<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) {
        deque<vector<int>> dq;
        bt(n, k, dq);
        return vector(dq.begin(), dq.end());
    }

    void bt(int n, int k, deque<vector<int>>& dq) {
        // cout << "k: " << k << endl;
        if (k == 1) {
            for (int i = 1; i <= n - k + 1; ++i) {
                // cout << "i: " << i << ' ';
                dq.push_back({i});
            }
            // cout << endl;
            return;
        }

        bt(n, k - 1, dq);

        auto len = dq.size();
        for (int i = 0; i < len; ++i) {
            auto temp = dq.front();
            dq.pop_front();
            for (int i = temp.back() + 1; i <= n; ++i) {
                temp.push_back(i);
                if (temp.size() == k)
                    dq.push_back(temp);
                temp.pop_back();
            }
        }
    }
};
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