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56 lines
1.6 KiB
Plaintext
56 lines
1.6 KiB
Plaintext
---
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title: '39. Combination Sum'
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description: Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order
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icon: dot
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topics:
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- { name: "Array", slug: "array" }
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- { name: "Backtracking", slug: "backtracking" }
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sidebar:
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label: 'Combination Sum'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `candidates = [2,3,6,7], target = 7`
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- Output: `[[2,2,3],[7]]`
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- Explanation: `2` and `3` are `candidates`, and `2 + 2 + 3 = 7`. Note that `2` can be used multiple times. `7` is a candidate, and `7 = 7`. These are the only two combinations.
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### Example 2:
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- Input: `candidates = [2,3,5], target = 8`
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- Output: `[[2,2,2,2],[2,3,3],[3,5]]`
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### Example 3:
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- Input: `candidates = [2], target = 1`
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- Output: `[]`
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### Constraints:
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- `1 <= candidates.length <= 30`
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- `2 <= candidates[i] <= 40`
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- All elements of `candidates` are distinct.
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- `1 <= target <= 40`
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## Solution
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```py
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class Solution:
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def combinationSum(self, candidates: list[int], target: int) -> list[list[int]]:
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candidates.sort()
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result, path = [], []
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def backtrack(start, remaining):
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if remaining == 0:
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result.append(path[:])
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return
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for i in range(start, len(candidates)):
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if candidates[i] > remaining:
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break
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path.append(candidates[i])
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backtrack(i, remaining - candidates[i])
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path.pop()
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backtrack(0, target)
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return result
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```
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