Reverse a number using stack
The problem is to reverse a given number using a stack. The goal is to reverse the digits of the number, maintaining the sign if it is a negative number, and ignoring leading zeroes if present in the original number.
Idea to Solve the Problem
To reverse the number using a stack, we will follow these steps:
 Create an empty custom stack
s
to store the digits of the number.  Extract the last digit of the number and push it onto the stack
s
.  Remove the last digit from the number.
 Repeat steps 2 and 3 until the number becomes zero.
 Pop digits from the stack
s
one by one and combine them to form the reversed number.
Algorithm
 Create a custom stack data structure with push, pop, isEmpty, and peek operations.
 Define a function
reverse
that takes an integerx
as input and returns the reversed number.  Create an empty stack
s
to store the digits of the number.  While
x
is not equal to zero, do the following: a. Get the last digit ofx
by takingx % 10
. b. Push the last digit onto the stacks
. c. Remove the last digit fromx
by updatingx = x / 10
.  Initialize a variable
multiply
to 1.  While the stack
s
is not empty, do the following: a. Pop the top element from the stacks
. b. Multiply the popped element bymultiply
and add it tox
. c. Multiplymultiply
by 10.  Return the reversed number
x
.
Pseudocode
FUNCTION reverse(x):
num = x
s = new MyStack()
WHILE num is not 0:
s.push(num % 10)
num = num / 10
multiply = 1
WHILE s is not empty:
num = s.pop() * multiply + num
multiply = multiply * 10
RETURN num
Code Solution

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Time Complexity Analysis
Let n be the number of digits in the input number. The time complexity of the reverse
function is
O(n) because we need to iterate through each digit of the number and perform stack operations.
Output Explanation
The output shows the number before and after reversing using the reverse
function. For example,
"Before 381725 After 527183" represents the original number 381725, and the reversed number is 527183.
Similarly, for other test cases, the numbers are reversed accordingly. Note that leading zeroes are ignored in
the reversed number, as shown in the output "Before 12340 After 4321".
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