<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 {

    vector<vector<int>> helper(int ind, vector<int>& nums){
        if(ind >= nums.size()){
            return {{}};
        }

        vector<vector<int>> resPerms;
        vector<vector<int>> perms;

        perms = helper(ind + 1, nums);

        for(auto p: perms){
            for(int i = 0; i <= p.size(); i++){
                vector<int> pCopy = p;
                pCopy.insert(pCopy.begin() + i, nums[ind]);
                resPerms.push_back(pCopy);
            }
        }

        return resPerms;
    }
public:
    vector<vector<int>> permute(vector<int>& nums) {
        return helper(0, nums);
    }
};
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