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Number Reverse
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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.
Code
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.
