algorithms/bit/subsets.py

"""
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)
}
"""
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
"""