Reverse Bits Leetcode Problem 190 [Python Solution]

Are you ready to delve into another exciting LeetCode problem? If you’re here, you’re most likely on a quest to master bit manipulation with Leetcode Problem 190.

The problem we’re tackling today is “Reverse Bits,” and it’s part of the Blind 75 list, a curated collection of common LeetCode problems.

Problem Overview

The problem is simple yet intriguing.

You are given a 32-bit unsigned integer, and your task is to reverse all the bits within it.

This means flipping the order of the bits from left to right.

For example, if you’re given the binary representation 00000010100101000001111010011100, you need to reverse it to 00111001011110000010100101000000.

Here’s a breakdown of the problem statement:

• Input: A 32-bit binary string.
• Output: A 32-bit integer with the bits reversed.

Before we dive into the solution, let’s clarify a few things.

While the problem is described in terms of unsigned integers, in languages like Java, there is no distinct unsigned integer type.

However, this distinction doesn’t affect our implementation because the binary representation remains the same.

Java compilers use two’s complement notation for signed integers.

So, whether it’s a signed or unsigned integer, the internal binary representation is identical.

Constraints

It’s essential to understand the constraints provided by the problem.

The key constraint here is the input, which must be a binary string of length 32. Now, let’s explore an efficient Python solution to this problem.

Efficient Python Code Solution

def reverseBits(self, n: int) -> int:
    result = 0
    for i in range(32):
        bit = (n >> i) & 1
        result += (bit << (31 - i))
    return result

The solution is elegant and operates in constant time (O(1)), making it efficient for any input size within the problem’s constraints.

It uses bit manipulation to achieve the bit reversal.

Here’s a step-by-step explanation of the code:

1. Initialize result to 0. This variable will store the reversed bits.
2. Iterate over the 32 bits of the input integer using a loop from i = 0 to i = 31.

These are the 32 bits you need to reverse.

3. Calculate bit by shifting the input n right by i positions and applying a bitwise AND operation with 1. This operation extracts the i-th bit of the input.
4. Update result by shifting bit left by (31 - i) positions and performing a bitwise OR operation with the current value of result.

This effectively places the i-th bit of the input in its reversed position in the result.

5. Repeat this process for all 32 bits, and you’ll have the desired reversed result.
6. Finally, return result as the reversed 32-bit integer.

This concise code leverages bit manipulation to provide an efficient solution to the problem.

Time and Space Complexity

Now, let’s discuss the time and space complexity of this solution.

Time Complexity: The solution operates in constant time, O(1), because it processes a fixed number of bits (32 bits) regardless of the input size.

This is the best possible time complexity for this problem.

Space Complexity: The solution uses a constant amount of additional space.

The only space required is for the result variable, making the space complexity O(1).

Reasoning Behind Our Approach

Understanding the bit manipulation operations involved in this solution is key to mastering the problem.

Here’s a breakdown of the crucial concepts used:

• Bitwise AND (&): This operation is used to extract a specific bit from the input.

If the bit at a particular position is 1, the result will be 1; otherwise, it will be 0.

• Bitwise Left Shift (<<): Shifting a bit to the left by n positions effectively moves it to a higher-order position in the binary representation.

This operation is used to position the bits correctly in the result.

• Bitwise OR (|): This operation is used to update the result by preserving the bits already in their reversed positions while placing the next bit in its correct position.

The efficient solution combines these bit manipulation operations to reverse the bits in constant time, making it a powerful and elegant approach.

Edge Cases a Valid Solution Must Consider

In the context of this problem, there are a few edge cases to consider:

1. All Zeros Input: If the input is 00000000000000000000000000000000, the output should be the same, as reversing all zeros results in all zeros.
2. All Ones Input: If the input is 11111111111111111111111111111111, the output should also be the same, as reversing all ones still results in all ones.
3. Mixed Ones and Zeros: For inputs that have a mix of ones and zeros, the solution should correctly reverse all the bits.
4. Input with Leading Zeros: Leading zeros in the input should not affect the result.

The solution should only consider the significant bits.

5. Max and Min 32-Bit Integers: The solution should handle the edge cases of the maximum and minimum 32-bit integers.
6. Performance: The solution should be efficient and operate within the problem’s constraints, as it’s designed for 32-bit integers.

The provided solution accounts for these edge cases and efficiently handles all scenarios.

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Conclusion

In this blog post, we tackled the Reverse Bits problem from LeetCode.

We learned how to reverse the bits of a 32-bit unsigned integer using bit manipulation operations.

The provided Python solution is both efficient and elegant, operating in constant time.

Mastering bit manipulation is a valuable skill, and this problem serves as a great exercise to enhance your understanding of binary operations.

Remember that practice is the key to mastery, so don’t hesitate to try more bit manipulation problems and explore different approaches to solidify your knowledge.

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Now that you’ve grasped the intricacies of reversing bits, why not take on more coding challenges?

Happy coding!

Question Link: Reverse Bits LeetCode Problem