LeetCode has a Medium coding Problem in Its’ Algorithm Section in Python “Reverse Integer”. Today We are going to solve this problem. LeetCode Link of the Problem is HERE

**Question**

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`

, then return ^{31}, 2^{31} - 1]`0`

.

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

**Example**s

Input: x = 123

Output: 321

Input: x = -123

Output: -321

Input: x = 120

Output: 21

**Constraints:**

`-2`

^{31}<= x <= 2^{31}- 1

**Solution** **to Python Reverse Integer**

Firstly, Check for the number either given input is -ve or +ve

Secondly, Declare one more variable reversed_number. Reversed Number storing the Previous Numbers + current Digit. It is the variable which we will return at the end of the program. Start a loop and run it until the given input reaches zero.

Thirdly, Check for the Reverse if it overflows. Remember that Our question condition is that our reverse should also be a 32-bit integer. If the reverse is not a 32-bit integer we should return 0.

Finally, Divide x again until it reaches 0 when x == 0 the Loop will break and we have to return the reversed Number. Check for the Negative flag too, if Negative flag is True return Negative of Number else return Positive.

Complete Solution:

class Solution(object):

def reverse(self, x):

“””

:type x: int

:rtype: int

“””

neg= False

if( x < 0):

neg = True

x = -x

Reversed_Number = 0

while( x != 0):

digit = x % 10

Reversed_Number = (Reversed_Number * 10) + digit

if(Reversed_Number >= 2147483647 or Reversed_Number <= -2147483647):

return 0

x = x // 10

if neg:

return -Reversed_Number

return Reversed_Number

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