# 65ms python solution with explanation

• For any given set S, each element in S can be only either selected or unselected.

If an S contains n elements, the solution will contains 2^n sets. Can be solution number from 0 -- 2^n - 1

Each element (eg. the i's element) be selected or not can be represented by the i's bit of the solution number

``````class Solution:
# @param S, a list of integer
# @return a list of lists of integer
def subsets(self, S):
solutions = []
S.sort()
bits = len(S)
for i in range(1 << bits):
solution = []
for j in range(bits):
if i & (1 << j):
solution.append(S[j])
solutions.append(solution)

return solutions``````

• Is the time complexity still n*2^n?

• Yes, because the lower bound of the question is O(N*2^N). For any set S, the number of subsets is 2^N. Since it wants all of it. And for generate any subset, you need to detect any element. This solution is just using iteration so that avoid of recursion.

• Thank you very much.

• This is so clear and smart! Better than my clumsy DP solution with the same time complexity!

I made two lists to progress from taking 1 elemnets subsets to taking n elements subsets, while list[i] means using only the first i+1 elements in the sorted input for the subset making, and each round I collect the last subsets from the list into results. This is so clumsy compared to your bit model!

``````class Solution:
# @param S, a list of integer
# @return a list of lists of integer
def subsets(self, S):
S=sorted(S)
n=len(S)
results=[[]]
baseList=[[[S[j]] for j in xrange(i+1)] for i in xrange(n)]
progList=[[] for i in xrange(n)]
for element in xrange(1,n):
results+=baseList[n-1]

for i in xrange(element,n):
progList[i]=progList[i-1]+[shortList+[S[i]] for shortList in baseList[i-1]]
baseList=progList
progList=[[] for i in xrange(n)]
results+=baseList[n-1]

return results
``````

• I find that it's hard for me to transform recursion to iteration. Thanks for your answer.

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