<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) {
        // k is the length of each vector<int>
        // n is the right end of the range of values that the values can be chosen from
        vector<vector<int>> res;
        vector<int> rem, cur;
        for (int i = 1; i <= n; i++) rem.push_back(i);
        findCombos(res, rem, cur, k);
        return res;
    }

    void findCombos(vector<vector<int>>& res, vector<int> rem, vector<int>& cur, int& k) {
        if (cur.size() == k) {
            res.push_back(cur);
            return;
        } else if (cur.size() + rem.size() < k) return;
        int n = rem.size();
        for (int i = 0; i < n; i++) {
            vector<int> next = cur;
            next.push_back(rem[0]);
            rem.erase(rem.begin());
            findCombos(res,rem,next,k);
        }
    }
};
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