# Program to check if an array is palindrome or not using Recursion

Given an array. The task is to determine whether an array is a palindrome or not using recursion.

Examples:

Input: arr[] = {3, 6, 0, 6, 3}
Output: Palindrome

Input: arr[] = {1, 2, 3, 4, 5}
Output: Not Palindrome

Approach:

1. Base case: If array has only one element i.e. begin == end then return 1, also if begin>end which means the array is palindrome then also return 1.
2. If the first and the last elements are equal then recursively call the function again but increment begin by 1 and decrement end by 1.
3. If the first and last element is not equal then return 0.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `// Recursive function that returns 1 if``// palindrome, 0 if not palindrome``int` `palindrome(``int` `arr[], ``int` `begin, ``int` `end)``{``    ``// base case``    ``if` `(begin >= end) {``        ``return` `1;``    ``}``    ``if` `(arr[begin] == arr[end]) {``        ``return` `palindrome(arr, begin + 1, end - 1);``    ``}``    ``else` `{``        ``return` `0;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 3, 6, 0, 6, 3 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);` `    ``if` `(palindrome(a, 0, n - 1) == 1)``        ``cout << ``"Palindrome"``;``    ``else``        ``cout << ``"Not Palindrome"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of above approach` `import` `java.io.*;` `class` `GFG {` `// Recursive function that returns 1 if``// palindrome, 0 if not palindrome``static` `int` `palindrome(``int` `arr[], ``int` `begin, ``int` `end)``{``    ``// base case``    ``if` `(begin >= end) {``        ``return` `1``;``    ``}``    ``if` `(arr[begin] == arr[end]) {``        ``return` `palindrome(arr, begin + ``1``, end - ``1``);``    ``}``    ``else` `{``        ``return` `0``;``    ``}``}` `// Driver code``    ``public` `static` `void` `main (String[] args) {``    ``int` `a[] = { ``3``, ``6``, ``0``, ``6``, ``3` `};``    ``int` `n = a.length;` `    ``if` `(palindrome(a, ``0``, n - ``1``) == ``1``)``        ``System.out.print( ``"Palindrome"``);``    ``else``        ``System.out.println( ``"Not Palindrome"``);``    ``}``}`

## Python 3

 `# Python 3 implementation of above approach` `# Recursive function that returns 1 if``# palindrome, 0 if not palindrome``def` `palindrome(arr, begin, end):` `    ``# base case``    ``if` `(begin >``=` `end) :``        ``return` `1``    ` `    ``if` `(arr[begin] ``=``=` `arr[end]) :``        ``return` `palindrome(arr, begin ``+` `1``,``                                 ``end ``-` `1``)``    ` `    ``else` `:``        ``return` `0` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ``a ``=` `[ ``3``, ``6``, ``0``, ``6``, ``3` `]``    ``n ``=` `len``(a)` `    ``if` `(palindrome(a, ``0``, n ``-` `1``) ``=``=` `1``):``        ``print``(``"Palindrome"``)``    ``else``:``        ``print``(``"Not Palindrome"``)` `# This code is contributed ``# by ChitraNayal`

## C#

 `// C# implementation of above approach``using` `System;` `class` `GFG``{` `// Recursive function that returns 1 ``// if palindrome, 0 if not palindrome``static` `int` `palindrome(``int` `[]arr, ``                      ``int` `begin, ``int` `end)``{``    ``// base case``    ``if` `(begin >= end) ``    ``{``        ``return` `1;``    ``}``    ``if` `(arr[begin] == arr[end]) ``    ``{``        ``return` `palindrome(arr, begin + 1,``                               ``end - 1);``    ``}``    ``else``    ``{``        ``return` `0;``    ``}``}` `// Driver code``public` `static` `void` `Main ()``{``    ``int` `[]a = { 3, 6, 0, 6, 3 };``    ``int` `n = a.Length;``    ` `    ``if` `(palindrome(a, 0, n - 1) == 1)``        ``Console.WriteLine(``"Palindrome"``);``    ``else``        ``Console.WriteLine(``"Not Palindrome"``);``}``}` `// This code is contributed by inder_verma`

## PHP

 `= ``\$end``) ``    ``{``        ``return` `1;``    ``}``    ``if` `(``\$arr``[``\$begin``] == ``\$arr``[``\$end``])``    ``{``        ``return` `palindrome(``\$arr``, ``\$begin` `+ 1, ``                                ``\$end` `- 1);``    ``}``    ``else``    ``{``        ``return` `0;``    ``}``}` `// Driver code``\$a` `= ``array``( 3, 6, 0, 6, 3 );``\$n` `= ``count``(``\$a``);` `if` `(palindrome(``\$a``, 0, ``\$n` `- 1) == 1)``    ``echo` `"Palindrome"``;``else``    ``echo` `"Not Palindrome"``;` `// This code is contributed ``// by inder_verma``?>`

## Javascript

 ``

Output:
`Palindrome`

Time complexity: O(N), where N is the size of the given array.
Auxiliary space: O(N), recursive stack space of size N will be required.

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