fun uniqueNum(nums: List<Int>): Int {
    return nums.reduce { a, b -> a xor b }
}

#Challenge

given a non empty array of integers with the guarantee that every number occurs twice except one. Find the number that occurs only once.

#Solution

This implementation takes advantage of the fact that $a\oplus b=c$ and $a\oplus c=b$. That is exclusive-or is a reflective operation. Each occurence of a number twice in nums cancels itself out in num ($a\oplus a=0$). Therefore if we xor every element, the bits remaining must correspond to the only element not cancelled out. I.E. The unique element.

#Test Cases

uniqueNum(listOf(1,2,3,2,1)) // 3
uniqueNum(listOf(1)) // 1
uniqueNum(listOf(2,2,1)) // 1
uniqueNum(listOf(4,1,2,1,2)) // 4