Write a program to reverse digits of a number

Write a program to reverse the digits of an integer.

Examples :

Input : num = 12345
Output: 54321

Input : num = 876
Output: 678

Flowchart:

ITERATIVE WAY

Algorithm:

Input:  num
(1) Initialize rev_num = 0
(2) Loop while num > 0
     (a) Multiply rev_num by 10 and add remainder of num  
          divide by 10 to rev_num
               rev_num = rev_num*10 + num%10;
     (b) Divide num by 10
(3) Return rev_num

Example:

num = 4562
rev_num = 0
rev_num = rev_num *10 + num%10 = 2
num = num/10 = 456
rev_num = rev_num *10 + num%10 = 20 + 6 = 26
num = num/10 = 45
rev_num = rev_num *10 + num%10 = 260 + 5 = 265
num = num/10 = 4
rev_num = rev_num *10 + num%10 = 2650 + 4 = 2654
num = num/10 = 0

Program:

C

#include <stdio.h>
 
/* Iterative function to reverse digits of num*/
int reverseDigits(int num)
{
    int rev_num = 0;
    while (num > 0) {
        rev_num = rev_num * 10 + num % 10;
        num = num / 10;
    }
    return rev_num;
}
 
/*Driver program to test reverseDigits*/
int main()
{
    int num = 4562;
    printf("Reverse of no. is %d", reverseDigits(num));
 
    getchar();
    return 0;
}

C++

#include <bits/stdc++.h>
 
using namespace std;
/* Iterative function to reverse digits of num*/
int reverseDigits(int num)
{
    int rev_num = 0;
    while (num > 0) {
        rev_num = rev_num * 10 + num % 10;
        num = num / 10;
    }
    return rev_num;
}
 
/*Driver program to test reverseDigits*/
int main()
{
    int num = 4562;
    cout << "Reverse of no. is " << reverseDigits(num);
    getchar();
    return 0;
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

Java

// Java program to reverse a number
 
class GFG {
    /* Iterative function to reverse
    digits of num*/
    static int reverseDigits(int num)
    {
        int rev_num = 0;
        while (num > 0) {
            rev_num = rev_num * 10 + num % 10;
            num = num / 10;
        }
        return rev_num;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int num = 4562;
        System.out.println("Reverse of no. is "
                           + reverseDigits(num));
    }
}
 
// This code is contributed by Anant Agarwal.

Python

# Python program to reverse a number
 
n = 4562
rev = 0
 
while(n > 0):
    a = n % 10
    rev = rev * 10 + a
    n = n // 10
 
print(rev)
 
# This code is contributed by Shariq Raza

C#

// C# program to reverse a number
using System;
 
class GFG {
    // Iterative function to
    // reverse digits of num
    static int reverseDigits(int num)
    {
        int rev_num = 0;
        while (num > 0) {
            rev_num = rev_num * 10 + num % 10;
            num = num / 10;
        }
        return rev_num;
    }
 
    // Driver code
    public static void Main()
    {
        int num = 4562;
        Console.Write("Reverse of no. is "
                      + reverseDigits(num));
    }
}
 
// This code is contributed by Sam007

PHP

<?php
// Iterative function to
// reverse digits of num
function reverseDigits($num)
{
    $rev_num = 0;
    while($num > 1)
    {
        $rev_num = $rev_num * 10 +
                        $num % 10;
        $num = (int)$num / 10;
    }
    return $rev_num;
}
 
// Driver Code
$num = 4562;
echo "Reverse of no. is ",
       reverseDigits($num);
 
// This code is contributed by aj_36
?>

Javascript

<script>
    let num = 4562;
    // Function to reverse digits of num
    function reverseDigits(num) {
        let rev_num = 0;
        while(num > 0)
        {
            rev_num = rev_num * 10 + num % 10;
            num = Math.floor(num / 10);
        }
        return rev_num;
    }
 // function call  
    document.write(reverseDigits(num));
     
// This code is contributed by Surbhi tyagi
 
</script>

Output

Reverse of no. is 2654

Time Complexity: O(log₁₀ n), where n is the input number.
Auxiliary Space: O(1)

RECURSIVE WAY :

