<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>> result;
    vector<int> r;

    void per(vector<int> a,vector<int> track){
    
    int flag=0;

    for(int i: track){
        if(i==0){
            flag=1;
            break;
        }
    }

    if(!flag){
        result.push_back(r);
        return;
    }

    for(int i=0;i<a.size();i++){

        if(track[i]==0){
            track[i]=1;
            r.push_back(a[i]);
            per(a,track);
            r.pop_back();
            track[i]=0;
        }
            
        
    }
    }
    
    vector<vector<int>> permute(vector<int>& nums) {
    
    vector<int> track(nums.size(),0);
    per(nums,track);    
    return result;

    }
};
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