<p>Given an integer array <code>nums</code> of <strong>unique</strong> elements, return <em>all possible</em> <span data-keyword="subset"><em>subsets</em></span> <em>(the power set)</em>.</p>

<p>The solution set <strong>must not</strong> contain duplicate subsets. Return the solution 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],[1,2],[3],[1,3],[2,3],[1,2,3]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0]
<strong>Output:</strong> [[],[0]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

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

class Solution {
// void solve(int idx, vector<int>& nums, vector<vector<int>> &result,vector<int> &temp){
//     if(idx >= nums.size()){
//         result.push_back(temp);
//         return;
//     }

//     temp.push_back(nums[idx]); //include
//     solve(idx+1,nums,result,temp);
//     temp.pop_back();            //exclude
//     solve(idx+1,nums,result,temp);

// }  

void solve(int idx, vector<int>& nums, vector<vector<int>> &result,vector<int> &temp){

        result.push_back(temp);//we dont have a condition becoz we have to put all combinations into result
                                //even empty subset
        for(int i = idx; i < nums.size(); i++) {
            temp.push_back(nums[i]);
            solve(i + 1,nums,result,temp); //i+1, ensures that each element is considered only once in each subset
            temp.pop_back();
        }
}

public:
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> result;
        vector<int> temp;
        solve(0,nums,result,temp);
        return result;
    }
};
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