# Permutations

{% embed url="<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**.
>
> &#x20;
>
> **Example 1:**
>
> <pre><code>Input: nums = [1,2,3]
> <strong>Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
> </strong></code></pre>
>
> **Example 2:**
>
> <pre><code>Input: nums = [0,1]
> <strong>Output: [[0,1],[1,0]]
> </strong></code></pre>
>
> **Example 3:**
>
> <pre><code>Input: nums = [1]
> <strong>Output: [[1]]
> </strong></code></pre>

### Pseudocode

```
- dfs with a for loop within
    - permutation with backtracking
    
- same as combination sum, don't understand this intuitively
```

### Solution

```javascript
// from solutions
// https://leetcode.com/problems/permutations/solutions/685868/dfs-backtracking-python-java-javascript-picture/
// good illustration on how to approach the problem
// time O(N^N), has to find every permutation
var permute = function (nums) {
  let res = [];
  dfs(nums, [], Array(nums.length).fill(false), res);
  return res;
};

function dfs(nums, path, used, res) {
  // console.log(path)
  if (path.length == nums.length) {
    // make a deep copy since otherwise we'd be append the same list over and over
    res.push(Array.from(path));
    return;
  }
  for (let i = 0; i < nums.length; i++) {
    // skip used nums
    if (used[i]) continue;
    // add letter to permutation, mark letter as used
    path.push(nums[i]);
    used[i] = true;
    dfs(nums, path, used, res);
    // remove letter from permutation, mark letter as unused
    path.pop();
    used[i] = false;
  }
}

```

### Time and Space Complexity

#### Time

* What did the code do
* Total -

#### Space

* What did the code do
* Total -


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