Input
- The code takes an integer input
n
from the user.
Initialization
- It initializes a variable
given_num
with the value ofn
. This variable is used to store the original value ofn
becausen
will be modified during the process.
check_palindrome
Function
- The
check_palindrome
function is defined to determine whether a number is a palindrome. - It initializes a variable
rev_num
to 0. This variable will be used to store the reverse ofn
.
Reversing n
- The code enters a
while
loop that continues as long asn
is greater than 0. - In each iteration of the loop:
- It calculates the last digit of
n
using the modulo operation (n % 10
) and stores it in thelast_digit
variable. - It updates
n
by removing the last digit using integer division (n //= 10
). - It updates
rev_num
by appending thelast_digit
to the right of the currentrev_num
. This is done by multiplyingrev_num
by 10 and addinglast_digit
to it (rev_num = (rev_num * 10) + last_digit
). - This process effectively reverses the digits of
n
and stores the result inrev_num
.
- It calculates the last digit of
Checking for Palindrome
- After the loop completes,
rev_num
contains the reversed value of the original numbern
. - It then checks whether
rev_num
is equal to thegiven_num
(the original number) using the equality operator (rev_num == given_num
). - The result of this comparison is a Boolean value (
True
ifn
is a palindrome,False
otherwise). - To match the expected output format mentioned in the problem statement, the Boolean value is first converted to a string using
str()
, and then it is converted to lowercase using.lower()
.
Printing the Result
- Finally, the code prints the result of the
check_palindrome
function, which will be either "true" (ifn
is a palindrome) or "false" (ifn
is not a palindrome).
The provided code is designed to check whether an input integer n
is a palindrome by reversing its digits. Let's analyze its time complexity:
-
Input: The code starts by taking an integer input
n
from the user, which is a constant-time operation and does not affect the overall time complexity. -
check_palindrome
Function:- The
check_palindrome
function contains awhile
loop for reversing the digits of the input integern
. - In each iteration of the loop, the following operations are performed:
- Calculation of the last digit of
n
using the modulo operation (n % 10
). - Integer division to remove the last digit from
n
(n //= 10
). - Update of
rev_num
by appending thelast_digit
to the right of the currentrev_num
using multiplication and addition (rev_num = (rev_num * 10) + last_digit
).
- Calculation of the last digit of
- The
-
The number of iterations in the
while
loop depends on the number of digits in the input integern
. For an integer withd
digits, the loop will run approximatelyd
times untiln
becomes zero. -
Therefore, the time complexity of the code is proportional to the number of digits in
n
, which can be represented as$O(log10(n))$ .
In summary, the time complexity of the given code is n
is the input integer. This reflects the logarithmic growth in the number of iterations as the input integer's digits increase in magnitude.
-
Initialization: The code initializes a single integer variable
revNum
to store the reversed number. This variable does not depend on the input sizen
, so its space complexity is constant (O(1)). -
Loop Variables:
- The code uses a loop to iterate through the digits of
n
. Inside the loop, it creates a few temporary variables:last_digit
: To hold the value of the last digit extracted in each iteration. This variable is also an integer and takes constant space (O(1)).- The loop does not create any new data structures or allocate memory based on the input size. Instead, it uses these variables to perform operations.
- The code uses a loop to iterate through the digits of
The space complexity of this code is primarily determined by the constant amount of memory used for the variables revNum
and last_digit
, both of which are integers.
Therefore, the space complexity is O(1), indicating that the memory requirements of the algorithm do not grow with the size of the input n
. The code uses a fixed amount of memory regardless of the value of n
.
In summary, the space complexity of this code is constant, denoted as O(1).