<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:
    def combine(self, n: int, k: int) -> List[List[int]]:
        # if choose only one number, return the original list
        # `[[1],[2],[3],[4]]`
        # if choose more than one number, sequentially concatenat the 
        # chosen number with combine the rest
        # e.g. choosing 3 numbers = choose 1 numbers and combine with 
        # (choosing 2 numbers from the rest)
        return self.chooseFrom(k, list(range(1, n+1)))
    def chooseFrom(self, k, nums):
        if k == 1:
            return [[x] for x in nums]
        i = 0
        ret = []
        while i < len(nums)-1:
            ret += [[nums[i]]+x for x in self.chooseFrom(k-1, nums[i+1:])]
            i += 1
        return ret
57225