LeetCode July - Day 5

Problem Statement

The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance.

Note: 0 ≤ x, y < 2^31.

Example:

Input: x = 1, y = 4 Output: 2 Explanation:

1   (0 0 0 1)
4   (0 1 0 0)
       ↑   ↑

The above arrows point to positions where the corresponding bits are different.

Inputs, Outputs, Constraints & Exceptions

• I: x {number}, y {number}
• O: number (number of bits that differ)
• C: 0 ≤ x, y < 2^31.
• E: x === y, x or y is === 0

PseudoCode

hammingDistance(x ,y):
  initialise var for count
  calculate x ^ y and set to val
  while val > 0:
    if (val & 1) > 0:
      count++
    rightshift val
  return

Code

const hammingDistance = (x, y) => {
  let count = 0, val = x ^ y;
  while(val > 0){
    if(val & 1){count++};
    val = val >> 1;
  }
  return count
}

Time complexity: O(1)
- we can assume this to be constant time since we iterate through at most 32 bits Space complexity: O(1)