# 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