Notebook

190. Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

Note:

Note that in some languages such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above, the input represents the signed integer -3 and the output represents the signed integer -1073741825.

Follow up:

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

Example 1:

Input: n = 00000010100101000001111010011100
Output:    964176192 (00111001011110000010100101000000)
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.

Example 2:

Input: n = 11111111111111111111111111111101
Output:   3221225471 (10111111111111111111111111111111)
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.

Constraints:

The input must be a binary string of length 32

Source

Code

In [1]:

def reverse_bits(n):
    """
    Time Complexity O(32)
    Space Complexity O(1)
    """
    result, power = 0, 31
    while n:
        result += (n & 1) << power  # returns n&1 with the bits shifted to the left by 'power' places.
                                    #  + info here: https://wiki.python.org/moin/BitwiseOperators
        n = n >> 1  # Returns n with the bits shifted to the right by 1 place
        power -= 1
    return result

In [2]:

def reverse_bits(n):
    """variations"""
    result = 0
    for _ in range(32):
        result = (result << 1) + (n & 1)
        n >>= 1
    return result

In [3]:

def reverse_bits(n):
    """variations"""
    out = 0
    for _ in range(32):
        out = (out << 1) ^ (n & 1)
        n >>= 1
    return out

In [4]:

def reverse_bits(n):
    """variations with O(16) Time Complexity."""
    result = 0
    for i in range(16):
        right_bit = (n >> i) & 1
        left_bit = (n >> (31-i)) & 1
        result |= right_bit << (31-i)  # |=  performs an in-place operation
        result |= left_bit << i        #     between pairs of objects
    return result