# Check if a number is Palindrome

Last Updated : 11 Sep, 2023

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

Method 1:

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 ` `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

 `#include ` `#include ` ` `  `// A function that returns true only  ` `// if num contains one digit ` `int` `oneDigit(``int` `num) ` `{ ` `    ``// Comparison operation is faster  ` `    ``// than division operation. ` `    ``// So using the 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 the 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 the 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 the i'th digit from the beginning, ` `    ``// then (*dupNum) % 10 contains the i'th digit from the end. ` `    ``return` `(num % 10 == (*dupNum) % 10); ` `} ` ` `  `// The main function that uses the recursive function ` `// isPalUtil() to find out whether num is palindrome or not. ` `bool` `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 the address dupNum don't change the input number. ` `    ``int` `dupNum = 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` `CheckPalindromeNumberRecursion { ` `  `  `    ``// 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 = ``12421``; ` `        ``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 `

## Javascript

 ` `

## PHP

 `= 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 ` `?> `

Output

```Yes
No
Yes
No
```

Time Complexity: O(log n)
Auxiliary Space: O(log n)

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

• When the number of digits of that number exceeds 1018, we canâ€™t take that number as an integer since the range of long long int doesnâ€™t satisfy the given number.
• 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 â€¦.If any character mismatches, the string wouldnâ€™t be a palindrome.

Below is the implementation of the above approach

## C++14

 `// C++ implementation of the above approach ` `#include ` `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

 ` `

Output

```Yes

```

Time Complexity: O(|str|)
Auxiliary Space: O(1)

Method 3:

Here is the simplest approach to check if a number is Palindrome or not . This approach can be used when the number of digits in the given number is less than 10^18 because if the number of digits of that number exceeds 10^18, we canâ€™t take that number as an integer since the range of long long int doesnâ€™t satisfy the given number.

To check whether the given number is palindrome or not we will just reverse the digits of the given number and check if the reverse of that number is equal to the original number or not . If reverse of number is equal to that number than the number will be Palindrome else it will not a Palindrome.

## C++

 `// C++ program to check if a number is Palindrome ` `#include ` `using` `namespace` `std; ` `// Function to check Palindrome ` `bool` `checkPalindrome(``int` `n) ` `{ ` `    ``int` `reverse = 0; ` `    ``int` `temp = n; ` `    ``while` `(temp != 0) { ` `        ``reverse = (reverse * 10) + (temp % 10); ` `        ``temp = temp / 10; ` `    ``} ` `    ``return` `(reverse ` `            ``== n); ``// if it is true then it will return 1; ` `                   ``// else if false it will return 0; ` `} ` `int` `main() ` `{ ` `    ``int` `n = 7007; ` `    ``if` `(checkPalindrome(n) == 1) { ` `        ``cout << ``"Yes\n"``; ` `    ``} ` `    ``else` `{ ` `        ``cout << ``"No\n"``; ` `    ``} ` `    ``return` `0; ` `} ` `// This code is contributed by Suruchi Kumari`

## Java

 `/*package whatever //do not write package name here */` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` `  ``// Java program to check if a number is Palindrome ` ` `  `  ``// Function to check Palindrome ` `  ``static` `boolean` `checkPalindrome(``int` `n) ` `  ``{ ` `    ``int` `reverse = ``0``; ` `    ``int` `temp = n; ` `    ``while` `(temp != ``0``) { ` `      ``reverse = (reverse * ``10``) + (temp % ``10``); ` `      ``temp = temp / ``10``; ` `    ``} ` `    ``return` `(reverse == n); ``// if it is true then it will return 1; ` `    ``// else if false it will return 0; ` `  ``} ` ` `  `  ``// Driver Code ` `  ``public` `static` `void` `main(String args[]) ` `  ``{ ` `    ``int` `n = ``7007``; ` `    ``if` `(checkPalindrome(n) == ``true``) { ` `      ``System.out.println(``"Yes"``); ` `    ``} ` `    ``else` `{ ` `      ``System.out.println(``"No"``); ` `    ``} ` `  ``} ` `} ` ` `  `// This code is contributed by shinjanpatra`

## Python3

 `# Python3 program to check if a number is Palindrome ` ` `  `# Function to check Palindrome ` `def` `checkPalindrome(n): ` ` `  `    ``reverse ``=` `0` `    ``temp ``=` `n ` `    ``while` `(temp !``=` `0``): ` `        ``reverse ``=` `(reverse ``*` `10``) ``+` `(temp ``%` `10``) ` `        ``temp ``=` `temp ``/``/` `10` `     `  `    ``return` `(reverse ``=``=` `n) ``# if it is true then it will return 1; ` `                   ``# else if false it will return 0; ` ` `  `# driver code ` `n ``=` `7007` `if` `(checkPalindrome(n) ``=``=` `1``): ` `    ``print``(``"Yes"``) ` ` `  `else``: ` `    ``print``(``"No"``) ` ` `  `# This code is contributed by shinjanpatra`

## C#

 `// C# program to check if a number is Palindrome ` ` `  `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to check Palindrome ` `    ``static` `bool` `checkPalindrome(``int` `n) ` `    ``{ ` `        ``int` `reverse = 0; ` `        ``int` `temp = n; ` `        ``while` `(temp != 0) { ` `            ``reverse = (reverse * 10) + (temp % 10); ` `            ``temp = temp / 10; ` `        ``} ` `        ``return` `( ` `            ``reverse ` `            ``== n); ``// if it is true then it will return 1; ` `        ``// else if false it will return 0; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``int` `n = 7007; ` `        ``if` `(checkPalindrome(n) == ``true``) { ` `            ``Console.WriteLine(``"Yes"``); ` `        ``} ` `        ``else` `{ ` `            ``Console.WriteLine(``"No"``); ` `        ``} ` `    ``} ` `} ` ` `  `// This code is contributed by phasing17`

## Javascript

 ``

Output

```Yes

```

Time Complexity : O(log10(n)) or O(Number of digits in a given number)
Auxiliary space : O(1) or constant