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Check if a number is Palindrome

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  • Difficulty Level : Medium
  • Last Updated : 27 Apr, 2022

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++




// A recursive C++ program to check
// whether a given number
// is palindrome or not
#include <iostream>
using namespace std;
 
// A function that returns 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
    // beginning, 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)
{
     
    // Check 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
    int* dupNum = new int(num);
 
    return isPalUtil(num, dupNum);
}
 
// Driver program to test
// above functions
int main()
{
    int n = 12321;
    isPal(n) ? cout <<"Yes\n":  cout <<"No" << endl;
 
    n = 12;
    isPal(n) ? cout <<"Yes\n": cout <<"No" << endl;
 
    n = 88;
    isPal(n) ? cout <<"Yes\n": cout <<"No" << endl;
 
    n = 8999;
    isPal(n) ? cout <<"Yes\n": cout <<"No";
    return 0;
}
 
// this code is contributed by shivanisinghss2110


C




// A recursive C program to check
// whether a given number
// is palindrome or not
#include <stdio.h>
 
// A function that returns 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
    // beginning, 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)
{
     
    // Check 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
    int* dupNum = new int(num);
 
    return isPalUtil(num, dupNum);
}
 
// Driver program to test
// above functions
int main()
{
    int n = 12321;
    isPal(n) ? printf("Yes\n") : printf("No\n");
 
    n = 12;
    isPal(n) ? printf("Yes\n") : printf("No\n");
 
    n = 88;
    isPal(n) ? printf("Yes\n") : printf("No\n");
 
    n = 8999;
    isPal(n) ? printf("Yes\n") : printf("No\n");
    return 0;
}


Java




// 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 returns 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


Python3




# A recursive Python3 program to check
# whether a given number is palindrome or not
 
# A function that returns 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[0]) % 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 not isPalUtil(num //10, dupNum):
        return False
 
    # The following statements are
    # executed when we move up the
    # recursion call tree
    dupNum[0] = dupNum[0] //10
 
    # At this point, if num%10
    # contains i'th digit from
    # beginning, then (*dupNum)%10
    # contains i'th digit from end
    return (num % 10 == (dupNum[0]) % 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):
    print("Yes")
else:
    print("No")
 
n = 12
if isPal(n) :
    print("Yes")
else:
    print("No")
 
n = 88
if isPal(n) :
    print("Yes")
else:
    print("No")
 
n = 8999
if isPal(n) :
    print("Yes")
else:
    print("No")
 
# This code is contributed by mits


C#




// A recursive C# program to
// check whether a given number
// is palindrome or not
using System;
 
class GFG
{
     
// A function that returns 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
    // beginning, 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


PHP




<?php
// A recursive PHP program to
// check whether a given number
// is palindrome or not
 
// A function that returns 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
    // beginning, 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
?>


Javascript




<script>
// A recursive javascript program to
// check whether a given number
// is palindrome or not
 
    // A function that returns true
    // only if num contains one digit
    function oneDigit(num) {
  
        if ((num >= 0) && (num < 10))
            return 1;
        else
            return 0;
    }
  
    function isPalUtil
    (num , dupNum) {
  
        // base condition to return once we
        // move past first digit
        if (num == 0) {
            return dupNum;
        } else {
            dupNum = isPalUtil(parseInt(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 parseInt(dupNum / 10);
        } else {
            // At position values are not
            // matching throw exception and exit.
            // no need to proceed further.
            throw e;
        }
  
    }
  
    function isPal(num)
    {
  
        if (num < 0)
            num = (-num);
  
        var dupNum = (num);
  
        return isPalUtil(num, dupNum);
    }
  
     
  
    var n = 1242;
    try {
        isPal(n);
        document.write("<br>Yes");
    } catch (e) {
        document.write("<br>No");
    }
    n = 1231;
    try {
        isPal(n);
        document.write("<br>Yes");
    } catch (e) {
        document.write("<br>No");
        }
  
        n = 12;
        try {
            isPal(n);
            document.write("<br>Yes");
    } catch (e) {
        document.write("<br>No");
        }
  
        n = 88;
        try {
            isPal(n);
            document.write("<br>Yes");
    } catch (e) {
        document.write("<br>No");
        }
  
        n = 8999;
        try {
            isPal(n);
            document.write("<br>Yes");
    } catch (e) {
        document.write("<br>No");
    }
 
// This code is contributed by Amit Katiyar
</script>


Output: 

Yes
No
Yes
No 
 

To check a number is palindrome or not without using any extra space
Method #2:Using string() method

  1. When the number of digits of that number exceeds 1018, we can’t take that number as an integer since the range of long long int doesn’t satisfy the given number.
  2. So take input as a string, Run a loop from starting to length/2 and check the first character(numeric) to the last character of the string and second to second last one, and so on ….If any character mismatches, the string wouldn’t be a palindrome.

