<p>Given an array <code>nums</code> of distinct integers, return <em>all the possible permutations</em>. You can return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [0,1]
<strong>Output:</strong> [[0,1],[1,0]]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> nums = [1]
<strong>Output:</strong> [[1]]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 6</code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
	<li>All the integers of <code>nums</code> are <strong>unique</strong>.</li>
</ul>

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        res = []

        def back(idx):
            #Exit condition
            if idx > len(nums):
                return
            
            if idx == len(nums):
                res.append(list(nums))
            
            for i in range(idx,len(nums)):
                nums[i],nums[idx] = nums[idx],nums[i]
                back(idx+1)
                nums[i],nums[idx] = nums[idx],nums[i]
        
        back(0)
        return res

        
        
16600