39. Combination Sum
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.
The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the
frequency
of at least one of the chosen numbers is different.
The test cases are generated such that the number of unique combinations that sum up to target is less than 150 combinations for the given input.
Example 1:
Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
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.
Example 2:
Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]
Example 3:
Input: candidates = [2], target = 1
Output: []
Constraints:
- 1 <= candidates.length <= 30
- 2 <= candidates[i] <= 40
- All elements of candidates are distinct.
- 1 <= target <= 40
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
ans = []
candidates.sort()
def dfs(nums: List[int], remain: int, res: List[int], idx: int):
for i in range(idx, len(nums)):
if nums[i] == remain:
ans.append(res[:] + [nums[i]])
return
elif nums[i] < remain:
dfs(nums, remain - nums[i], res[:] + [nums[i]], i)
dfs(candidates, target, [], 0)
return ans
To figure out combinations where elemets are unique and their sum equals target, we should check each elements to see if it can be included in the combinations while iterating throguh each element. Before checking every elements, we have to sort candidates in ascending order since a combination consists of a unique list. It means that we should not use elements that precede the current one in the sorted array.
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