Is Combination Sum III easy or hard?

Combination Sum III is generally rated Medium difficulty. The constraints are small, but the problem requires understanding how to generate combinations with recursion and prune invalid paths efficiently.

Combination Sum III Python/Java solution

Python and Java implementations typically use recursive backtracking. Start from a candidate number, append it to the current path, reduce the remaining sum, and recursively continue with the next number. When the path length equals k and the remaining sum is zero, add the combination to the result list.

How to solve Combination Sum III in O(n)?

An O(n) solution is not possible because the problem requires exploring combinations of numbers. The optimal approach is backtracking with pruning, which effectively explores a small subset of the 2^9 possible combinations. This keeps the runtime manageable despite being exponential in theory.

What is the best approach for Combination Sum III?

Backtracking is the best approach because it generates combinations incrementally and prunes invalid states early. By stopping when the sum exceeds the target or the combination size exceeds k, the search space stays small. The effective complexity is about O(2^9) since the numbers are limited to 1 through 9.

Is Combination Sum III asked at Google/Amazon/Meta?

Combination-style backtracking problems frequently appear in interviews at companies like Amazon, Google, and Meta. Variants such as Combination Sum, Subsets, and Combination Sum II are common. This problem specifically tests recursion, pruning, and combinatorial generation.

What data structure is used in Combination Sum III?

The solution mainly uses a dynamic list or array to store the current combination while exploring possibilities. Recursion manages the search tree, and results are stored in a list of lists. The algorithmic technique is backtracking rather than a specialized data structure.

What is the time complexity of Combination Sum III?

The typical backtracking solution runs in about O(2^9) time because each number from 1 to 9 can either be included or skipped. The recursion depth is at most k, so auxiliary space is O(k). A bitmask enumeration approach takes O(2^9 * 9) time because each subset checks up to nine elements.

Description Solution Video 23:00

216. Combination Sum III

Medium73.6% acceptance4 companies

Find all valid combinations of k numbers that sum up to n such that the following conditions are true:

Only numbers 1 through 9 are used.
Each number is used at most once.

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

Example 1:

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

Example 2:

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

Example 3:

Input: k = 4, n = 1
Output: []
Explanation: 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 > 1, there are no valid combination.

Constraints: