LeetCode July - Day 23

Problem Statement

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

Example:

Input:  [1,2,1,3,2,5]
Output: [3,5]

Note:

The order of the result is not important. So in the above example, [5, 3] is also correct.

Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

Inputs, Outputs, Constraints & Exceptions

• I: number[]
• O: number[]
• C: time O(n), space O(1)
• E: only two elements in array

Code (using a hashmap)

/**
 * @param {number[]} nums
 * @return {number[]}
 */
function singleNumber(nums) {
  if(nums.length < 3) return [...nums];
  
  const hm = new Map()
  for(let num of nums){
    if(hm.get(num)) hm.delete(num);
    else hm.set(num, 1);
  }
 
  return [...hm.keys()]
};

Time complexity: O(N)
- where N is the number of items in nums
Space complexity: O(N)

Code (using xor)

/**
 * @param {number[]} nums
 * @return {number[]}
 */
function singleNumber(nums) {
  if(nums.length < 3) return [...nums];
  
  const xy = nums
    .reduce((acc, curr) => acc ^ curr, 0)
  const mostSigBit = xy.toString(2).length - 1
  const x = nums
    .filter(num => (num >> mostSigBit & 1) === 1)
    .reduce((acc, curr) => acc ^ curr, 0)

  return [x, xy ^ x]
};

Time complexity: O(n)
Space complexity: O(1)