[NeetCode 150] Single Number

Single Number

You are given a non-empty array of integers nums. Every integer appears twice except for one.

Return the integer that appears only once.

You must implement a solution with $O(n)$ runtime complexity and use only $O(1)$ extra space.

Example 1:

Input: nums = [3,2,3]

Output: 2

Example 2:

Input: nums = [7,6,6,7,8]

Output: 8

Constraints:

1 <= nums.length <= 10000
-10000 <= nums[i] <= 10000

Solution

The key idea of this problem is to find an operation that can transform 2 identical numbers into a constant, while when the constant is operated with another number, the value of the number doesn’t change.

Operation XOR $\oplus$ has such properties that
$$\begin{gathered} a\oplus a=0 \\ 0\oplus a=a \end{gathered}$$
When XOR operation is applied to the whole list, each identical number pair will cancel each other. Therefore, the remaining number with be the only single one.

Code

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