## # 39. Combination Sum

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]
]
``````

Approach 1: DFS.

### # Code (Python)

Approach 1:

``````class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
if not candidates or min(candidates) > target:
return []
result = []
current = []
#self._combination_sum(candidates, current, 0, target, result)
self._combination_sum2(candidates, current, 0, target, result)
return result

# method 1
def _combination_sum(self, candidates, current, index, target, result):
if target == 0:
result.append([_ for _ in current])
return
if target < 0 or index >= len(candidates):
return
# do not use candidate[index]
self._combination_sum(candidates, current, index + 1, target, result)
# or use candidate[index], and can use again
current.append(candidates[index])
self._combination_sum(candidates, current, index, target - candidates[index], result)
current.pop()

# method 2
def _combination_sum2(self, candidates, current, index, target, result):
if target == 0:
result.append([_ for _ in current])
return
if target < 0 or index >= len(candidates):
return
# # use i to limit duplicates -- limit the rest of the combination to using only the numbers on or after i
for i in range(index, len(candidates)):
self._combination_sum2(candidates, current + [candidates[i]], i, target - candidates[i], result)
``````

### # Code (C++)

Approach 1:

``````class Solution {
private:
vector<vector<int>> solutionSet;
void combinationSum(vector<int>& candidates, int head, int target, vector<int>& solution) {
if (target == 0)
{
solutionSet.push_back(solution);
return;
}
for (int i = head; i < candidates.size(); ++i)
{
if (candidates[i] > target)
break;
solution.push_back(candidates[i]);
combinationSum(candidates, i, target - candidates[i], solution);
solution.pop_back();
}
}
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<int> solution;
std::sort(candidates.begin(), candidates.end());
combinationSum(candidates, 0, target, solution);
return solutionSet;
}
};
``````