Question
Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
- All numbers (including target) will be positive integers.
- The solution set must not contain duplicate combinations.
Example 1:
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
Example 2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
Solution
Use backtracking. A partial solution is 0 or more candidates with a sum smaller or equal to target. Each recursion adds a number larger or equal to the last iteration to eliminate duplication.
For more information on backtracking, see this note.
Python objects are passed by reference. Thus, to make a copy of a list when a solution is found, use list(a_list)
or a_list[:]
to copy a new reference before adding to the result set. Or create a new list for every recursion with progress + [c]
.
class Solution:
def combinationSum(self, candidates, target):
"""
:type candidates: List[int]
:type target: int
:rtype: List[List[int]]
"""
solutions = list()
progress = list()
candidates.sort()
size = len(candidates)
def backtrack(last_index, remaining):
if remaining < 0:
return
if remaining == 0:
solutions.append(list(progress)) # copy progress
return
if remaining > 0:
for idx in range(last_index, size):
c = candidates[idx]
if c <= remaining:
progress.append(c)
backtrack(idx, remaining - c)
progress.pop()
return
backtrack(0, target)
return solutions