Following the explanation in this java solution. To quote, "The basic idea is to use XOR operation. We all know that a^b^b =a, which means two xor operations with the same number will eliminate the number and reveal the original number. In this solution, I apply XOR operation to both the index and value of the array. In a complete array with no missing numbers, the index and value should be perfectly corresponding( nums[index] = index), so in a missing array, what left finally is the missing number."
def missingNumber(self, nums): xor = 0 for i, e in enumerate(nums): xor = xor ^ i ^ e return xor ^ len(nums)