Check if a number is Palindrome

Given an integer, write a function that returns true if the given number is palindrome, else false. For example, 12321 is palindrome, but 1451 is not palindrome.

Let the given number be num. A simple method for this problem is to first reverse digits of num, then compare the reverse of num with num. If both are same, then return true, else false.



Following is an interesting method inspired from method#2 of this post. The idea is to create a copy of num and recursively pass the copy by reference, and pass num by value. In the recursive calls, divide num by 10 while moving down the recursion tree. While moving up the recursion tree, divide the copy by 10. When they meet in a function for which all child calls are over, the last digit of num will be ith digit from the beginning and the last digit of copy will be ith digit from the end.

C++

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// A recursive C++ program to check whether a given number is
// palindrome or not
#include <stdio.h>
  
// A function that reurns true only if num contains one digit
int oneDigit(int num)
{
    // comparison operation is faster than division operation.
    // So using following instead of "return num / 10 == 0;"
    return (num >= 0 && num < 10);
}
  
// A recursive function to find out whether num is palindrome
// or not. Initially, dupNum contains address of a copy of num.
bool isPalUtil(int num, int* dupNum)
{
    // Base case (needed for recursion termination): This statement
    // mainly compares the first digit with the last digit
    if (oneDigit(num))
        return (num == (*dupNum) % 10);
  
    // This is the key line in this method. Note that all recursive
    // calls have a separate copy of num, but they all share same copy
    // of *dupNum. We divide num while moving up the recursion tree
    if (!isPalUtil(num/10, dupNum))
        return false;
  
    // The following statements are executed when we move up the
    // recursion call tree
    *dupNum /= 10;
  
    // At this point, if num%10 contains i'th digit from beiginning,
    // then (*dupNum)%10 contains i'th digit from end
    return (num % 10 == (*dupNum) % 10);
}
  
// The main function that uses recursive function isPalUtil() to
// find out whether num is palindrome or not
int isPal(int num)
{
    // If num is negative, make it positive
    if (num < 0)
       num = -num;
  
    // Create a separate copy of num, so that modifications made
    // to address dupNum don't change the input number.
    int *dupNum = new int(num); // *dupNum = num
  
    return isPalUtil(num, dupNum);
}
  
// Driver program to test above functions
int main()
{
    int n = 12321;
    isPal(n)? printf("Yesn"): printf("Non");
  
    n = 12;
    isPal(n)? printf("Yesn"): printf("Non");
  
    n = 88;
    isPal(n)? printf("Yesn"): printf("Non");
  
    n = 8999;
    isPal(n)? printf("Yesn"): printf("Non");
    return 0;
}

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Java

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// A recursive Java program to 
// check whether a given number 
// is palindrome or not
import java.io.*;
import java.util.*;
   
public class CheckPallindromNumberRecursion {
   
    // A function that reurns true
    // only if num contains one digit
    public static int oneDigit(int num) {
   
        if ((num >= 0) && (num < 10))
            return 1;
        else
            return 0;
    }
   
    public static int isPalUtil
    (int num, int dupNum) throws Exception {
   
        // base condition to return once we 
        // move past first digit
        if (num == 0) {
            return dupNum;
        } else {
            dupNum = isPalUtil(num / 10, dupNum);
        }
   
        // Check for equality of first digit of
        // num and dupNum
        if (num % 10 == dupNum % 10) {
            // if first digit values of num and 
            // dupNum are equal divide dupNum
            // value by 10 to keep moving in sync
            // with num.
            return dupNum / 10;
        } else {
            // At position values are not 
            // matching throw exception and exit. 
            // no need to proceed further.
            throw new Exception();
        }
   
