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🟩 Reverse Bits (#190)

🔗 Problem Link​

📋 Problem Statement​

Reverse bits of a given 32 bits unsigned integer.

💡 Examples​

Example 1​

Input: n = 00000010100101000001111010011100
Output: 964176192 (00111001011110000010100101000000)

Example 2​

Input: n = 11111111111111111111111111111101
Output: 3221225471 (10111111111111111111111111111111)

🔑 Key Insights & Approach​

Core Observation: Build result by extracting rightmost bit of n and adding to leftmost position of result.

Why Bit-by-Bit?

  • O(32) = O(1) time
  • Extract each bit and place in reversed position
  • Simple and efficient

Pattern: "Bit Reversal" pattern.

🐍 Solution: Python​

Approach: Bit-by-Bit​

Time Complexity: O(1) | Space Complexity: O(1)

class Solution:
    def reverseBits(self, n: int) -> int:
        result = 0

        for i in range(32):
            bit = (n >> i) & 1
            result |= (bit << (31 - i))

        return result

Approach 2: Shift and Build​

Time Complexity: O(1) | Space Complexity: O(1)

class Solution:
    def reverseBits(self, n: int) -> int:
        result = 0

        for _ in range(32):
            result = (result << 1) | (n & 1)
            n >>= 1

        return result

🔵 Solution: Golang​

Approach: Bit-by-Bit​

Time Complexity: O(1) | Space Complexity: O(1)

func reverseBits(num uint32) uint32 {
    var result uint32

    for i := 0; i < 32; i++ {
        bit := (num >> i) & 1
        result |= (bit << (31 - i))
    }

    return result
}

📚 References​