C

// C program to reverse digits of a number
#include <stdio.h>
 
/* Recursive function to reverse digits of num*/
int reverseDigits(int num)
{
    static int rev_num = 0;
    static int base_pos = 1;
    if (num > 0) {
        reverseDigits(num / 10);
        rev_num += (num % 10) * base_pos;
        base_pos *= 10;
    }
    return rev_num;
}
 
/*Driver program to test reverse Digits*/
int main()
{
    int num = 4562;
    printf("Reverse of no. is %d", reverseDigits(num));
 
    getchar();
    return 0;
}

C++

// C++ program to reverse digits of a number
#include <bits/stdc++.h>
using namespace std;
/* Recursive function to reverse digits of num*/
int reverseDigits(int num)
{
    static int rev_num = 0;
    static int base_pos = 1;
    if (num > 0) {
        reverseDigits(num / 10);
        rev_num += (num % 10) * base_pos;
        base_pos *= 10;
    }
    return rev_num;
}
 
// Driver Code
int main()
{
    int num = 4562;
    cout << "Reverse of no. is " << reverseDigits(num);
 
    return 0;
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

Java

// Java program to reverse digits of a number
 
// Recursive function to
// reverse digits of num
class GFG {
    static int rev_num = 0;
    static int base_pos = 1;
    static int reverseDigits(int num)
    {
        if (num > 0) {
            reverseDigits(num / 10);
            rev_num += (num % 10) * base_pos;
            base_pos *= 10;
        }
        return rev_num;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int num = 4562;
        System.out.println(reverseDigits(num));
    }
}
 
// This code is contributed by mits

Python3

# Python 3 program to reverse digits
# of a number
rev_num = 0
base_pos = 1
 
# Recursive function to reverse
# digits of num
 
 
def reverseDigits(num):
    global rev_num
    global base_pos
    if(num > 0):
        reverseDigits((int)(num / 10))
        rev_num += (num % 10) * base_pos
        base_pos *= 10
    return rev_num
 
 
# Driver Code
num = 4562
print("Reverse of no. is ",
      reverseDigits(num))
 
# This code is contributed by Rajput-Ji

C#

// C# program to reverse digits of a number
 
// Recursive function to
// reverse digits of num
using System;
class GFG {
    static int rev_num = 0;
    static int base_pos = 1;
    static int reverseDigits(int num)
    {
        if (num > 0) {
            reverseDigits(num / 10);
            rev_num += (num % 10) * base_pos;
            base_pos *= 10;
        }
        return rev_num;
    }
 
    // Driver Code
    public static void Main()
    {
        int num = 4562;
        Console.WriteLine(reverseDigits(num));
    }
}
 
// This code is contributed
// by inder_verma

PHP

<?php
// PHP program to reverse digits of a number
$rev_num = 0;
$base_pos = 1;
 
/* Recursive function to
reverse digits of num*/
function reversDigits($num)
{
    global $rev_num;
    global $base_pos;
    if($num > 0)
    {
        reverseDigits((int)($num / 10));
        $rev_num += ($num % 10) *
                     $base_pos;
        $base_pos *= 10;
    }
    return $rev_num;
}
 
// Driver Code
$num = 4562;
echo "Reverse of no. is ",
       reverseDigits($num);
 
// This code is contributed by ajit
?>

Javascript

<script>
 
// Javascript program to reverse digits of a number
 
/* Recursive function to reverse digits of num*/
var rev_num = 0;
var base_pos = 1;
function reversDigits(num)
{
 
    if(num > 0)
    {
        reverseDigits(Math.floor(num/10));
        rev_num += (num%10)*base_pos;
        base_pos *= 10;
    }
    return rev_num;
}
 
// Driver Code
    let num = 4562;
    document.write("Reverse of no. is "
        + reverseDigits(num));
 
// This code is contributed
// by Mayank Tyagi
 
 
</script>

Output

Reverse of no. is 2654

Time Complexity: O(log(n))
Auxiliary Space: O(log(n)), where n is the input number.

