1+ # Given an array of distinct integers candidates and a target integer target,
2+ # return a list of all unique combinations of candidates where the chosen
3+ # numbers sum to target. You may return the combinations in any order.
4+ 5+ # The same number may be chosen from candidates an unlimited number of times.
6+ # Two combinations are unique if the
7+ # frequency
8+ # of at least one of the chosen numbers is different.
9+ 10+ # The test cases are generated such that the number of unique combinations that
11+ # sum up to target is less than 150 combinations for the given input.
12+ 13+ 14+ # Example 1:
15+ # Input: candidates = [2,3,6,7], target = 7
16+ # Output: [[2,2,3],[7]]
17+ # Explanation:
18+ 19+ # 2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
20+ # 7 is a candidate, and 7 = 7.
21+ # These are the only two combinations.
22+ 23+ # Example 2:
24+ # Input: candidates = [2,3,5], target = 8
25+ # Output: [[2,2,2,2],[2,3,3],[3,5]]
26+ 27+ # Example 3:
28+ # Input: candidates = [2], target = 1
29+ # Output: []
30+ 31+ 32+ # Constraints:
33+ 34+ # 1 <= candidates.length <= 30
35+ # 2 <= candidates[i] <= 40
36+ # All elements of candidates are distinct.
37+ # 1 <= target <= 40
38+ 39+ 40+ from typing import List
41+ class Solution :
42+ def combinationSum (self , candidates : List [int ], target : int ) -> List [List [int ]]:
43+ ans = []
44+ def dfs (s , t , path : list [int ]):
45+ if t < 0 :
46+ return
47+ if t == 0 :
48+ ans .append (path .copy ())
49+ for i in range (s , len (candidates )):
50+ path .append (candidates [i ])
51+ dfs (i , t - candidates [i ], path )
52+ path .pop ()
53+ dfs (0 , target , [])
54+ return ans
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