LeetCode in Kotlin

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

class Solution {
    fun singleNumber(nums: IntArray): IntArray {
        var xorSum = 0
        for (num in nums) {
            // will give xor of required nos
            xorSum = xorSum xor num
        }
        // find rightmost bit which is set
        val rightBit = xorSum and -xorSum
        var a = 0
        for (num in nums) {
            // xor only those number whose rightmost bit is set
            if (num and rightBit != 0) {
                a = a xor num
            }
        }
        return intArrayOf(a, a xor xorSum)
    }
}

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