Below is the implementation of the above approach

C++14




// C++ implementation of the above approach
#include <iostream>
using namespace std;
 
// Function to check palindrome
int checkPalindrome(string str)
{
    // Calculating string length
    int len = str.length();
   
    // Traversing through the string
    // upto half its length
    for (int i = 0; i < len / 2; i++) {
       
        // Comparing i th character
        // from starting and len-i
        // th character from end
        if (str[i] != str[len - i - 1])
            return false;
    }
   
    // If the above loop doesn't return then it is
    // palindrome
    return true;
}
 
// Driver Code
int main()
{ // taking number as string
    string st
        = "112233445566778899000000998877665544332211";
    if (checkPalindrome(st) == true)
        cout << "Yes";
    else
        cout << "No";
    return 0;
}
// this code is written by vikkycirus


Java




// Java implementation of the above approach
import java.io.*;
 
class GFG{
 
// Function to check palindrome
static boolean checkPalindrome(String str)
{
     
    // Calculating string length
    int len = str.length();
 
    // Traversing through the string
    // upto half its length
    for(int i = 0; i < len / 2; i++)
    {
         
        // Comparing i th character
        // from starting and len-i
        // th character from end
        if (str.charAt(i) !=
            str.charAt(len - i - 1))
            return false;
    }
 
    // If the above loop doesn't return then
    // it is palindrome
    return true;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Taking number as string
    String st = "112233445566778899000000998877665544332211";
     
    if (checkPalindrome(st) == true)
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by subhammahato348


Python3




# Python3 implementation of the above approach
 
# function to check palindrome
def checkPalindrome(str):
   
    # Run loop from 0 to len/2
    for i in range(0, len(str)//2):
        if str[i] != str[len(str)-i-1]:
            return False
           
    # If the above loop doesn't
    #return then it is palindrome
    return True
 
 
# Driver code
st = "112233445566778899000000998877665544332211"
if(checkPalindrome(st) == True):
    print("it is a palindrome")
else:
    print("It is not a palindrome")


C#




// C# implementation of the above approach
using System;
 
class GFG{
 
// Function to check palindrome
static bool checkPalindrome(string str)
{
     
    // Calculating string length
    int len = str.Length;
 
    // Traversing through the string
    // upto half its length
    for(int i = 0; i < len / 2; i++)
    {
         
        // Comparing i th character
        // from starting and len-i
        // th character from end
        if (str[i] != str[len - i - 1])
            return false;
    }
 
    // If the above loop doesn't return then
    // it is palindrome
    return true;
}
 
// Driver Code
public static void Main()
{
     
    // Taking number as string
    string st = "112233445566778899000000998877665544332211";
 
    if (checkPalindrome(st) == true)
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by subhammahato348


Javascript




<script>
 
// Javascript implementation of the above approach
 
// Function to check palindrome
function checkPalindrome(str)
{
    // Calculating string length
    var len = str.length;
    
    // Traversing through the string
    // upto half its length
    for (var i = 0; i < len / 2; i++) {
        
        // Comparing ith character
        // from starting and len-ith
        // character from end
        if (str[i] != str[len - i - 1])
            return false;
    }
    
    // If the above loop doesn't return then it is
    // palindrome
    return true;
}
  
// Driver Code
 // taking number as string
    let st
        = "112233445566778899000000998877665544332211";
    if (checkPalindrome(st) == true)
        document.write("Yes");
    else
        document.write("No");
         
// This code is contributed by Mayank Tyagi
 
</script>


Output

Yes

Time Complexity: O(|str|)

Method #3:Using split(), reverse() and join() methods simultaneously 

  1. When user inputs an integer, it is further passed inside a method which will evaluate the result or the actual logic part.
  2. Logic part inside the method focuses on using several important JavaScript methods simultaneously.
  3. First task is to convert the integer passed in into the string using toString() method.
  4. Then split(), reverse() and join() method will be applied to obtain the reverse of that string.
  5. Thereafter using the triple equality operator (“===”), reverse string and original string will be compared and based on that result will be printed on the console.

 Below is the implementation of the above approach

Python3




def checkPalindrome(x):
    convertedNumber = str(x)
    reverseString = convertedNumber[::-1]
    return "Yes" if reverseString == convertedNumber else "No"
 
# Some Testcases...
num = 12321
print(checkPalindrome(num)) # Yes
 
number = 456
print(checkPalindrome(number)) # No
 
# This code is contributed by shinjanpatra


Javascript




function checkPalindrome(x) {
    let convertedNumber = x.toString();
    let reverseString = convertedNumber.split("").reverse().join("");
    return reverseString === convertedNumber ? "Yes" : "No";
}
 
// Some Testcases...
let num = 12321;
console.log(checkPalindrome(num)); // Yes
 
let number = 456;
console.log(checkPalindrome(number)); // No
 
// This code is contributed by Aman Singla....


Output

Yes
No

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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