78. Subsets

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:

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


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  -> Don't take 3 , Dont take 2 , Dont take 1 = { } 
1) 0 0 1  -> Don't take 3 , Dont take 2 ,   take 1       =  {1 } 
2) 0 1 0  -> Don't take 3 ,    take 2       , Dont take 1 = { 2 } 
3) 0 1 1  -> Don't 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>& nums) {
        sort(nums.begin(), nums.end());

        int num_len = nums.size();
        int subSet_len = pow(2, num_len);

        vector<vector<int>> res(subSet_len, vector<int>());

        for(int i = 0; i < num_len; i++)
            for(int j = 0; j < subSet_len; j++){
                if((j >> i) & 1){
                    res[j].push_back(nums[i]);
                }
            }

        return res;
    }
};

每次从nums挑选一个数，加入已有的集合，从而构成新的集合。

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> res(1, vector<int>());
        sort(nums.begin(), nums.end());

        for(int i = 0; i < nums.size(); i++){
            int len = res.size();
            for(int j = 0; j < len; j++){
                res.push_back(res[j]);
                res.back().push_back(nums[i]);
            }
        }
        return res;
    }
};