103 lines
2 KiB
Markdown
103 lines
2 KiB
Markdown
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# Leetcode Subsets
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#### 2022-07-22 16:16
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> ##### Algorithms:
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> #algorithm #backtrack
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> ##### Data structures:
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> #DS #vector
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> ##### Difficulty:
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> #coding_problem #difficulty-medium
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> ##### Additional tags:
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> #leetcode
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> ##### Revisions:
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> N/A
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##### Related topics:
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```expander
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tag:#backtrack
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```
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##### Links:
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- [Link to problem](https://leetcode.com/problems/subsets/)
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___
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### Problem
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Given an integer array `nums` of **unique** elements, return _all possible subsets (the power set)_.
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The solution set **must not** contain duplicate subsets. Return the solution in **any order**.
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#### Examples
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**Example 1:**
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```
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**Input:** nums = [1,2,3]
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**Output:** [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]
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```
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**Example 2:**
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```
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**Input:** nums = [0]
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**Output:** [[],[0]]
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```
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#### Constraints
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- `1 <= nums.length <= 10`
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- `-10 <= nums[i] <= 10`
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- All the numbers of `nums` are **unique**.
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### Thoughts
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> [!summary]
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> This is a super simple #backtrack problem.
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It asked for combinations, so this is a backtrack problem.
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go from index 0 to last one, in each iteration, you have two choices:
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- Add this number
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- Don't add this number
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And we try different combinations at the same level, aka. backtracking.
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#### Pseudo code:
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- When loc == size, append combination to answer
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- Else, start trying different solutions:
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- Append this element to combs and backtrack
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- pop this from combs and backtrack
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### Solution
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```cpp
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class Solution {
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vector<vector<int>> ans;
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vector<int> combs;
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void backtrack(vector<int> &nums, int nextLoc) {
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// We don't require min amount of location
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if (nextLoc == nums.size()) {
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ans.push_back(combs);
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} else {
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combs.push_back(nums[nextLoc]);
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backtrack(nums, nextLoc + 1);
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combs.pop_back();
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backtrack(nums, nextLoc + 1);
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}
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}
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public:
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vector<vector<int>> subsets(vector<int> &nums) {
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// Backtracking for accumulating combinations.
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ans = {};
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combs = {};
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backtrack(nums, 0);
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return ans;
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}
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};
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```
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