Reverse Integer

Solution
class Solution:
    def reverse(self, x: int) -> int:
        sign = [1, -1][x < 0]
        rev, x = 0, abs(x)
        while x:
            x, mod = divmod(x, 10)
            rev = rev * 10 + mod
            if rev > 2**31 - 1:
                return 0
        return sign * rev

Time Complexity

O(log(x)

Space Complexity

O(1)