<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 {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector < vector < int > > ans;
        vector < bool > fr(nums.size(), false);
        vector < int > aux;
        bkt(nums, ans, 1, fr, aux);
        return ans;
    }
private:
    void bkt(vector < int > nums, vector < vector < int > >& ans, int step, vector < bool > fr, vector < int > aux) {

        for (int i = 0 ; i < nums.size() ; i++) {
            if (!fr[i]) {
                fr[i] = true;
                aux.push_back(nums[i]);
                if (step == nums.size()) {
                    ans.push_back(aux);
                } else {
                    bkt(nums, ans, step + 1, fr, aux);
                }
                fr[i] = false;
                aux.pop_back();
            }
        }
    }
};
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