LeetCode 3827 - Count Monobit Integers

LeetCode Problem 3827

Difficulty: 🟢 Easy
Topics: Bit Manipulation, Enumeration

Solution

Problem Understanding

The problem asks us to count Monobit integers from 0 to n. A Monobit integer is one where all bits in its binary representation are the same. This means there are only two possibilities for a valid Monobit integer: all zeros or all ones. For instance, 0 (0b0), 1 (0b1), 3 (0b11), 7 (0b111), and so on.

The input is a single integer n with a constraint of 0 <= n <= 1000. The output is a single integer representing the count of numbers in [0, n] whose binary representation consists entirely of the same bit.

Key points from the constraints are that n is small, which allows us to consider solutions that may generate numbers explicitly rather than relying solely on bit manipulation formulas. Edge cases include n = 0 (only 0 is valid) and small values of n that fall between two Monobit numbers.

Approaches

The brute-force approach is straightforward. We iterate through all numbers from 0 to n, convert each to binary, and check if all bits are the same. This works correctly but involves unnecessary string operations and checks, which is acceptable for small n but not elegant.

The optimal approach leverages the mathematical property of Monobit integers. Monobit integers are either 0 or numbers that are all ones in binary (like 1, 3, 7, 15, etc.). Each of these numbers can be expressed as (1 << k) - 1 where k >= 1. We can generate Monobit numbers by starting from 1 and doubling the previous number and adding 1 until the number exceeds n. This is highly efficient and avoids checking each number individually.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n log n)	O(1)	Iterate through numbers and check if all bits are the same
Optimal	O(log n)	O(1)	Generate Monobit numbers directly using (1 << k) - 1 pattern

Algorithm Walkthrough

1. Initialize a counter count to track Monobit numbers.

2. Include 0 as a valid Monobit number and increment count.

3. Initialize a variable num to 1. This represents the smallest nonzero Monobit integer.

4. While num is less than or equal to n:

5. Increment count.

6. Update num = (num << 1) | 1 to generate the next Monobit integer in sequence.

7. Return count.

Why it works: The formula (1 << k) - 1 produces all Monobit integers sequentially without missing any. Starting from 1 and doubling while adding 1 ensures we only generate numbers with all ones in their binary form. Counting 0 separately ensures completeness.

Python Solution

class Solution:
    def countMonobit(self, n: int) -> int:
        count = 1  # include 0
        num = 1
        while num <= n:
            count += 1
            num = (num << 1) | 1
        return count

Explanation: We initialize the counter with 1 to include 0. The variable num starts at 1, and the loop generates the next Monobit number by left-shifting num by 1 and adding 1 (num << 1 | 1). We continue until num exceeds n and then return the total count.

Go Solution

func countMonobit(n int) int {
    count := 1 // include 0
    num := 1
    for num <= n {
        count++
        num = (num << 1) | 1
    }
    return count
}

Go notes: The implementation is nearly identical to Python. We use integer variables for count and num and perform the same left-shift and OR operation to generate Monobit numbers. Go handles integer overflow safely within the constraint n <= 1000.

Worked Examples

Example 1: n = 1

Step	num	count	Condition
Initialize	1	1	include 0
Loop 1	1 <= 1	2	increment count
Update num	num = (1 << 1)	3	3 > 1, exit loop
Return	count = 2		correct

Example 2: n = 4

Step	num	count	Condition
Initialize	1	1	include 0
Loop 1	1 <= 4	2	increment count
Update num	3		still <=4
Loop 2	3 <= 4	3	increment count
Update num	7		7>4, exit loop
Return	count = 3		correct

Complexity Analysis

Measure	Complexity	Explanation
Time	O(log n)	Each iteration doubles num, generating a new Monobit number; loop runs O(log n) times
Space	O(1)	Only a few integer variables are used; no extra data structures

The complexity is very efficient due to the sequential generation of Monobit integers rather than checking each number.

Test Cases

# Provided examples
assert Solution().countMonobit(1) == 2  # 0,1
assert Solution().countMonobit(4) == 3  # 0,1,3

# Boundary cases
assert Solution().countMonobit(0) == 1  # only 0
assert Solution().countMonobit(7) == 4  # 0,1,3,7
assert Solution().countMonobit(8) == 4  # 0,1,3,7

# Random small cases
assert Solution().countMonobit(15) == 5  # 0,1,3,7,15
assert Solution().countMonobit(31) == 6  # 0,1,3,7,15,31
assert Solution().countMonobit(1000) == 10  # 0,1,3,7,15,31,63,127,255,511

Test	Why
n = 1	Smallest nonzero n
n = 4	Tests between Monobit numbers
n = 0	Boundary case with only zero
n = 7	Exact Monobit integer
n = 1000	Upper bound stress test

Edge Cases

1. n = 0: Only 0 is valid. This could be missed if the implementation only generates numbers starting from 1. By initializing count = 1, we handle this correctly.
2. n between two Monobit integers: For example, n = 4 lies between 3 (0b11) and 7 (0b111). The algorithm correctly stops before exceeding n.
3. n is exactly a Monobit integer: If n is 7, the algorithm counts 0,1,3,7. The inclusive condition num <= n ensures we include n if it is Monobit.

This implementation covers all possible cases efficiently and accurately.