Tags: "leetcode", "bit-manipulation", access_time 1-min read

Single Number III

Created: September 14, 2018 by [lek-tin]

Last updated: September 14, 2018

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.

For example: Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note

1. The order of the result is not important. So in the above example, [5, 3] is also correct.
2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

Credit

Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

Solution

class Solution {
public:
  vector<int> singleNumber(vector<int>& nums) {
    // Pass 1 : 
    // Get the XOR of the two numbers we need to find
    int twoSinglesXORed = accumulate(nums.begin(), nums.end(), 0, bit_xor<int>());
    // Get its last set bit
    twoSinglesXORed &= ~(twoSinglesXORed-1);

    // Pass 2 :
    vector<int> singles = {0, 0}; // this vector stores the two numbers we will return
    for (int num : nums)
    {
        if ((num & twoSinglesXORed) == 0) // the bit is not set
        {
            singles[0] ^= num;
        }
        else // the bit is set
        {
            singles[1] ^= num;
        }
    }
    return singles;
  }
};

Think beyond

How come twoSinglesXORed &= ~(twoSinglesXORed-1); equals twoSinglesXORed &= -twoSinglesXORed-1

Explanation

1. https://leetcode.com/problems/single-number-iii/discuss/68901/Sharing-explanation-of-the-solution
2. https://leetcode.com/problems/single-number-iii/discuss/68900/Accepted-C++Java-O(n)-time-O(1)-space-Easy-Solution-with-Detail-Explanations