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