<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 combinationgenerator(vector<vector<int>> &res, vector<int> &subarr, int n, int k, int i, int layer, int &resind)
    {
        if (layer > k)
        {
            res[resind++] = subarr;
        }
        else
        {
            for (int j = i; j <= n; j++)
            {
                subarr[layer - 1] = j;
                combinationgenerator(res, subarr, n, k, j + 1, layer + 1, resind);
            }
        }
    }
    long nCr(int n, int k)
    {
        long result = 1;
        for (int i = 0; i < k; i++)
        {
            result *= (n - i);
            result /= (i + 1);
        }
        return result;
    }
    vector<vector<int>> combine(int n, int k)
    {
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
        vector<vector<int>> res(nCr(n, k));
        vector<int> subarr(k, 0);
        int resind = 0;
        combinationgenerator(res, subarr, n, k, 1, 1, resind);
        return res;
    }
};
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