90. Subsets II

📝 Problem Details

• Title: 90 Subsets II
• Link: https://leetcode.com/problems/subsets-ii/
• Difficulty: Medium
• Tags/Categories: Backtracking

input array nums can contain duplicates return all possible subsets (powerset) most not contain duplicate sets

💭What Were My Initial Thoughts?

- difference between the original subsets problem and this problem is that the input array can contain duplicates
- how do we manage duplicates when generating unique subsets?
	- by sorting the input array first, we can check the previous element to see if we've already processed that number

💡 Explanation of Solution

- sort the input array
- create a DFS function that has the function signature:
	- nums
	- vector<vector<int>> result
	- vector<int> curr
	- int index

- push the current vector into result (as each recursive call is a valid subset)
- iterate over nums, either choose to include or exclude the current element, reducing the index by 1 each recursive call
	- if the previous element is the same as the current, then continue / skip
	- push the current element into curr
	- recrusively call dfs , incrementing index by 1
	- undo the last choice (pop from curr)

⌛ Complexity Analysis

Time Complexity: O(2^n)
Space Complexity: O(2^n)

💻 Implementation of Solution

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<vector<int>> result;
        vector<int> current;
        sort(nums.begin(), nums.end());
        dfs(nums, 0, current, result);
        return result;
    }
 
    void dfs(vector<int>& nums, int index, vector<int>& current, vector<vector<int>>& result) {
        // add the current subset to the result
        result.push_back(current);
 
        // explore further by including each remaining element
        for(int i = index; i < nums.size(); i++) {
 
            // skip duplicates
            if(i > index && nums[i] == nums[i - 1]) continue;
 
            // choose
            current.push_back(nums[i]);
 
            // explore
            dfs(nums, i + 1, current, result);
 
            // backtack (undo the choice)
            current.pop_back();
        }
    }
 
};