Subsets

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: nums = [1,2,3] Output: [ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]

Solution: Iterative

• We don't need to sort the input array
• We can push an empty vector to the result vector first
• Iterate through the input vector
• Append the number to all subsets in vector and push it back to result vector
• We need to remember the size of the current vector first

    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> ans(1, vector<int>());

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

Solution: Backtracking

• First we need to sort the input arry

    vector<vector<int>> subsets(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> subs;
        vector<int> sub;  
        genSubsets(nums, 0, sub, subs);
        return subs; 
    }
    void genSubsets(vector<int>& nums, int start, vector<int>& sub, vector<vector<int>>& subs) {
        subs.push_back(sub);
        for (int i = start; i < nums.size(); i++) {
            sub.push_back(nums[i]);
            genSubsets(nums, i + 1, sub, subs);
            sub.pop_back();
        }
    }