Subsets

Given a set of distinct integers, nums, return all possible subsets.

Note:
Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,3], a solution is:

[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]

解：

class Solution {public:    vector<vector<int>> subsets(vector<int>& nums) {        sort(nums.begin(),nums.end(),less<int>());        return subsets(nums,nums.size()-1);    }    vector<vector<int>> subsets(vector<int>& nums,size_t n)    {        if(n==-1)return vector<vector<int>>{vector<int>()};        vector<vector<int>> res=subsets(nums,n-1);        vector<vector<int>> merge=res;        for(int i=0;i<res.size();++i)        {            res[i].push_back(nums[n]);        }        copy(res.begin(),res.end(),back_inserter(merge));        return merge;    }};

bit manipulation
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

class Solution {public:    vector<vector<int> > subsets(vector<int> &S) {        sort (S.begin(), S.end());        int elem_num = S.size();        int subset_num = pow (2, elem_num);        vector<vector<int> > subset_set (subset_num, vector<int>());        for (int i = 0; i < elem_num; i++)            for (int j = 0; j < subset_num; j++)                if ((j >> i) & 1)                    subset_set[j].push_back (S[i]);        return subset_set;    }};