Given a set of distinct integers, nums, return all possible subsets. Note: The solution set must not contain duplicate subsets. For example, If nums  1,2,3, a solution is:  1, 2, 1, 3, 1,, 2,, 3,, 1, 2, 3, , 2, 3  :param nums: Listint :return: Settuple this explanation is from leetnik  leetcode This is an amazing solution. Learnt a lot. Number of subsets for 1 , 2 , 3   23 . why ? case    possible outcomes for the set of subsets 1             Take or dont take  2 2             Take or dont take  2 3             Take or dont take  2 therefore, total  222  23  , 1, 2, 3, 1,2, 1,3, 2,3, 1,2,3 Lets assign bits to each outcome   First bit to 1 , Second bit to 2 and third bit to 3 Take  1 Dont take  0 0 0 0 0   Dont take 3 , Dont take 2 , Dont take 1    1 0 0 1   Dont take 3 , Dont take 2 ,   take 1      1  2 0 1 0   Dont take 3 ,    take 2   , Dont take 1   2  3 0 1 1   Dont take 3 ,    take 2   ,   take 1      1 , 2  4 1 0 0      take 3   , Dont take 2 , Dont take 1   3  5 1 0 1      take 3   , Dont take 2 ,   take 1      1 , 3  6 1 1 0      take 3   ,    take 2   , Dont take 1   2 , 3  7 1 1 1      take 3   ,    take 2   ,   take 1      1 , 2 , 3  In the above logic ,Insert Si only if ji1 true  j E  0,1,2,3,4,5,6,7    i  ith element in the input array  element 1 is inserted only into those places where 1st bit of j is 1 if j  0 1    for above above eg. this is true for sl.no. j  1 , 3 , 5 , 7 element 2 is inserted only into those places where 2nd bit of j is 1 if j  1 1    for above above eg. this is true for sl.no. j   2 , 3 , 6 , 7 element 3 is inserted only into those places where 3rd bit of j is 1 if j  2  1    for above above eg. this is true for sl.no. j   4 , 5 , 6 , 7 Time complexity : On2n , for every input element loop traverses the whole solution set length i.e. 2n
def subsets(nums):
    """
    :param nums: List[int]
    :return: Set[tuple]
    """
    n = len(nums)
    total = 1 << n
    res = set()

    for i in range(total):
        subset = tuple(num for j, num in enumerate(nums) if i & 1 << j)
        res.add(subset)

    return res
"""
this explanation is from leet_nik @ leetcode
This is an amazing solution. Learnt a lot.

Number of subsets for {1 , 2 , 3 } = 2^3 .
why ?
case    possible outcomes for the set of subsets
  1   ->          Take or dont take = 2
  2   ->          Take or dont take = 2
  3   ->          Take or dont take = 2

therefore,
total = 2*2*2 = 2^3 = {{}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}}

Lets assign bits to each outcome  ->
First bit to 1 , Second bit to 2 and third bit to 3
Take = 1
Dont take = 0

0) 0 0 0  -> Dont take 3 , Dont take 2 , Dont take 1 = { }
1) 0 0 1  -> Dont take 3 , Dont take 2 ,   take 1    = { 1 }
2) 0 1 0  -> Dont take 3 ,    take 2   , Dont take 1 = { 2 }
3) 0 1 1  -> Dont take 3 ,    take 2   ,   take 1    = { 1 , 2 }
4) 1 0 0  ->    take 3   , Dont take 2 , Dont take 1 = { 3 }
5) 1 0 1  ->    take 3   , Dont take 2 ,   take 1    = { 1 , 3 }
6) 1 1 0  ->    take 3   ,    take 2   , Dont take 1 = { 2 , 3 }
7) 1 1 1  ->    take 3   ,    take 2   ,   take 1    = { 1 , 2 , 3 }

In the above logic ,Insert S[i] only if (j>>i)&1 ==true
{ j E { 0,1,2,3,4,5,6,7 }   i = ith element in the input array }

element 1 is inserted only into those places where 1st bit of j is 1
if( j >> 0 &1 )  ==> for above above eg.
this is true for sl.no.( j )= 1 , 3 , 5 , 7

element 2 is inserted only into those places where 2nd bit of j is 1
if( j >> 1 &1 )  == for above above eg.
this is true for sl.no.( j ) = 2 , 3 , 6 , 7

element 3 is inserted only into those places where 3rd bit of j is 1
if( j >> 2 & 1 )  == for above above eg.
this is true for sl.no.( j ) = 4 , 5 , 6 , 7

Time complexity : O(n*2^n) , for every input element loop traverses
the whole solution set length i.e. 2^n
"""
