---
title: '78. Subsets'
description: Given an integer array nums of unique elements, return all possible subsets (the power set)
icon: dot
topics:
  - { name: "Array", slug: "array" }
  - { name: "Backtracking", slug: "backtracking" }
  - { name: "Bit Manipulation", slug: "bit-manipulation" }
issue: "https://github.com/prdlk/leetcode/issues/149"
companies:
  - { name: "Bloomberg", period: "Thirty Days", frequency: 69.8 }
  - { name: "Amazon", period: "Thirty Days", frequency: 61.3 }
  - { name: "Agoda", period: "Three Months", frequency: 75.3 }
  - { name: "Microsoft", period: "Three Months", frequency: 48.3 }
  - { name: "Meta", period: "Three Months", frequency: 43.3 }
  - { name: "Google", period: "Three Months", frequency: 23.9 }
  - { name: "TikTok", period: "Six Months", frequency: 64.9 }
sidebar:
  label: 'Subsets'
  badge: 'Medium'
---

### 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.

## Solution

```py
class Solution:
    def subsets(self, nums: list[int]) -> list[list[int]]:
        res = []

        def backtrack(i, path):
            res.append(path[:])
            for num_idx in range(i, len(nums)):
                path.append(nums[num_idx])
                backtrack(num_idx + 1, path)
                path.pop()

        backtrack(0, [])
        return res
```