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: []
1 <= candidates.length <= 302 <= candidates[i] <= 40- All elements of
candidatesare distinct. 1 <= target <= 40
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
def backtrack(start, path, remaining):
# Base case: if the remaining target is 0, add the path to the result
if remaining == 0:
result.append(path.copy())
return
# Recursive case: try adding each candidate to the path
for i in range(start, len(candidates)):
if candidates[i] <= remaining:
path.append(candidates[i])
backtrack(i, path, remaining - candidates[i])
path.pop()
result = []
backtrack(0, [], target)
return resultThe time complexity of this solution is O(N^(T/M + 1)), where N is the number of candidates, T is the target value, and M is the minimal value among the candidates. This is because, in the worst case, we might need to explore each candidate up to T/M times.
The space complexity is O(T/M) for the recursion stack, as in the worst case, we might have a combination where we use the smallest candidate T/M times.