    }
   
    public static int isPal(int num) 
    throws Exception {
   
        if (num < 0)
            num = (-num);
   
        int dupNum = (num);
   
        return isPalUtil(num, dupNum);
    }
   
    public static void main(String args[]) {
   
        int n = 1242;
        try {
            isPal(n);
            System.out.println("Yes");
        } catch (Exception e) {
            System.out.println("No");
        }
        n = 1231;
        try {
            isPal(n);
            System.out.println("Yes");
        } catch (Exception e) {
            System.out.println("No");
        }
   
        n = 12;
        try {
            isPal(n);
            System.out.println("Yes");
        } catch (Exception e) {
            System.out.println("No");
        }
   
        n = 88;
        try {
            isPal(n);
            System.out.println("Yes");
        } catch (Exception e) {
            System.out.println("No");
        }
   
        n = 8999;
        try {
            isPal(n);
            System.out.println("Yes");
        } catch (Exception e) {
            System.out.println("No");
        }
    }
}
   
// This code is contributed
// by Nasir J

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Python3

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# A recursive Pyhton3 program to check
# whether a given number is palindrome or not
  
# A function that reurns true 
# only if num contains one digit
def oneDigit(num):
      
    # comparison operation is faster 
    # than division operation. So 
    # using following instead of 
    # "return num / 10 == 0;"
    return ((num >= 0) and
            (num < 10));
  
# A recursive function to find 
# out whether num is palindrome
# or not. Initially, dupNum 
# contains address of a copy of num.
def isPalUtil(num, dupNum):
      
    # Base case (needed for recursion
    # termination): This statement
    # mainly compares the first digit
    # with the last digit
    if (oneDigit(num)):
        return (num == (dupNum) % 10);
  
    # This is the key line in this 
    # method. Note that all recursive
    # calls have a separate copy of
    # num, but they all share same
    # copy of *dupNum. We divide num
    # while moving up the recursion tree
    if (isPalUtil(int(num / 10), dupNum) == False):
        return -1;
  
    # The following statements are
    # executed when we move up the
    # recursion call tree
    dupNum = int(dupNum / 10);
  
    # At this point, if num%10 
    # contains i'th digit from 
    # beiginning, then (*dupNum)%10 
    # contains i'th digit from end
    return (num % 10 == (dupNum) % 10);
  
# The main function that uses 
# recursive function isPalUtil()
# to find out whether num is 
# palindrome or not
def isPal(num):
    # If num is negative, 
    # make it positive
    if (num < 0):
        num = (-num);
  
    # Create a separate copy of 
    # num, so that modifications 
    # made to address dupNum 
    # don't change the input number.
    dupNum = (num); # *dupNum = num
  
    return isPalUtil(num, dupNum);
  
# Driver Code
n = 12321;
if(isPal(n) == 0):
    print("Yes");
else:
    print("No");
  
n = 12;
if(isPal(n) == 0):
    print("Yes");
else:
    print("No");
  
n = 88;
if(isPal(n) == 1):
    print("Yes");
else:
    print("No");
  
n = 8999;
if(isPal(n) == 0):
    print("Yes");
else:
    print("No");
  
# This code is contributed by mits

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C#

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// A recursive C# program to 
// check whether a given number 
// is palindrome or not
using System;
  
class GFG
{
      
// A function that reurns true 
// only if num contains one digit
public static int oneDigit(int num)
{
    // comparison operation is 
    // faster than division 
    // operation. So using 
    // following instead of 
    // "return num / 10 == 0;"
    if((num >= 0) &&(num < 10))
    return 1;
    else
    return 0;
}
  
// A recursive function to 
// find out whether num is 
// palindrome or not. 
// Initially, dupNum contains
// address of a copy of num.
public static int isPalUtil(int num, 
                            int dupNum)
{
    // Base case (needed for recursion
    // termination): This statement
    // mainly compares the first digit
    // with the last digit
    if (oneDigit(num) == 1)
        if(num == (dupNum) % 10)
        return 1;
        else
        return 0;
  
    // This is the key line in 
    // this method. Note that 
    // all recursive calls have 
    // a separate copy of num,
    // but they all share same
    // copy of *dupNum. We divide 
    // num while moving up the 
    // recursion tree
    if (isPalUtil((int)(num / 10), dupNum) == 0)
        return -1;
  
