How do you solve Reverse Bits on LeetCode?

Reverse Bits (LeetCode #190) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Bit manipulation can be more efficient than string manipulation.

What is the brute force solution for Reverse Bits?

The brute force approach for Reverse Bits is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves converting the integer to its binary representation, reversing that string, and then converting it back to an integer. This is simple but inefficient. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Reverse Bits?

Reverse Bits uses the following concepts: Divide and Conquer, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Two Pointers.

What are the interview tips for Reverse Bits?

Explain your thought process clearly while coding. Discuss potential edge cases before diving into the solution. Practice bit manipulation problems to build confidence.

What are common mistakes when solving Reverse Bits?

Confusing signed and unsigned integers when reversing bits. Not considering edge cases like 0.

#190

Reverse Bits

Easy

Divide and ConquerBit ManipulationBit ManipulationTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(1)

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Intuition

Time O(1)Space O(1)

The optimal solution uses bit manipulation to reverse the bits directly without converting to a string. This is more efficient and avoids unnecessary overhead.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to hold the reversed bits.
2Step 2: For each of the 32 bits, extract the least significant bit from n and add it to the reversed variable.
3Step 3: Shift n right by 1 and the reversed variable left by 1.
4Step 4: Repeat this for all 32 bits.

solution.py8 lines

1# Full working Python code
2
3def reverseBits(n):
4    result = 0
5    for _ in range(32):
6        result = (result << 1) | (n & 1)
7        n >>= 1
8    return result

ℹ

Complexity note: This is constant time complexity because we always perform a fixed number of operations (32) regardless of the input value.

1Bit manipulation can be more efficient than string manipulation.
2Understanding how to manipulate bits directly is crucial for optimizing performance.

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