Using Recursion, without extra variable

C++

#include <iostream>
#include <math.h>
using namespace std;
 
// util functions
int len(int number)
{
    // returns the length of a given number
    return log10(number) + 1;
}
 
// reverse a given number
int rev_number(int& number)
{
    if ((number % 10) == number)
        return number;
    int last = number % 10;
    int remaining = number / 10;
    int l = len(remaining);
    return last * pow(10, l) + rev_number(remaining);
}
 
int main()
{
    int number = 123456;
    cout << rev_number(number) << endl;
    return 0;
}

Java

import java.lang.Math;
import java.util.*;
 
class GFG {
 
    public static int len(int number)
    {
        // returns the length of a given number
        return (int)(Math.log10(number)) + 1;
    }
 
    // reverse a given number
    public static int rev_number(int number)
    {
        if (number % 10 == number) {
            return number;
        }
 
        int last = number % 10;
        int remaining = number / 10;
        int l = len(remaining);
        return last * (int)Math.pow(10, l)
            + rev_number(remaining);
    }
 
    public static void main(String[] args)
    {
        int number = 123456;
        System.out.println(rev_number(number));
    }
}
 
// This code is contributed by talktoanmol.

Python3

# util functions
def number_length(num):
 
  # Return length of given number
    return len(str(num))
 
# reverse a given number
 
 
def rev_number(number):
    if (number % 10) == number:
        return number
 
    last = number % 10
    remaining = number // 10
    l = number_length(remaining)
 
    return last*pow(10, l) + rev_number(remaining)
 
 
def main():
    number = 123456
    print(rev_number(number))
 
 
# driver code
if __name__ == "__main__":
    main()
 
    # This code is contributed by talktoanmol

C#

// C# program to reverse a number
using System;
 
public class GFG {
 
    public static int length(int num)
    {
        // returns the length of a given number
        return (int)Math.Log10(num) + 1;
    }
 
    // reverse a given number
    static int rev_number(int num)
    {
        if ((num % 10) == num)
            return num;
        int last = num % 10;
        int remaining = num / 10;
        int l = length(remaining);
        return last * (int)Math.Pow(10, l)
            + rev_number(remaining);
    }
 
    // Driver code
    static public void Main()
    {
        int num = 123456;
        Console.Write(rev_number(num));
    }
}
 
// This Code is contributed by Susobhan Akhuli

Javascript

<script>
// Javascript program to reverse digits of a number
 
// Function to reverse digits of num
 
function length(number){
    // returns the length of a given number
    return ~~(Math.log10(number))+1;
}
 
// reverse a given number
function rev_number(num){
    if ((num % 10) == num)
        return num;
    let last = ~~(num % 10);
    let remain = ~~(num / 10);
    let l = length(remain);
    return last * ~~(Math.pow(10, l)) + rev_number(remain);
}
 
 
let num = 123456;
// function call
document.write(rev_number(num));
     
// This code is contributed by Susobhan Akhuli
 
</script>

PHP

<?php
// PHP program to reverse digits of a number
 
// Function to reverse digits of num
function length($num){
    // returns the length of a given number
    return (int)(log10($num))+1;
}
 
// reverse a given number
function rev_number($num){
    if (($num % 10) == $num)
        return $num;
    $last = ($num % 10);
    $remain = (int)($num / 10);
    $l = length($remain);
    return $last * (int)(pow(10, $l)) + rev_number($remain);
}
 
// Driver Code
$num = 123456;
// function call
echo rev_number($num);
 
// This code is contributed by Susobhan Akhuli
?>

Output

Time Complexity: O(log₁₀n) where n is the given input number.
Auxiliary Space: O(log₁₀n) for recursive stack space.

Using String in java

We will convert the number to a string using StringBuffer after this, we will reverse that string using the reverse() method

corner case

Input: 32100
So for the above input if we try to solve this by reversing the string, then the output will be 00123.
So to deal with this situation we again need to convert the string to integer so that our output will be 123

C

// C program to reverse a number
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
// reverse string function
void reverse(char* begin, char* end)
{
    char temp;
    while (begin < end) {
        temp = *begin;
        *begin++ = *end;
        *end-- = temp;
    }
}
 
// Function to reverse words
void reverseWords(char* s)
{
    char* word_begin = s;
 
    // Word boundary
    char* temp = s;
 
    // Reversing individual words as
    // explained in the first step
    while (*temp) {
        temp++;
        if (*temp == '\0') {
            reverse(word_begin, temp - 1);
        }
        else if (*temp == ' ') {
            reverse(word_begin, temp - 1);
            word_begin = temp + 1;
        }
    }
 
    // Reverse the entire string
    reverse(s, temp - 1);
}
 
int reverseDigits(int num)
{
    char strin[100];
    // converting number to string
    sprintf(strin, "%d", num);
 
