<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>> answer;
    vector<int> path;
    int k;
    int n;

    void doPath(int current_minimal_value){
        // 1. put values

        for (int current_value = current_minimal_value; current_value <= n; current_value++){
            path.push_back(current_value);
            if (path.size() == k) {
                answer.push_back(path);
            }
            else if (current_value < n){
                doPath(current_value + 1);
            }
            path.pop_back();
        };
    }

    vector<vector<int>> combine(int n, int k) {
        
        int numAnswer = 1 / ((n + 1) * beta(n - k + 1, k + 1));
        answer.reserve(numAnswer);
        path.reserve(k);

        this->k = k;
        this->n = n;
        if (k > 0 && n > 0) doPath(1);

        return answer;
    }
};
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