# Easy Python O(n) - O(1) solution

• ``````class Solution(object):
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
xor = 0
a = 0
b = 0
for num in nums:
xor ^= num
for num in nums:
a ^= num
else:
b ^= num
return [a, b]``````

• class Solution(object):
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
nums_map, result = collections.defaultdict(list), []
for num in nums:
nums_map[num].append(1)

``````    for key in nums_map:
if len(nums_map[key]) == 1:
result.append(key)

return result``````

• Your idea is quite delicate but somehow not easy to understand.

Here is how I understand your thought (let's take nums = [1, 2, 1, 3, 2, 5] as example):

1. The binary mask has only one digit equals to one (in this case, mask = 2 = 0010)
2. xor = 3^5 = 6 = 0110.
3. for all numbers in nums other than 3, 5, each pair will counteract themselves no matter this pair belongs to 'num&mask ==0' or 'num&mask !=0', what you wanna do is put 3 and 5 into these two different statements, respectively.
4. thus, for xor = 0110, each '1' digit comes from 3 (0011) or 5 (0101) according to the property of operator '^'.
5. based on the while loop, xor&mask != 0 means there is only one common digit of both xor and mask that equals to 1 (xor = 0110, mask = 0010) and this '1' digit either comes from 3 or 5.
6. Thus, 3&mask and 5&mask must have different results (0011&0010 = 0010, 0101&0010 = 0000), which means 'a ^=num' won't contain both the cases of (num=3, num=5).
7. based on the steps above, the 3 and 5 will be sure to be assigned to a and b, respectively.

• Great idea!
You first xor all numbers and then find a bit that is different in those two distinct numbers. This bit is used to divide all numbers into two group. Finally xor them individually, the two numbers left are the two distinct numbers.