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

Difficulty Level : Basic
Last Updated : 01 Apr, 2021

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:

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.
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.
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 <iostream>
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

<?php
// PHP implementation of above approach
 
// Recursive function that returns 1
// if palindrome, 0 if not palindrome
function 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
$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

<script>
    // Javascript implementation of above approach
     
    // Recursive function that returns 1 if
    // palindrome, 0 if not palindrome
    function 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
    let a = [ 3, 6, 0, 6, 3 ];
    let n = a.length;
   
    if (palindrome(a, 0, n - 1) == 1)
        document.write("Palindrome");
    else
        document.write("Not Palindrome");
         
        // This code is contributed by divyeshrabadiya07.
     
</script>

Output:

Palindrome

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