:orphan:

Reverse Integer
===============

.. highlight:: none

Problem
-------
https://leetcode.com/problems/reverse-integer/

Given a signed 32-bit integer ``x``, return ``x`` *with its digits
reversed*. If reversing ``x`` causes the value to go outside the signed
32-bit integer range
``[-2``\ :sup:```31```\ ``, 2``\ :sup:```31```\ ``- 1]``, then return
``0``.

**Assume the environment does not allow you to store 64-bit integers
(signed or unsigned).**

 

**Example 1:**

::


   Input: x = 123
   Output: 321

**Example 2:**

::


   Input: x = -123
   Output: -321

**Example 3:**

::


   Input: x = 120
   Output: 21

 

**Constraints:**

- ``-2``\ :sup:```31```\ ``<= x <= 2``\ :sup:```31```\ ``- 1``

.. highlight:: python

Pattern
-------
Math

Solution
--------
Extract the sign of ``x``. We can use modulus and division to get the digits of
``x``. Reverse the list of digits. Then check that the digits are in the range
:math:`[2^{31} - 1, -2^{31}]`. If so, calculate the reversed integer using
the expansion

.. math::

    r = \\sum_{i=0}^{n-1} 10^i d_i.

Code
----

.. literalinclude:: ../problems/medium/reverse-integer/reverse_integer__approach_1.py
    :language: python
    :lines: 12-

Test
----
>>> from reverse_integer__approach_1 import reverse
>>> reverse(123)
321
>>> reverse(-123)
-321
>>> reverse(120)
21

Complexity
----------
| Time: :math:`O(\log_{10} x)` — extract each digit, reverse list of digits, and calculate reversed integer
| Auxiliary Space: :math:`O(\log_{10} x)` — digits list

.. autofunction:: reverse_integer__approach_1.reverse

.. autofunction:: reverse_integer__approach_1.too_large