<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>

class Solution:
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        if n == 0 or k == 0:
            return []
        ans = []
        def backtrack(curr,i, total):
            if total == n and len(curr) ==k:
                ans.append(curr[:])
                return
            
            for j in range(i,10):
                if total + j <= n:
                    curr.append(j)
                    backtrack(curr,j+1,total+j)
                    curr.pop()
            
            
        backtrack([],1,0)
        return ans
16100