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Problem Link

https://leetcode.com/problems/subsets-ii/description/

Description

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

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

Solution

Since this problem is seeking Subset not Extreme Value, dynamic programming is not an ideal solution. Other approaches should be taken into our consideration.

Actually, there is a general approach to solve problems similar to this one -- backtracking. Given a Code Template here, it demonstrates how backtracking works with varieties of problems. Apart from current one, many problems can be solved by such a general approach. For more details, please check the Related Problems section below.

Given a picture as followed, let's start with problem-solving ideas of this general solution.

backtrack

See Code Template details below.

Key Points

  • Backtrack Approach
  • Backtrack Code Template/ Formula

Code

  • Supported Language：JS，C++，Python3

JavaScript Code：

/*
 * @lc app=leetcode id=90 lang=javascript
 *
 * [90] Subsets II
 *
 * https://leetcode.com/problems/subsets-ii/description/
 *
 * algorithms
 * Medium (41.53%)
 * Total Accepted:    197.1K
 * Total Submissions: 469.1K
 * Testcase Example:  '[1,2,2]'
 *
 * Given a collection of integers that might contain duplicates, nums, return
 * all possible subsets (the power set).
 * 
 * Note: The solution set must not contain duplicate subsets.
 * 
 * Example:
 * 
 * 
 * Input: [1,2,2]
 * Output:
 * [
 * ⁠ [2],
 * ⁠ [1],
 * ⁠ [1,2,2],
 * ⁠ [2,2],
 * ⁠ [1,2],
 * ⁠ []
 * ]
 * 
 * 
 */
function backtrack(list, tempList, nums, start) {
    list.push([...tempList]);
    for(let i = start; i < nums.length; i++) {
        //nums can be duplicated, which is different from Problem 78 - subsets
        //So the situation should be taken into consideration
        if (i > start && nums[i] === nums[i - 1]) continue;
        tempList.push(nums[i]);
        backtrack(list, tempList, nums, i + 1)
        tempList.pop();
    }
}
/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var subsetsWithDup = function(nums) {
    const list = [];
    backtrack(list, [], nums.sort((a, b) => a - b), 0, [])
    return list;
};

C++ Code：

class Solution {
private:
    void subsetsWithDup(vector<int>& nums, size_t start, vector<int>& tmp, vector<vector<int>>& res) {
        res.push_back(tmp);
        for (auto i = start; i < nums.size(); ++i) {
            if (i > start && nums[i] == nums[i - 1]) continue;
            tmp.push_back(nums[i]);
            subsetsWithDup(nums, i + 1, tmp, res);
            tmp.pop_back();
        }
    }
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        auto tmp = vector<int>();
        auto res = vector<vector<int>>();
        sort(nums.begin(), nums.end());
        subsetsWithDup(nums, 0, tmp, res);
        return res;
    }
};

Python Code:

class Solution:
    def subsetsWithDup(self, nums: List[int], sorted: bool=False) -> List[List[int]]:
        """Backtrack Approach: by sorting parameters first to avoid repeting sort later"""
        if not nums:
            return [[]]
        elif len(nums) == 1:
            return [[], nums]
        else:
            # Sorting first to filter duplicated numbers
            # Note，this problem takes higher time complexity
            # So, it could greatly improve time efficiency by adding one parameter to avoid repeting sort in following procedures 
            if not sorted:
                nums.sort()
            # Backtrack Approach
            pre_lists = self.subsetsWithDup(nums[:-1], sorted=True)
            all_lists = [i+[nums[-1]] for i in pre_lists] + pre_lists
            # distinct elements
            result = []
            for i in all_lists:
                if i not in result:
                    result.append(i)
            return result

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