90. Subsets II

Problem Link

Description

---

Given an integer array nums that may contain duplicates, return all possible subsets (the power set).

The solution set must not contain duplicate subsets. Return the solution in any order.

 

Example 1:

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

Example 2:

Input: nums = [0]
Output: [[],[0]]

 

Constraints:

• 1 <= nums.length <= 10
• -10 <= nums[i] <= 10

Solution

---

Python3

class Solution:
    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
        
        n = len(nums)
        nums.sort()
        ans = [[]]
        start = 1
        
        for i in range(n):
            temp = []
            for j, num in enumerate(ans):
                if i > 0 and nums[i] == nums[i-1] and j < start: continue
                    
                c = num + [nums[i]]
                temp.append(c)
            
            start = len(ans)
            ans += temp
        
        return ans
                
        

C++

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<vector<int>> res;
        
        for (int i=0; i < 1<<n; ++i){
            vector<int> c;
            bool illegal = false;
            for (int j = 0; j < n; ++j)
                if (i >> j&1){
                    if (j > 0 && nums[j]==nums[j-1] && (i>>(j-1)&1) == 0){
                        illegal = true;
                        break;
                    }else{
                        c.push_back(nums[j]);
                    }
                }
                    
            if (!illegal){
                res.push_back(c);
            }
        }
        
        return res;
    }
};

Python

class Solution:
    def subsetsWithDup(self, nums):
        if not nums:
            return []
        nums.sort()
        res, cur = [[]], []
        for i in range(len(nums)):
            if i > 0 and nums[i] == nums[i-1]:
                cur = [item + [nums[i]] for item in cur]
            else:
                cur = [item + [nums[i]] for item in res]
            res += cur
        return res