Combination Sum | 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.
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
1 <= candidates[i] <= 200
- All elements of
candidates
are distinct. 1 <= target <= 500
Combination Sum Solutions
✅Time:
✅Space: O(target)
C++
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> ans;
sort(begin(candidates), end(candidates));
dfs(candidates, 0, target, {}, ans);
return ans;
}
private:
void dfs(const vector<int>& A, int s, int target, vector<int>&& path,
vector<vector<int>>& ans) {
if (target < 0)
return;
if (target == 0) {
ans.push_back(path);
return;
}
for (int i = s; i < A.size(); ++i) {
path.push_back(A[i]);
dfs(A, i, target - A[i], move(path), ans);
path.pop_back();
}
}
};
Java
class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> ans = new ArrayList<>();
Arrays.sort(candidates);
dfs(0, candidates, target, new ArrayList<>(), ans);
return ans;
}
private void dfs(int s, int[] candidates, int target, List<Integer> path,
List<List<Integer>> ans) {
if (target < 0)
return;
if (target == 0) {
ans.add(new ArrayList<>(path));
return;
}
for (int i = s; i < candidates.length; ++i) {
path.add(candidates[i]);
dfs(i, candidates, target - candidates[i], path, ans);
path.remove(path.size() - 1);
}
}
}
Python
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
ans = []
def dfs(s: int, target: int, path: List[int]) -> None:
if target < 0:
return
if target == 0:
ans.append(path.clone())
return
for i in range(s, len(candidates)):
path.append(candidates[i])
dfs(i, target - candidates[i], path)
path.pop()
candidates.sort()
dfs(0, target, [])
return ans
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