Given an integer array nums, in which exactly two elements appear only once and all the other elemen

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To understand this solution there are a couple of important bitwise xor properties that you must keep in mind. Skip to the bottom if know these already.

a ⊕ b = b ⊕ a and a ⊕ (b ⊕ c) = (a ⊕ b) ⊕ c
1.xor is commutative and associative.

a ⊕ b = c => a ⊕ c = b => b ⊕ c = a
2.xor is the inverse of itself.

x ⊕ x = 0
3.Therefore, we can say that if a number x appeares an even number of times ⊕ operation will cancel it out.
Formally, x ⊕ x ⊕ x ⊕ x .... n times = 0, where n is even.

y = x & (x-1)
4.This is more of a trick rather than a property to strip the lowest set bit of a number. You would've already known this If you're familliar with BIT, Binary lifting or Binary exponentation.

Finally, to solve this problem:

Find the xor of the two numbers a and b that only occur once by xor-ing the entire array. Property 3
The two numbers a and b must differ by at least one bit, otherwise their xor would've been 0. Lets say b is the number that has the lowest set bit and a has the lowest unset bit.

Find a by xor-ing all those numbers that have the lowest unset bit. Property 4

Once you have a we can find b by xor-ing with xor of the entire array. Property 2
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class Solution:
    def singleNumber(self, nums):
        res=0
        for i in nums:
            res=res^i
        last_setbit=res&(-res)
        ans=[0,0]
        for i in nums:
            if i& last_setbit==0:
                ans[0]=ans[0]^i
            else:
                ans[1]=ans[1]^i
        ans.sort()
        return ans