260. Single Number III - Explanation

Description

You are given an integer array 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. You can return the answer in any order.

You must write an algorithm that runs in linear runtime complexity and uses only constant extra space.

Example 1:

Input: nums = [1,1,2,3,2,4]

Output: [3,4]

Example 2:

Input: nums = [2,1]

Output: [2,1]

Example 3:

Input: nums = [-50,1,2,3,3,2,1,10]

Output: [-50,10]

Constraints:

2 <= nums.length <= 30,000
(-(2^31)) <= nums[i] <= ((2^31)-1)
Each integer in nums will appear twice, only two integers will appear once.

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1. Brute Force

class Solution:
    def singleNumber(self, nums: List[int]) -> List[int]:
        n, res = len(nums), []

        for i in range(n):
            flag = True
            for j in range(n):
                if i != j and nums[i] == nums[j]:
                    flag = False
                    break

            if flag:
                res.append(nums[i])
                if len(res) == 2:
                    break

        return res

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(1)$ extra space.

2. Hash Map

class Solution:
    def singleNumber(self, nums: List[int]) -> List[int]:
        count = {}
        for num in nums:
            count[num] = 1 + count.get(num, 0)

        return [k for k in count if count[k] == 1]

Time & Space Complexity

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

3. Hash Set

class Solution:
    def singleNumber(self, nums: List[int]) -> List[int]:
        seen = set()
        for num in nums:
            if num in seen:
                seen.remove(num)
            else:
                seen.add(num)
        return list(seen)

Time & Space Complexity

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

4. Sorting

class Solution:
    def singleNumber(self, nums: List[int]) -> List[int]:
        res, n = [], len(nums)
        nums.sort()

        for i in range(n):
            if ((i > 0 and nums[i] == nums[i - 1]) or
                (i + 1 < n and nums[i] == nums[i + 1])):
                continue
            res.append(nums[i])

        return res

Time & Space Complexity

Time complexity: $O(n \log n)$
Space complexity: $O(1)$ or $O(n)$ depending on the sorting algorithm.

5. Bitwise XOR (Least Significant Bit)

class Solution:
    def singleNumber(self, nums: List[int]) -> List[int]:
        xor = 0
        for num in nums:
            xor ^= num

        diff_bit = 1
        while not (xor & diff_bit):
            diff_bit <<= 1

        a = b = 0
        for num in nums:
            if diff_bit & num:
                a ^= num
            else:
                b ^= num
        return [a, b]

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$ extra space.

6. Bitwise XOR (Most Significant Bit)

class Solution:
    def singleNumber(self, nums: List[int]) -> List[int]:
        xor = 0
        for num in nums:
            xor ^= num

        diff_bit = xor & (-xor)

        a = b = 0
        for num in nums:
            if diff_bit & num:
                a ^= num
            else:
                b ^= num
        return [a, b]

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$ extra space.