Reverse Bits

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Given a 32-bit unsigned integer n, return the integer obtained by reversing the order of its bits.

Example

Key Pointers

• Build the answer bit by bit rather than reversing a string form of the number.
• Shifting the accumulator left makes room for the next bit; OR-ing in n's lowest bit places it correctly.
• The loop always runs exactly 32 times — the width is fixed, unlike problems where you loop "while n is nonzero".

Step-by-Step Analysis

1. Initialize result = 0.
2. Repeat 32 times: shift result left by 1, then OR in n's current lowest bit (n & 1).
3. Shift n right by 1 to expose the next bit for the following iteration.
4. After 32 iterations, result holds n's bits in reverse order — return it.

Complexity

Time: O(1) — always 32 iterations — Space: O(1)

Code

def reverseBits(n: int) -> int:
    result = 0
    for _ in range(32):
        result = (result << 1) | (n & 1)
        n >>= 1
    return result

Code Explanation

How the solution above works, piece by piece.

1. Process exactly 32 bits, including leading zeros, since they become trailing bits of the result.
2. n & 1 reads the lowest bit of n.
3. result = (result << 1) | bit shifts the result left and appends that bit.
4. n >>= 1 moves to the next bit. After 32 rounds the order is fully reversed.
5. O(32) = O(1) time and space.

Visual Walkthrough

Running it on n = 43261596:

Video Explanation

NeetCode — Binary, LeetCode 190