# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 ` `

Output:

```abab
4
abcdabcd
8
```

Time Complexity: O(1).

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