Second LeetCode Blog Post

October 4, 2024

Second LeetCode Blog Post

Today is another easy problem, and it is focused on bit manipulation. Let's dive on in.

The Problem

Today's problem is 190. Reverse Bits. The problem is as follows: Reverse bits of a given 32 bits unsigned integer.

The Solution

I like to format my answers using the UMPIRE method:

• Understand
  • We are given a 32-bit unsigned integer, and we are to reverse the bits.
• Match
  • We can match this problem to a bit manipulation problem.
• Plan
  • We can write a function which casts the number to binary, then a string, and finally reverses it before casting back to binary.
• Implement

Copy
1def reverseBits(self, n: int) -> int:
2        b = bin(n)
3        s = str(b)[2:].zfill(32)
4        s = s[::-1]
5        return int(s, 2)

• Review
  • The code is clean and concise. We used descriptive variable names. We have implemented a bit manipulation.
  • We can test our code using the examples provided.
  • n = 43261596
    • b = '0b10100101000001111010011100'
    • s = '00000010100101000001111010011100'
    • s = '00111001011110000010100101000000'

Copy
1def test_reverseBits(self):
2        assert self.reverseBits(43261596) == 964176192
3        assert self.reverseBits(4294967293) == 3221225471

• Evaluate
  • Our code runs in O(n) time complexity due to our list slicing.
  • Our space complexity is O(1) as we use the same amount of space regardless of input size.

My First LeetCode Blog Post

LeetCode as Procrastination