How do you solve Count Monobit Integers on LeetCode?

Count Monobit Integers (LeetCode #3827) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(1) space complexity. Key insight: Monobit integers are 0 and numbers of the form 2^k - 1.

What is the time complexity of Count Monobit Integers?

The optimal solution for Count Monobit Integers runs in O(log n) time and uses O(1) space. The loop runs logarithmically with respect to n, as we double the power each iteration.

What is the brute force solution for Count Monobit Integers?

The brute force approach for Count Monobit Integers is Brute Force, with O(n) time complexity and O(1) space complexity. We can check each integer from 0 to n and determine if its binary representation consists of all identical bits. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Count Monobit Integers?

Count Monobit Integers uses the following concepts: Bit Manipulation, Enumeration. The recommended patterns to study are: Bit Manipulation, Mathematical Patterns.

What are the interview tips for Count Monobit Integers?

Explain your thought process clearly. Use examples to illustrate your approach. Be prepared to optimize your solution.

What are common mistakes when solving Count Monobit Integers?

Overlooking 0 as a Monobit integer. Miscounting powers of 2.

#3827

Count Monobit Integers

Easy

Bit ManipulationEnumerationBit ManipulationMathematical Patterns

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(log n)
Space	O(1)	O(1)

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Intuition

Time O(log n)Space O(1)

Monobit integers are either 0 or powers of 2 minus 1. We can efficiently count these by recognizing the pattern.

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Algorithm

3 steps

1Step 1: Initialize a count to 0.
2Step 2: Count 0 as a Monobit integer.
3Step 3: For each power of 2 (1, 3, 7, ...), check if it's less than or equal to n and increment the count.

solution.py7 lines

1def countMonobitIntegers(n):
2    count = 1  # for 0
3    power = 1
4    while power <= n:
5        count += 1
6        power = (power << 1) | 1
7    return count

ℹ

Complexity note: The loop runs logarithmically with respect to n, as we double the power each iteration.

1Monobit integers are 0 and numbers of the form 2^k - 1.
2Recognizing patterns in binary helps simplify counting.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.