Combination Sum III
<p>Find all valid combinations of <code>k</code> numbers that sum up to <code>n</code> such that the following conditions are true:</p>

<ul>
	<li>Only numbers <code>1</code> through <code>9</code> are used.</li>
	<li>Each number is used <strong>at most once</strong>.</li>
</ul>

<p>Return <em>a list of all possible valid combinations</em>. The list must not contain the same combination twice, and the combinations may be returned in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> k = 3, n = 7
<strong>Output:</strong> [[1,2,4]]
<strong>Explanation:</strong>
1 + 2 + 4 = 7
There are no other valid combinations.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> k = 3, n = 9
<strong>Output:</strong> [[1,2,6],[1,3,5],[2,3,4]]
<strong>Explanation:</strong>
1 + 2 + 6 = 9
1 + 3 + 5 = 9
2 + 3 + 4 = 9
There are no other valid combinations.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> k = 4, n = 1
<strong>Output:</strong> []
<strong>Explanation:</strong> There are no valid combinations.
Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10 and since 10 &gt; 1, there are no valid combination.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= k &lt;= 9</code></li>
	<li><code>1 &lt;= n &lt;= 60</code></li>
</ul>
Array; Backtracking
class Solution:
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        def dfs(i: int, s: int):
            if s == 0:
                if len(t) == k:
                    ans.append(t[:])
                return
            if i > 9 or i > s or len(t) >= k:
                return
            t.append(i)
            dfs(i + 1, s - i)
            t.pop()
            dfs(i + 1, s)

        ans = []
        t = []
        dfs(1, n)
        return ans
