---
title: '46. Permutations'
description: Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order
icon: dot
topics:
  - { name: "Array", slug: "array" }
  - { name: "Backtracking", slug: "backtracking" }
issue: "https://github.com/prdlk/leetcode/issues/151"
companies:
  - { name: "Bloomberg", period: "Thirty Days", frequency: 69.8 }
  - { name: "Amazon", period: "Thirty Days", frequency: 48.9 }
  - { name: "Microsoft", period: "Three Months", frequency: 42.1 }
  - { name: "Google", period: "Three Months", frequency: 32.7 }
  - { name: "Infosys", period: "Six Months", frequency: 67.9 }
  - { name: "LinkedIn", period: "Six Months", frequency: 54.1 }
  - { name: "Meta", period: "Six Months", frequency: 43.4 }
sidebar:
  label: 'Permutations'
  badge: 'Medium'
---

### Example 1:
- Input: `nums = [1,2,3]`
- Output: `[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]`

### Example 2:
- Input: `nums = [0,1]`
- Output: `[[0,1],[1,0]]`

### Example 3:
- Input: `nums = [1]`
- Output: `[[1]]`

### Constraints:

- `1 <= nums.length <= 6`
- `-10 <= nums[i] <= 10`
- All the integers of `nums` are unique.

## Solution

```py
class Solution:
    def permute(self, nums: list[int]) -> list[list[int]]:
        result, path = [], []
        used = [False] * len(nums)

        def backtrack():
            if len(path) == len(nums):
                result.append(path[:])
                return
            for i in range(len(nums)):
                if used[i]:
                    continue
                used[i] = True
                path.append(nums[i])
                backtrack()
                path.pop()
                used[i] = False

        backtrack()
        return result
```