LeetCode 405: Convert a Number to Hexadecimal ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2022-08-23

Submissions: Runtime Error ×2 → Wrong Answer ×2 → Accepted

Approach
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Each hex digit is 4 bits (Bit Manipulation; also base conversion from Math Template 7). Negatives use 32-bit two’s complement, and num & 0xffffffff (adding 2^32) gives the matching non-negative value. Then repeatedly look up num & 15 and shift num >>= 4 until 0. For num = 0 the loop never runs, so return "0".

My Attempts
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Attempt 1 looped while num > 15, dropping the last digit, and returned an empty string for num = 0.
After adding the 0 case, negatives were unhandled: -9 returned an empty string instead of "fffffff7". Adding num += 2**32 passed.

Solution
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class Solution:
    def toHex(self, num: int) -> str:
        if num == 0:
            return "0"
        num &= 0xFFFFFFFF
        digits = "0123456789abcdef"
        res = []
        while num:
            res.append(digits[num & 15])
            num >>= 4
        return "".join(reversed(res))

Complexity
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Time O(8), space O(1).

Mistakes to Avoid
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Convert negatives to 32-bit two’s complement first.
Loop while num > 0; num > 15 drops the last digit.
0 returns "0".

Interview Explanation
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I AND negatives with 0xffffffff to get the two’s-complement value, then repeatedly map the lowest 4 bits and shift by 4, reversing at the end; 0 returns directly. At most 8 digits.