<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:
    vector<vector<int>> combine(int n, int k) {
        int total = 1;
        for (int i = 0; i < k; ++i)
        {
            total *= (n-i);
            total /= (i+1);
        }
        
        answer = std::vector<std::vector<int>>(total, std::vector<int>(k));
        index = 0;

        std::vector<int> nums(k, 0);
        combine_rec(1, n, k, nums);

        return answer;
    }

private:
    void combine_rec(int start, int n, int k, std::vector<int>& nums)
    {
        if (k == 1)
        {
            for (int i = start; i <= n; ++i)
            {
                answer[index] = nums;
                answer[index][0] = i;
                ++index;
            }
            return;
        }

        for (int i = start; i <= n; ++i)
        {
            nums[k-1] = i;
            combine_rec(i+1, n, k-1, nums);
        }
    }

    std::vector<std::vector<int>> answer;
    int index;
};
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