half);

    // Print the double string
    echo $first_half . "" .
         $second_half . "\n";

    // Print the length of the double string
    if ($l % 2 == 0)
        echo $l . "\n";
    else
        echo ($l - 1) . "\n";
}

// Driver Code
$n = "abba";
lenDoubleString($n);

$n = "abcdedcba";
lenDoubleString($n);

// This code is contributed by mits
?>

Javascript

<script>
// Javascript implementation of the
// above approach

// Function to return the required position
function lenDoubleString(s)
{

    var l = s.length;
    var first_half = s.substr(0, l / 2);
    var second_half = "";

    if (l % 2 == 0)
        second_half = s.substr(l / 2);
    else
        second_half = s.substr(l / 2 + 1);

    second_half = second_half.split("").reverse().join("");

    // Print the double string
    document.write(first_half);
    document.write(second_half+"<br>");

    // Print the length of the double string
    if (l % 2 == 0)
        document.write(l+"<br>");
    else
        document.write(l - 1+"<br>");

}

  var n = "abba";
  lenDoubleString(n);

  n = "abcdedcba";
  lenDoubleString(n);

// This code is contributed by ShubhamSingh10
</script>

Output:

abab
4
abcdabcd
8

Time Complexity: O(N), as we are using built-in functions reverse which will cost O(N) time.

Auxiliary Space: O(N), as we are using extra space.