7. Reverse Integer

Problem¶

Reverse the decimal digits of a signed 32-bit integer and keep the sign; return 0 if the reversed value is outside \([-2^{31}, 2^{31} - 1]\).

Example:

x = -120  ->  reverse = -21
x = 1534236469  ->  0   # reversed value overflows 32-bit signed int

Key trick¶

Detect overflow while building the reversed number digit by digit.

Pop the last digit with % 10 and // 10.
Before doing rev = rev * 10 + digit, check whether that would exceed 32-bit bounds.

Trap¶

Forgetting the overflow check.
Using string reversal and only checking overflow at the end, which ignores the "no 64-bit storage" constraint.
Mishandling negatives because Python % and // behave differently for negative numbers.

Why is it interesting?¶

It tests careful integer manipulation.

You must handle sign, trailing zeros, and overflow cleanly.
The intended solution is language-aware and avoids "easy but not portable" shortcuts.

Python solution¶

class Solution:
    def reverse(self, x: int) -> int:
        INT_MAX = 2**31 - 1

        sign = -1 if x < 0 else 1
        x = abs(x)
        rev = 0

        while x:
            x, digit = divmod(x, 10)

            # If rev * 10 + digit would overflow 32-bit signed range, stop.
            if rev > (INT_MAX - digit) // 10:
                return 0

            rev = rev * 10 + digit

        return sign * rev

Time: \(O(d)\)
Space: \(O(1)\)

Comment on my solution¶

Your solution is correct for the shown examples, but it misses the required overflow handling.

1534236469 should return 0, but your code returns 9646324351.
It uses string conversion, which is fine in Python practice, but not the intended interview trick here.
lstrip("0") or "0" is neat and correctly handles inputs like 0 and 120.

class Solution:
    def reverse(self, x: int) -> int:
        sign = -1 if x < 0 else 1
        x_str_reversed = "".join(reversed(str(abs(x)))).lstrip("0") or "0"
        return sign * int(x_str_reversed)

# "00012300".lstrip("0") # '12300'
# "000".lstrip("0") # ''
# "000".lstrip("0") or "0" # "0"

## Test

import pytest

@pytest.mark.parametrize(
    ("x", "expected"),
    [(123, 321), (-123, -321), (120, 21)]
)
def test_reverse_integer(x, expected):
    assert Solution().reverse(x) == expected