Approach: XOR then split by lowest differing bit

def singleNumber(nums):
    x = 0
    for v in nums:
        x ^= v
    diff = x & (-x)
    a = b = 0
    for v in nums:
        if v & diff:
            a ^= v
        else:
            b ^= v
    return sorted([a, b])

Complexity: O(n) time, O(1) space.

Full Solution — all languages

class Solution:
    def singleNumber(self, nums):
        x = 0
        for v in nums:
            x ^= v
        diff = x & (-x)
        a = b = 0
        for v in nums:
            if v & diff:
                a ^= v
            else:
                b ^= v
        return sorted([a, b])

import java.util.*;
class Solution {
    public int[] singleNumber(int[] nums) {
        int xorAll = 0;
        for (int v : nums) xorAll ^= v;
        int diff = xorAll & (-xorAll);
        int a = 0, b = 0;
        for (int v : nums) {
            if ((v & diff) != 0) a ^= v;
            else b ^= v;
        }
        return new int[]{Math.min(a, b), Math.max(a, b)};
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: vector<int> singleNumber(vector<int>& nums) {
        int xorAll = 0;
        for (int v : nums) xorAll ^= v;
        int diff = xorAll & (-xorAll);
        int a = 0, b = 0;
        for (int v : nums) {
            if (v & diff) a ^= v;
            else b ^= v;
        }
        return {
            min(a, b), max(a, b)
        };
    }
};

/**
 * @param {number[]} nums
 * @return {number[]}
 */ var singleNumber = function(nums) {
    let xorAll = 0;
    for (const v of nums) xorAll ^= v;
    const diff = xorAll & (-xorAll);
    let a = 0, b = 0;
    for (const v of nums) {
        if (v & diff) a ^= v;
        else b ^= v;
    }
    return [Math.min(a, b), Math.max(a, b)];
};

Single Number III

Example 1

Example 2

Constraints