Reverse Bits Python Solution with Tests

Tested Python solution for LeetCode 190 with 16 pytest cases. Generate a practice environment with lcpy.

LeetCode 190, Easy. Topics: Divide and Conquer, Bit Manipulation. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 16 parametrized pytest cases, and a playground notebook:

lcpy gen -n 190   # by problem number
lcpy gen -s reverse_bits   # by problem name

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Problem

Reverse bits of a given 32 bits signed integer.

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Examples

Input: n = 43261596
Output: 964176192
Explanation:

| Integer    | Binary                              |
|------------|-------------------------------------|
| 43261596   | 00000010100101000001111010011100    |
| 964176192  | 00111001011110000010100101000000    |

Input: n = 2147483644
Output: 1073741822
Explanation:

| Integer     | Binary                              |
|-------------|-------------------------------------|
| 2147483644  | 01111111111111111111111111111100    |
| 1073741822  | 00111111111111111111111111111110    |

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Constraints

• 0 <= n <= 2^31 - 2
• n is even.

Follow up: If this function is called many times, how would you optimize it?

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(1) - always 32 iterations
    # Space: O(1) - only using constant extra space
    def reverse_bits(self, n: int) -> int:
        """
        Reverse the bits of a 32-bit unsigned integer.

        Algorithm:
        1. Initialize result to 0
        2. For each of the 32 bits:
           - Extract the rightmost bit of n using (n & 1)
           - Add it to the result at the appropriate position
           - Right shift n to get the next bit
           - Left shift result to make room for the next bit
        3. Return the result

        This approach is optimal for single calls. For multiple calls,
        we could use a lookup table for optimization.
        """
        result = 0
        for _ in range(32):
            result = (result << 1) | (n & 1)
            n >>= 1
        return result

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Complexity

Time	Space
O(1) - always 32 iterations	O(1) - only using constant extra space

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Tags

Grind, Blind 75, NeetCode 150, NeetCode 250, NeetCode All.