137. Single Number II

Given an integer array nums where every element appears three times except for one, which appears exactly once. Find the single element and return it.

You must implement a solution with a linear runtime complexity and use only constant extra space.

Example 1:

Input: nums = [2,2,3,2] Output: 3

Example 2:

Input: nums = [0,1,0,1,0,1,99] Output: 99

Constraints:

• 1 <= nums.length <= 3 * 104
• -231 <= nums[i] <= 231 - 1
• Each element in nums appears exactly three times except for one element which appears once.

Code

class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        map = {}

        for num in nums:
            map[num] = map.get(num, 0) + 1

        for key, value in map.items():
            if value == 1:
                return key

class Solution:
    def singleNumber(self, nums):
        ans = 0

        for i in range(32):
            bit_sum = 0
            for num in nums:
                if num < 0:
                    num = num & (2**32-1)
                bit_sum += (num >> i) & 1
            bit_sum %= 3
            ans |= bit_sum << i

        if ans >= 2**31:
            ans -= 2**32

        return ans

class Solution:
    def singleNumber(self, nums):
        ones = 0
        twos = 0

        for num in nums:
            ones ^= (num & ~twos)
            twos ^= (num & ~ones)

        return ones

function singleNumber(nums: number[]): number {
  const map = {};

  for (let num of nums) {
    const value = map[num];
    if (value) {
      map[num] = value + 1;
    } else {
      map[num] = 1;
    }
  }

  for (const [key, value] of Object.entries(map)) {
    if (value == 1) {
      return +key;
    }
  }

  return 0;
}

function singleNumber(nums: number[]): number {
  let ones = 0;
  let twos = 0;

  for (let num of nums) {
    ones ^= num & ~twos;
    twos ^= num & ~ones;
  }

  return ones;
}