## Number Reverse

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### Description

The program is written here to reverse a given number.

• First, the number is taken as input.
• Using the modulus operator each time the last digit of the number at that moment is taken out and appended to the new number in reverse order.
• The number is now gets divided by 10 so that the last digit is cut.

The process is continued till the original number becomes 0.

For example, if a given number is 1234 then the reverse of this number is 4321.

### Algorithm

INPUT: A number

OUTPUT: The reverse of the number.

Step 1: [taking the input]

Read n [the number to be reversed]

Step 2: [reversing the number]

Set rev<-0

While n>0 repeat

Set rev<-rev*10+(n mod 10)

Set n<-n/10

[End of ‘while’ loop]

Step 3: Print “The reverse of the number is: rev”

Step 4: Stop.

## TIME COMPLEXITY:

while(n>0)----------------------------------O(m)

{

rev=rev*10+(n%10);

n=n/10;

}

The time complexity is O(m) where ‘m’ is the number of digits of the given input.

The time complexity can also be represented as O(log10n) where n is the given input. There is almost log10n number of digits in a number n.

## SPACE COMPLEXITY:

The space complexity of this program is O(1) as it requires a constant number of memory spaces to execute the program for any given input.