39. Combination Sum
MediumLeetCodeGiven 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
How to solve the problem
- Backtracking (DFS)
Code
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
path = []
result = []
candidates.sort() # Timsort
def backtracking(startIndex):
if sum(path) == target:
result.append(path[:])
return # Prune
if sum(path) > target: # Prune
return
for i in range(startIndex, len(candidates)):
path.append(candidates[i])
backtracking(i) # Recurse with the same index (since the same number can be used unlimited times)
path.pop()
backtracking(0)
return result
Complexity
- Time complexity: O(n^(target / min))
- Space complexity: O(target / min)
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