<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 backtrack(int start, vector<int> visited, int depth, vector<vector<int>> &ans, int k) {
        int n = visited.size();
        if(depth == k) {
            vector<int> l;
            for(int i = 0; i < visited.size(); i++) {
                if(visited[i] == 1)
                l.push_back(i+1);
            }
            ans.push_back(l);
            return;
        }
        for(int i = start + 1; i < n; i++) {
            if(visited[i] == 1) {
                continue;
            }
            visited[i] = 1;
            backtrack(i, visited, depth + 1, ans, k);
            visited[i] = 0;
        }
    }

    vector<vector<int>> combine(int n, int k) {
        vector<int> v;
        for(int i = 1; i < n + 1; i++) {
            v.push_back(0);
        }
        vector<vector<int>> ans;
        for(int i = 0; i < n; i++) {
            v[i] = 1;
            backtrack(i, v, 1, ans, k);
            v[i] = 0;
        }
        return ans;
    }
};
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