LeetCode 3827: Count Monobit Integers

Meta Data

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Difficulty: easy First Attempt: 2026-02-01

• Submissions: Accepted (first try)

Approach

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All bits equal: all zeros is only 0, and all ones is 1, 3, 7, 15, ..., i.e. 2^k - 1 (Bit Manipulation Template 3). 2^k - 1 <= n means 2^k <= n + 1, so k runs from 1 to (n + 1).bit_length() - 1; adding 0 gives (n + 1).bit_length() in total.

My Solution

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class Solution:
    def countMonobit(self, n: int) -> int:
        return (n + 1).bit_length()

I originally looped, adding 2^power starting from 1 to generate 1, 3, 7, …, counting those <= n plus 0; same result.

Complexity

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

Mistakes to Avoid

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• Do not forget that 0 counts.

Interview Explanation

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Numbers with all-equal bits are just 0 and 2^k - 1. There are (n + 1).bit_length() - 1 values of k with 2^k - 1 <= n, plus 0. Time O(1).