checking integer overflow in python

Question

class Solution(object):
    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """
        negative = False
        if(x < 0):
            x = x * -1
            negative = True
        else:
            x = x
        sum = 0
        dig = 1
        strX = str(x)
        lst = list(strX)
        for i in lst:
            sum += int(i) * dig
            dig *= 10

        if(abs(sum) > 2 ** 32):
            return 0
        elif(negative == True):
            return sum * -1
        else:
            return sum

This is a leetcode problem that asks us to reverse an integer. I know it's a dirty code but it still works but it does not return 0 when the reversed integer overflows. I tried to check that on the line

        if(abs(sum) > 2 ** 32):
            return 0

but one of the test cases gives me:

Input: 1563847412

---

Output: 2147483651

---

Expected: 0

First, I am not sure why this overflows, and I am not sure how to fix this.

Thanks!

Answer 1

change if(abs(sum) > 2 ** 32): to if(abs(sum) > (2 ** 31 - 1)): or abs(sum) > (1 << 31) - 1): The largest 32 bit signed interger is actually not 2^32 but (2 ^ (31)) -1). because we need one bit reserve as the sign bit.

Read about it here of why The number 2,147,483,647 (or hexadecimal 7FFF,FFFF) is the maximum positive value for a 32-bit signed binary integer

Answer 2

I guess some thing light weight like below could perhaps achieve the same logic, For someone else looking , the main overflow check after reversed 32 bit int is

if(abs(n) > (2 ** 31 - 1)):
                    return 0    

Full code below

def reverse(self, x):

            neg = False
            if x < 0:
                neg = True
                x = x * -1

            s = str(x)[::-1]
            n = int(s)
            if neg:
                n = n*-1
            if(abs(n) > (2 ** 31 - 1)):
                return 0
            return n