Mahdi Shamlou | Solving LeetCode #7: Reverse Integer — My Math-Based Reversal with Overflow Safety

Source: Dev.to

Hey everyone! Mahdi Shamlou here — continuing my LeetCode classic problems series 🚀

After solving #6 Zigzag Conversion, today we’re tackling Problem #7 — Reverse Integer — a very common easy/medium question that looks simple at first… until you remember the 32‑bit integer overflow trap!

Problem Statement

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 [-2³¹, 2³¹ − 1], then return 0.

Assume the environment does not allow 64‑bit integers.

Examples

Input:  x = 123
Output: 321

Input:  x = -123
Output: -321

Input:  x = 120
Output: 21

Input:  x = 0
Output: 0

My Solution

Digit‑by‑digit reversal with overflow check. I avoided string conversion and performed the reversal mathematically — popping digits with % 10, building the reversed number, and checking for overflow before multiplying by 10 and adding the digit.

class Solution:
    def reverse(self, x: int) -> int:
        reversed_x = 0

        # Handle the sign separately
        sign = 1 if x >= 0 else -1
        x = abs(x)

        while x > 0:
            digit = x % 10
            x //= 10

            # Check for overflow before adding the new digit
            # 2**31 - 1 = 2147483647
            # -2**31    = -2147483648
            if reversed_x > (2147483647 - digit) // 10:
                return 0
            if reversed_x < (-2147483648 + digit) // 10:
                return 0

            reversed_x = reversed_x * 10 + digit

        return sign * reversed_x

The code passes all test cases and feels clean because we catch overflow before it actually happens.

You can find all my solutions in the repository:

GitHub – mahdi0shamlou/LeetCode: Solve LeetCode Problems

Time & Space Complexity

• Time: O(log |x|) — we process each digit once.
• Space: O(1) — only a few integer variables are used.

Result

After submitting, the overflow cases correctly return 0 and the normal cases reverse perfectly. Here’s a snapshot of the successful submission: