<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:
    void solve(int ind,vector<int>nums,unordered_map<int,bool>&mp,vector<vector<int>>&ans,vector<int>temp,int n){
           
        if(ind == n){
            ans.push_back(temp);
            return;
        }
        for(int i = 0;i<n;i++){
            if(!mp[i]){
                mp[i] = true;
                temp.push_back(nums[i]);
                solve(ind+1,nums,mp,ans,temp,n);
                temp.pop_back();
                mp[i] = false;
            }
        }
         
    }
    vector<vector<int>> permute(vector<int>& nums) {
         
        vector<vector<int>>ans;
        unordered_map<int,bool>mp;
        solve(0,nums,mp,ans,{},nums.size());
        return ans;
    }
};
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