Reverse a number in C

reverse number in C
Reverse a number in C

Reverse a number in C.

Problem Statement: Given a number N reverse the number and print it.

Examples:

Example 1:
Input: N = 123
Output: 321
Explanation: The reverse of 123 is 321

Example 2:
Input: N = 234
Output: 432
Explanation: The reverse of 234 is 432

DisclaimerDon’t jump directly to the solution, try it out yourself first.

The idea is to extract digits from the end of the given number and create a new number in reverse order.

How to extract digits from the end of a number?

To extract the last digit, if you divide a number by 10, then the remainder will be the last digit. We can simply use the modulo(%) operator to do this, for example the last digit of 123 will be (123 % 10), which is 3.

To reduce the number by one digit from the end, simply divide the number by 10. example: to reduce 123 to 12, simply do (123/10) which is equal to 12.

To create a number from digits: The idea is to start with 0, and for every digit, multiply the number generated so far by 10, and add the digit to it.

For example, to create a number from digits: [1,2,3]:

Consider the number, num = 0.

Then,

For first digit: 1
num = num*10 + 1 = 0*10 + 1 = 1;

For second digit: 2
num = num*10 + 2 = 1*10 + 2 = 12;

For third digit: 3
num = num*10 + 2 = 12*10 + 3 = 123;

Approach:

  • Run a while loop until the given number N is equal to zero while(N!=0)
  • Initialize a variable reverse = 0;
  • now in each step take the remainder of the given number N and store it as a variable digit, digit = N % 10
  • Also, Divide the number by 10. N= N / 10
  • in each step, the variable reverse get updated as reverse = reverse*10+digit.

Code:

C Program

#include<stdio.h>

int main()
{
    int N = 123;
    int num = N;
    int reverse = 0;
    while(N!=0)
    {
        int digit = N%10;
        reverse = reverse*10+digit;
        N = N/10;
    }
    printf("The reverse of the %d is %d",num,reverse);
}

Output: The reverse of the 123 is 321

Time Complexity: O(n), where n is the length of the given number

Space Complexity: O(1)

Python Code

if __name__ == '__main__':
    N = 123
    num = N
    reverse = 0
    while N != 0:
        digit = N % 10
        reverse = reverse * 10 + digit
        N = N // 10
    print("The reverse of the {} is {}".format(num, reverse))

Output: The reverse of the 123 is 321

Time Complexity: O(n), where n is the length of the given number

Space Complexity: O(1)

Special thanks to Subhrajit Das for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article