<p>Given two integers <code>n</code> and <code>k</code>, return <em>all possible combinations of</em> <code>k</code> <em>numbers chosen from the range</em> <code>[1, n]</code>.</p>

<p>You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 4, k = 2
<strong>Output:</strong> [[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
<strong>Explanation:</strong> There are 4 choose 2 = 6 total combinations.
Note that combinations are unordered, i.e., [1,2] and [2,1] are considered to be the same combination.
</pre>

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

<pre>
<strong>Input:</strong> n = 1, k = 1
<strong>Output:</strong> [[1]]
<strong>Explanation:</strong> There is 1 choose 1 = 1 total combination.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 20</code></li>
	<li><code>1 &lt;= k &lt;= n</code></li>
</ul>

class Solution {
public:


    void fun(int n, int r, vector<vector<int>> &ans, int idx, int curr_count, vector<int> curr_vector)
{
    if(idx== n+1)return;
    if(curr_count==r)
    {
        ans.push_back(curr_vector);
        return;
    }
    
    for(int i=idx; i<n+1;i++)
    {
        curr_vector.push_back(i);
        fun(n,r,ans,i+1,curr_count+1,curr_vector);
        curr_vector.pop_back();
        
    }
    return; 

}

    vector<vector<int>> combine(int n, int k) {
        vector<vector<int>> ans;
     vector<int>curr_vector;
     fun(n+1,k,ans,1,0,curr_vector);
     return ans;
        
        
    }
};
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