LeetCode-in-Java

260. Single Number III

Medium

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 = [1,2,1,3,2,5]

Output: [3,5] **Explanation: ** [5, 3] is also a valid answer.

Example 2:

Input: nums = [-1,0]

Output: [-1,0]

Example 3:

Input: nums = [0,1]

Output: [1,0]

Constraints:

• 2 <= nums.length <= 3 * 104
• -231 <= nums[i] <= 231 - 1
• Each integer in nums will appear twice, only two integers will appear once.

Solution

public class Solution {
    public int[] singleNumber(int[] nums) {
        int xorSum = 0;
        for (int num : nums) {
            // will give xor of required nos
            xorSum ^= num;
        }
        // find rightmost bit which is set
        int rightBit = xorSum & -xorSum;
        int a = 0;
        for (int num : nums) {
            // xor only those number whose rightmost bit is set
            if ((num & rightBit) != 0) {
                a ^= num;
            }
        }
        return new int[] {a, a ^ xorSum};
    }
}

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