167 lines
4.6 KiB
Markdown
167 lines
4.6 KiB
Markdown
## Problem Link
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https://leetcode.com/problems/subsets-ii/description/
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## Description
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```
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Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).
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Note: The solution set must not contain duplicate subsets.
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Example:
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Input: [1,2,2]
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Output:
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[
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[2],
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[1],
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[1,2,2],
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[2,2],
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[1,2],
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[]
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]
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```
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## Solution
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Since this problem is seeking `Subset` not `Extreme Value`, dynamic programming is not an ideal solution. Other approaches should be taken into our consideration.
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Actually, there is a general approach to solve problems similar to this one -- backtracking. Given a [Code Template](https://leetcode.com/problems/combination-sum/discuss/16502/A-general-approach-to-backtracking-questions-in-Java-(Subsets-Permutations-Combination-Sum-Palindrome-Partitioning)) 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.
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Given a picture as followed, let's start with problem-solving ideas of this general solution.
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See Code Template details below.
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## Key Points
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- Backtrack Approach
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- Backtrack Code Template/ Formula
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## Code
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* Supported Language：JS，C++，Python3
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JavaScript Code：
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```js
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/*
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* @lc app=leetcode id=90 lang=javascript
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*
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* [90] Subsets II
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*
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* https://leetcode.com/problems/subsets-ii/description/
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*
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* algorithms
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* Medium (41.53%)
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* Total Accepted: 197.1K
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* Total Submissions: 469.1K
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* Testcase Example: '[1,2,2]'
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*
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* Given a collection of integers that might contain duplicates, nums, return
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* all possible subsets (the power set).
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*
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* Note: The solution set must not contain duplicate subsets.
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*
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* Example:
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*
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*
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* Input: [1,2,2]
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* Output:
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* [
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* [2],
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* [1],
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* [1,2,2],
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* [2,2],
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* [1,2],
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* []
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* ]
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*
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*
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*/
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function backtrack(list, tempList, nums, start) {
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list.push([...tempList]);
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for(let i = start; i < nums.length; i++) {
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//nums can be duplicated, which is different from Problem 78 - subsets
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//So the situation should be taken into consideration
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if (i > start && nums[i] === nums[i - 1]) continue;
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tempList.push(nums[i]);
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backtrack(list, tempList, nums, i + 1)
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tempList.pop();
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}
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}
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/**
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* @param {number[]} nums
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* @return {number[][]}
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*/
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var subsetsWithDup = function(nums) {
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const list = [];
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backtrack(list, [], nums.sort((a, b) => a - b), 0, [])
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return list;
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};
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```
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C++ Code：
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```C++
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class Solution {
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private:
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void subsetsWithDup(vector<int>& nums, size_t start, vector<int>& tmp, vector<vector<int>>& res) {
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res.push_back(tmp);
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for (auto i = start; i < nums.size(); ++i) {
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if (i > start && nums[i] == nums[i - 1]) continue;
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tmp.push_back(nums[i]);
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subsetsWithDup(nums, i + 1, tmp, res);
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tmp.pop_back();
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}
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}
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public:
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vector<vector<int>> subsetsWithDup(vector<int>& nums) {
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auto tmp = vector<int>();
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auto res = vector<vector<int>>();
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sort(nums.begin(), nums.end());
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subsetsWithDup(nums, 0, tmp, res);
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return res;
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}
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};
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```
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Python Code:
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```Python
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class Solution:
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def subsetsWithDup(self, nums: List[int], sorted: bool=False) -> List[List[int]]:
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"""Backtrack Approach: by sorting parameters first to avoid repeting sort later"""
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if not nums:
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return [[]]
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elif len(nums) == 1:
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return [[], nums]
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else:
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# Sorting first to filter duplicated numbers
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# Note，this problem takes higher time complexity
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# So, it could greatly improve time efficiency by adding one parameter to avoid repeting sort in following procedures
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if not sorted:
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nums.sort()
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# Backtrack Approach
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pre_lists = self.subsetsWithDup(nums[:-1], sorted=True)
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all_lists = [i+[nums[-1]] for i in pre_lists] + pre_lists
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# distinct elements
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result = []
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for i in all_lists:
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if i not in result:
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result.append(i)
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return result
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```
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## Related Problems
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- [39.combination-sum](./39.combination-sum.md)(chinese)
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- [40.combination-sum-ii](./40.combination-sum-ii.md)(chinese)
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- [46.permutations](./46.permutations.md)(chinese)
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- [47.permutations-ii](./47.permutations-ii.md)(chinese)
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- [78.subsets](./78.subsets-en.md)
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- [113.path-sum-ii](./113.path-sum-ii.md)(chinese)
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- [131.palindrome-partitioning](./131.palindrome-partitioning.md)(chinese)
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