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

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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 <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

 

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.



Last Updated : 05 Sep, 2022
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