    // reversing the string
    reverseWords(strin);
 
    // converting string to integer
    num = atoi(strin);
 
    // returning integer
    return num;
}
 
int main()
{
    int num = 123456;
    printf("Reverse of no. is %d", reverseDigits(num));
    return 0;
}
 
// This Code is contributed by Susobhan Akhuli

C++

// C++ program to reverse a number
#include <bits/stdc++.h>
using namespace std;
 
int reverseDigits(int num)
{
    // converting number to string
    string strin = to_string(num);
 
    // reversing the string
    reverse(strin.begin(), strin.end());
 
    // converting string to integer
    num = stoi(strin);
 
    // returning integer
    return num;
}
int main()
{
    int num = 4562;
    cout << "Reverse of no. is " << reverseDigits(num);
    return 0;
}
 
// This Code is contributed by ShubhamSingh10

Java

// Java program to reverse a number
 
public class GFG {
    static int reversDigits(int num)
    {
        // converting number to string
        StringBuffer string
            = new StringBuffer(String.valueOf(num));
 
        // reversing the string
        string.reverse();
 
        // converting string to integer
        num = Integer.parseInt(String.valueOf(string));
 
        // returning integer
        return num;
    }
    public static void main(String[] args)
    {
        int num = 4562;
        System.out.println("Reverse of no. is "
                           + reversDigits(num));
    }
}

Python3

# Python 3 program to reverse a number
 
 
def reversDigits(num):
 
    # converting number to string
    string = str(num)
 
    # reversing the string
    string = list(string)
    string.reverse()
    string = ''.join(string)
 
    # converting string to integer
    num = int(string)
 
    # returning integer
    return num
 
 
# Driver code
if __name__ == "__main__":
 
    num = 4562
    print("Reverse of no. is ", reversDigits(num))
 
    # This code is contributed by ukasp.

C#

// C# program to reverse a number
using System;
 
public class GFG {
 
    public static string ReverseString(string s)
    {
        char[] array = s.ToCharArray();
        Array.Reverse(array);
        return new string(array);
    }
 
    static int reversDigits(int num)
    {
        // converting number to string
        string strin = num.ToString();
 
        // reversing the string
        strin = ReverseString(strin);
 
        // converting string to integer
        num = int.Parse(strin);
 
        // returning integer
        return num;
    }
 
    // Driver code
    static public void Main()
    {
        int num = 4562;
        Console.Write("Reverse of no. is "
                      + reversDigits(num));
    }
}
 
// This Code is contributed by ShubhamSingh10

Javascript

<script>
 
// Javascript program to reverse a number
 
    function reversDigits(num)
    {
        // converting number to string
        let str
            = num.toString().split("").reverse().join("");
         
 
        // converting string to integer
        num = parseInt(str);
 
 
        // returning integer
        return str;
    }
 
// Driver Code
     
    let num = 4562;
    document.write("Reverse of no. is "
                           + reversDigits(num));
 
</script>

PHP

<?php
 
// PHP program to reverse a number
 
function reverseDigits($num)
{
    // converting number to string
    $strin = strval($num);
 
    // reversing the string
    $strin = strrev($strin);
 
    // converting string to integer
    $num = (int)($strin);
 
    // returning integer
    return $num;
}
 
$num = 123456;
echo "Reverse of no. is ". reverseDigits($num);
 
// This Code is contributed by Susobhan Akhuli
?>

Output

Reverse of no. is 123456

Time Complexity: O(log₁₀n) where n is the input number
Auxiliary Space: O(1)

Reverse digits of an integer with overflow handled

Note that the above program doesn’t consider leading zeroes. For example, for 100 programs will print 1. If you want to print 001 then see this comment from Maheshwar.

Using Slicing in Python:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
int reverseDigits(int num)
{
 
    // converting number to string
    string str = to_string(num);
 
    // reversing the string
    reverse(str.begin(), str.end());
 
    // converting string to integer
    num = stoll(str);
 
    // returning integer
    return num;
}
 
// Driver Code
int main()
{
    int num = 4562;
    cout << "Reverse of no. is " << reverseDigits(num);
 
    return 0;
}
 
// This code is contributed by sanjoy_62.

