Subsets II LeetCode Solution

Problem – Subsets II LeetCode Solution

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

Subsets II LeetCode Solution in C++

    class Solution {
public:
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        vector<vector<int> > totalset = {{}};
        sort(S.begin(),S.end());
        for(int i=0; i<S.size();){
            int count = 0; // num of elements are the same
            while(count + i<S.size() && S[count+i]==S[i])  count++;
            int previousN = totalset.size();
            for(int k=0; k<previousN; k++){
                vector<int> instance = totalset[k];
                for(int j=0; j<count; j++){
                    instance.push_back(S[i]);
                    totalset.push_back(instance);
                }
            }
            i += count;
        }
        return totalset;
        }
};

Subsets II LeetCode Solution in Java

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<>();
        helper(res,new ArrayList<>(),nums,0);
        return res;
    }
    
    public void helper(List<List<Integer>> res, List<Integer> ls, int[] nums, int pos) {
        res.add(new ArrayList<>(ls));
        for(int i=pos;i<nums.length;i++) {
            if(i>pos&&nums[i]==nums[i-1]) continue;
            ls.add(nums[i]);
            helper(res,ls,nums,i+1);     
            ls.remove(ls.size()-1);
        }
    }
}

Subsets II LeetCode Solution in Python

class Solution:
    # @param num, a list of integer
    # @return a list of lists of integer
    def subsetsWithDup(self, S):
        res = [[]]
        S.sort()
        for i in range(len(S)):
            if i == 0 or S[i] != S[i - 1]:
                l = len(res)
            for j in range(len(res) - l, len(res)):
                res.append(res[j] + [S[i]])
        return res
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