Solution
Approach #1 List operation [Time Limit Exceeded]
Algorithm
 Iterate over all the elements in
 If some number in is new to array, append it
 If some number is already in the array, remove it
Complexity Analysis

Time complexity : . We iterate through , taking time. We search the whole list to find whether there is duplicate number, taking time. Because search is in the
for
loop, so we have to multiply both time complexities which is . 
Space complexity : . We need a list of size to contain elements in .
Approach #2 Hash Table [Accepted]
Algorithm
We use hash table to avoid the time required for searching the elements.
 Iterate through all elements in
 Try if has the key for
pop
 If not, set up key/value pair
 In the end, there is only one element in , so use
popitem
to get it
Complexity Analysis

Time complexity : . Time complexity of
for
loop is . Time complexity of hash table(dictionary in python) operationpop
is . 
Space complexity : . The space required by is equal to the number of elements in .
Approach #3 Math [Accepted]
Concept
Complexity Analysis

Time complexity : .
sum
will callnext
to iterate through . We can see it assum(list(i, for i in nums))
which means the time complexity is because of the number of elements() in . 
Space complexity : .
set
needs space for the elements innums
Approach #4 Bit Manipulation [Accepted]
Concept
 If we take XOR of zero and some bit, it will return that bit
 If we take XOR of two same bits, it will return 0
So we can XOR all bits together to find the unique number.
Complexity Analysis

Time complexity : . We only iterate through , so the time complexity is the number of elements in .

Space complexity : .
Analysis written by: @Ambition_Wang