Question

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

Example 1:

Input: [3,0,1]
Output: 2

Example 2:

Input: [9,6,4,2,3,5,7,0,1]
Output: 8

Note:

Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

Solution 1

The problem can be reduced to Single Number by XOR’ing all inputs together and XOR’ing with 0, 1, 2, ..., n.

class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        res = 0
        for i in range(1, len(nums) + 1):
            res ^= i

        for n in nums:
            res ^= n

        return res

Solution 2

We know the sum of 0, 1, 2, ..., n is \(n(n + 1) / 2\). Thus, we subtract every number in nums from this sum and the result is the missing number.

class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) == 0:
            res = 0
        if len(nums) == 1:
            res = 1
        else:
            res = len(nums) * (len(nums) + 1) // 2

        for n in nums:
            res -= n

        return res