Java

// Java program for the above approach
 
import java.util.*;
 
class GFG {
 
    static int reversDigits(int num)
    {
 
        // converting number to string
        String str = String.valueOf(num);
 
        // reversing the string
        str = new StringBuilder(str).reverse().toString();
 
        // converting string to integer
        num = Integer.valueOf(str);
 
        // returning integer
        return num;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int num = 4562;
        System.out.println("Reverse of no. is "
                           + reversDigits(num));
    }
}
 
// This code is contributed by phasing17

Python3

# Python 3 program to reverse a number
 
 
def reversDigits(num):
 
    # converting number to string
    string = str(num)
 
    # reversing the string
    string = string[::-1]
 
    # converting string to integer
    num = int(string)
 
    # returning integer
    return num
 
 
# Driver code
if __name__ == "__main__":
 
    num = 4562
    print("Reverse of no. is ", reversDigits(num))
 
    # This code is contributed by Susobhan Akhuli

C#

// C# program for the above approach
using System;
class GFG {
 
    static int reversDigits(int num)
    {
 
        // converting number to string
        string str = Convert.ToString(num);
 
        // reversing the string
        char[] charArray = str.ToCharArray();
        Array.Reverse(charArray);
        str = new string(charArray);
 
        // converting string to integer
        num = Convert.ToInt32(str);
 
        // returning integer
        return num;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int num = 4562;
        Console.Write("Reverse of no. is "
                      + reversDigits(num));
    }
}
 
// This code is contributed by phasing17

Javascript

// JavaScript program for the above approach
 
function reversDigits(num){
 
  // converting number to string
  let str = num.toString().split("");
 
  // reversing the string
  str.reverse();
   
 
  // converting string to integer
  num = parseInt(str.join(""))
 
  // returning integer
  return num;
}
 
// Driver Code
let num = 4562;
console.log("Reverse of no. is " + reversDigits(num));
 
 
// This code is contributed by phasing17

Output

Reverse of no. is 2654

Time Complexity: O(n), where n is the input number
Auxiliary Space: O(1)

Try extensions of above functions that should also work for floating-point numbers.

Divide and Conquer:

Algorithm:

Input: n
(1) Initialize rev1=0, rev2=0
(2) Compute no of digits in given input n and store it in size variable.
(3) Divide the number n into two parts i.e first_half and second_half.
(4) Loop for i in range(mid):
    (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1.
             rem=first_half%10
             rev1=10*rev1+rem
    (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2.
             rem=second_half%10
             rev2=10*rev2+rem
    (c) Divide first_half and second_half with 10.
             first_half=first_half//10
             second_half=second_half//10
 (5)if size is even.
             return rev2*10**mid+rev1
    otherwise 
             return (rev2*10**mid+rev1)*10+first_half

C++

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    long int n,first_half,second_half,rem,rev1,rev2,size,mid;
    n=1234;
    rev1=0,rev2=0;
 
//Find no of digits in number store it in size and divide size into two parts.
    size=floor(log10(n)+1);
    mid=size/2;
 
//Divide the given number in to two parts.
    first_half=n/pow(10,mid);
    second_half=n%(long int)(pow(10,mid)+0.5);
 
//Iterate the loop upto half no of digits in number.
    for(int i=0; i<mid; i++)
    {
//Reverse the First half digits and second half digits concurrently
        rem=first_half%10;
        rev1=10*rev1+rem;
        rem=second_half%10;
        rev2=10*rev2+rem;
        first_half=first_half/10;
        second_half=second_half/10;
    }
cout<<"Original no: "<<n<<endl;
//Concate Last half with First Half
    if(size%2==0)
    {
        cout<<"Reversed no: "<<rev2*(long int)(pow(10,mid)+0.5)+rev1;
    }
    else{
        cout<<"Reversed no: "<<(rev2*(long int)(pow(10,mid)+0.5)+rev1)*10+first_half;
    }
    return 0;
}
 
//This code contributed by SR.DHANUSH

Java

import java.lang.Math;
class Main
{
    public static void main (String[] args)
    {
    int n,first_half,second_half,rem,rev1=0,rev2=0,size,mid;
    n=1234;
//Find no of digits in number store it in size and divide size into two parts.
    size=(int) Math.log10(n)+1;
    mid=(int) size/2;
 
//Divide the given number in to two parts.
    first_half=(int) (n/Math.pow(10,mid));
    second_half=n%(int) (Math.pow(10,mid));  
     
