# Leetcode Permutations

#### 2022-07-20 14:16

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

##### Related topics:
```expander
tag:#backtrack
```
 
 

##### Links:
- [Link to problem](https://leetcode.com/problems/permutations/)
___
### Problem

Given an array `nums` of distinct integers, return _all the possible permutations_. You can return the answer in **any order**.

#### Examples

**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

**Constraints:**

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

### Thoughts

> [!summary]
> This is a #backtrack problem

Backtrack problem, see alse [[Leetcode Combinations]]

### Solution

```cpp
class Solution {
  void backtrack(vector<vector<int>> &ans, vector<int> &combs, bool unused[],
                 vector<int> &nums, int n, int k) {
    // if the end is hit, append it to ans.
    if (k == 0) {
      ans.push_back(combs);
    } else {
      // iterate over the unused vector
      for (int i = 0; i < n; i++) {
        if (!unused[i]) {
          continue;
        }
        unused[i] = false;
        combs.push_back(nums[i]);

        backtrack(ans, combs, unused, nums, n, k - 1);

        combs.pop_back();
        unused[i] = true;
      }
    }
  }

public:
  vector<vector<int>> permute(vector<int> &nums) {
    // This can be modified from combinations
    vector<vector<int>> ans = {};
    vector<int> combs = {};
    int size = nums.size();

    bool unused[size];
    for (int i = 0; i < size; i++) {
      unused[i] = true;
    }

    backtrack(ans, combs, unused, nums, size, size);
    return ans;
  }
};
```