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.
It is guaranteed that the number of unique combinations that sum up to target is less than 150 combinations for the given input.
Test Cases
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]]
Solution
class Solution {
public:
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
sort(candidates.begin(), candidates.end());
vector<vector<int> > res;
vector<int> combination;
combinationSum(candidates, target, res, combination, 0);
return res;
}
private:
void combinationSum(vector<int> &candidates, int target, vector<vector<int> > &res, vector<int> &combination, int begin) {
if (!target) {
res.push_back(combination);
return;
}
for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i) {
combination.push_back(candidates[i]);
combinationSum(candidates, target - candidates[i], res, combination, i);
combination.pop_back();
}
}
};class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<>();
backtrack(candidates, target, 0, new ArrayList<>(), res);
return res;
}
private void backtrack(int[] candidates, int target, int pos, List<Integer> combo, List<List<Integer>> res) {
if (pos == candidates.length || target < 0) return;
if (target == 0) {
res.add(new ArrayList<>(combo));
return;
}
backtrack(candidates, target, pos+1, combo, res);
combo.add(candidates[pos]);
backtrack(candidates, target - candidates[pos], pos, combo, res);
combo.remove(combo.size() - 1);
}
}class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
res = []
def backtrack(sum: int, pos: int, arr: List[int]):
if sum <0 or pos >= len(candidates):
return
if sum == 0:
res.append(list(arr))
return
backtrack(sum, pos+1, arr)
arr.append(candidates[pos])
backtrack(sum-candidates[pos], pos, arr)
arr.pop()
backtrack(target, 0, [])
return res