//Iterate the loop upto half no of digits in number.
    for(int i=0; i<mid; i++)
    {
//Reverse the First half digits and second half digits concurrently
        rem=first_half%10;
        rev1=10*rev1+rem;
        rem=second_half%10;
        rev2=10*rev2+rem;
        first_half=first_half/10;
        second_half=second_half/10;
    }
System.out.println("Original no: "+n);
//Concate Last half with First Half
    if(size%2==0)
    {
        System.out.println("Reversed no: "+(rev2*(int)(Math.pow(10,mid))+rev1));
    }
    else{
        System.out.println("Reversed no: "+((rev2*(int)(Math.pow(10,mid)+0.5)+rev1)*10+first_half));
    }
//This code contributed by SR.DHANUSH
}
}

Python3

# Python program to reverse a number.
import math
n = 1234
rev1 = 0
rev2 = 0
 
# Find no of digits in number store it in size and divide size into two parts.
size = int(math.log10(n))+1
mid = size//2
 
# Divide the given number in to two parts.
first_half = n//10**mid
second_half = n % 10**mid
 
# Iterate the loop upto half no of digits in number.
for i in range(mid):
# Reverse the First half digits and second half digits concurrently
    rem = first_half % 10
    rev1 = 10*rev1+rem
    rem = second_half % 10
    rev2 = 10*rev2+rem
    first_half = first_half//10
    second_half = second_half//10
     
# Concate Last half with First Half
print('Original no:', n)
if(size % 2 == 0):
    print('Reversed no:', rev2*10**mid+rev1)
else:
    print('Reversed no:', (rev2*10**mid+rev1)*10+first_half)
     
#This code contributed by SR DHANUSH

C#

// C# program to reverse a number.
using System;
 
class Program
{
    static void Main(string[] args)
    {
        int first_Half, second_Half, rem, rev1 = 0, rev2 = 0, size, mid;
        int n = 1234;
        //Find no of digits in number store it in size and divide size into two parts.
        size = (int)Math.Log10(n) + 1;
        mid = size / 2;
         
        //Divide the given number in to two parts.
        first_Half = (int)(n / Math.Pow(10, mid));
        second_Half = n % (int)(Math.Pow(10, mid));
         
        //Iterate the loop upto half no of digits in number.
        for (int i = 0; i < mid; i++)
        {
            //Reverse the First half digits and second half digits concurrently
            rem = first_Half % 10;
            rev1 = 10 * rev1 + rem;
            rem = second_Half % 10;
            rev2 = 10 * rev2 + rem;
            first_Half /= 10;
            second_Half /= 10;
        }
 
        Console.WriteLine("Original no: " + n);
         
        //Concate Last half with First Half
        if (size % 2 == 0)
        {
            Console.WriteLine("Reversed no: " + (rev2 * (int)(Math.Pow(10, mid)) + rev1));
        }
        else
        {
            Console.WriteLine("Reversed no: " + ((rev2 * (int)(Math.Pow(10, mid) + 0.5) + rev1) * 10 + first_Half));
        }
    }
}
 
// This code is contributed by Aman Kumar

Javascript

// JS code to implement the approach
 
// Find the number of digits in the number and divide it in
// half
let n = 1234;
let rev1 = 0;
let rev2 = 0;
let size = Math.floor(Math.log10(n)) + 1;
let mid = Math.floor(size / 2);
 
// Divide the number into two parts
let firstHalf = Math.floor(n / 10 ** mid);
let secondHalf = n % 10 ** mid;
 
// Reverse the first half and second half digits
// concurrently
for (let i = 0; i < mid; i++) {
    let rem = firstHalf % 10;
    rev1 = rev1 * 10 + rem;
    let rem2 = secondHalf % 10;
    rev2 = rev2 * 10 + rem2;
    firstHalf = Math.floor(firstHalf / 10);
    secondHalf = Math.floor(secondHalf / 10);
}
 
// Concatenate the last half with the first half
console.log("Original no:", n);
if (size % 2 == 0) {
    console.log("Reversed no:", rev2 * 10 ** mid + rev1);
}
else {
    console.log("Reversed no:",
                (rev2 * 10 ** mid + rev1) * 10 + firstHalf);
}
 
// This code is contributed by phasing17