notes/OJ notes/pages/Leetcode Subsets.md

# Leetcode Subsets

#### 2022-07-22 16:16

> ##### Algorithms:
>
> #algorithm #backtrack
>
> ##### Data structures:
>
> #DS #vector
>
> ##### Difficulty:
>
> #coding_problem #difficulty-medium
>
> ##### Additional tags:
>
> #leetcode
>
> ##### Revisions:
>
> N/A

##### Related topics:

##### Links:

- [Link to problem](https://leetcode.com/problems/subsets/)

---

### Problem

Given an integer array `nums` of **unique** elements, return _all possible subsets (the power set)_.

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

#### Examples

**Example 1:**

```
**Input:** nums = [1,2,3]
**Output:** [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]
```

**Example 2:**

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

#### Constraints

- `1 <= nums.length <= 10`
- `-10 <= nums[i] <= 10`
- All the numbers of `nums` are **unique**.

### Thoughts

> [!summary]
> This is a super simple #backtrack problem.

It asked for combinations, so this is a backtrack problem.

go from index 0 to last one, in each iteration, you have two choices:

- Add this number
- Don't add this number
  And we try different combinations at the same level, aka. backtracking.

#### Pseudo code:

- When loc == size, append combination to answer
- Else, start trying different solutions:
  - Append this element to combs and backtrack
  - pop this from combs and backtrack

### Solution

```cpp
class Solution {
  vector<vector<int>> ans;
  vector<int> combs;
  void backtrack(vector<int> &nums, int nextLoc) {
    // We don't require min amount of location

    if (nextLoc == nums.size()) {
      ans.push_back(combs);
    } else {
      combs.push_back(nums[nextLoc]);
      backtrack(nums, nextLoc + 1);

      combs.pop_back();
      backtrack(nums, nextLoc + 1);
    }
  }

public:
  vector<vector<int>> subsets(vector<int> &nums) {
    // Backtracking for accumulating combinations.
    ans = {};
    combs = {};
    backtrack(nums, 0);

    return ans;
  }
};
```