    // The following statements 
    // are executed when we move
    // up the recursion call tree
    dupNum = (int)(dupNum / 10);
  
    // At this point, if num%10 
    // contains i'th digit from 
    // beiginning, then (*dupNum)%10 
    // contains i'th digit from end
    if(num % 10 == (dupNum) % 10) 
        return 1; 
    else
        return 0;
}
  
// The main function that uses 
// recursive function isPalUtil()
// to find out whether num is 
// palindrome or not
public static int isPal(int num)
{
    // If num is negative, 
    // make it positive
    if (num < 0)
    num = (-num);
  
    // Create a separate copy 
    // of num, so that modifications 
    // made to address dupNum 
    // don't change the input number.
    int dupNum = (num); // *dupNum = num
  
    return isPalUtil(num, dupNum);
}
  
// Driver Code
public static void Main()
{
int n = 12321;
if(isPal(n) == 0)
    Console.WriteLine("Yes");
else
    Console.WriteLine("No");
  
n = 12;
if(isPal(n) == 0)
    Console.WriteLine("Yes");
else
    Console.WriteLine( "No");
  
n = 88;
if(isPal(n) == 1)
    Console.WriteLine("Yes");
else
    Console.WriteLine("No");
  
n = 8999;
if(isPal(n) == 0)
    Console.WriteLine("Yes");
else
    Console.WriteLine("No");
}
}
  
// This code is contributed by mits

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PHP

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<?php
// A recursive PHP program to 
// check whether a given number 
// is palindrome or not
  
// A function that reurns true 
// only if num contains one digit
function oneDigit($num)
{
    // comparison operation is faster 
    // than division operation. So 
    // using following instead of 
    // "return num / 10 == 0;"
    return (($num >= 0) && 
            ($num < 10));
}
  
// A recursive function to find 
// out whether num is palindrome
// or not. Initially, dupNum 
// contains address of a copy of num.
function isPalUtil($num, $dupNum)
{
    // Base case (needed for recursion
    // termination): This statement
    // mainly compares the first digit
    // with the last digit
    if (oneDigit($num))
        return ($num == ($dupNum) % 10);
  
    // This is the key line in this 
    // method. Note that all recursive
    // calls have a separate copy of
    // num, but they all share same
    // copy of *dupNum. We divide num
    // while moving up the recursion tree
    if (!isPalUtil((int)($num / 10), 
                         $dupNum))
        return -1;
  
    // The following statements are
    // executed when we move up the
    // recursion call tree
    $dupNum = (int)($dupNum / 10);
  
    // At this point, if num%10  
    // contains i'th digit from 
    // beiginning, then (*dupNum)%10 
    // contains i'th digit from end
    return ($num % 10 == ($dupNum) % 10);
}
  
// The main function that uses 
// recursive function isPalUtil()
// to find out whether num is 
// palindrome or not
function isPal($num)
{
    // If num is negative, 
    // make it positive
    if ($num < 0)
    $num = (-$num);
  
    // Create a separate copy of 
    // num, so that modifications 
    // made to address dupNum 
    // don't change the input number.
    $dupNum = ($num); // *dupNum = num
  
    return isPalUtil($num, $dupNum);
}
  
// Driver Code
$n = 12321;
if(isPal($n) == 0)
    echo "Yes\n";
else
    echo "No\n";
  
$n = 12;
if(isPal($n) == 0)
    echo "Yes\n";
else
    echo "No\n";
  
$n = 88;
if(isPal($n) == 1)
    echo "Yes\n";
else
    echo "No\n";
  
$n = 8999;
if(isPal($n) == 0)
    echo "Yes\n";
else
    echo "No\n";
  
// This code is contributed by m_kit
?>

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Output:

Yes
No
Yes
No

To check a number is palindrome or not without using any extra space

This article is compiled by Aashish Barnwal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



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