In this guide, you will get 216. Combination Sum III LeetCode Solution with the best time and space complexity. The solution to Combination Sum III problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Find all valid combinations of k numbers that sum up to n such that the following conditions are true:
Only numbers 1 through 9 are used.
Each number is used at most once.
Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.
Example 1:
Input: k = 3, n = 7
Output: [[1,2,4]]
Explanation:
1 + 2 + 4 = 7
There are no other valid combinations.
Example 2:
Input: k = 3, n = 9
Output: [[1,2,6],[1,3,5],[2,3,4]]
Explanation:
1 + 2 + 6 = 9
1 + 3 + 5 = 9
2 + 3 + 4 = 9
There are no other valid combinations.
Example 3:
Input: k = 4, n = 1
Output: []
Explanation: There are no valid combinations.
Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10 and since 10 > 1, there are no valid combination.
Constraints:
2 <= k <= 9
1 <= n <= 60
Complexity Analysis
Time Complexity: O(C(9, k)) = O(9^k)
Space Complexity: O(n)
216. Combination Sum III LeetCode Solution in C++
class Solution {
public:
vector<vector<int>> combinationSum3(int k, int n) {
vector<vector<int>> ans;
dfs(k, n, 1, {}, ans);
return ans;
}
private:
void dfs(int k, int n, int s, vector<int>&& path, vector<vector<int>>& ans) {
if (k == 0 && n == 0) {
ans.push_back(path);
return;
}
if (k == 0 || n <= 0)
return;
for (int i = s; i <= 9; ++i) {
path.push_back(i);
dfs(k - 1, n - i, i + 1, std::move(path), ans);
path.pop_back();
}
}
}; /* code provided by PROGIEZ */
216. Combination Sum III LeetCode Solution in Java
class Solution {
public List<List<Integer>> combinationSum3(int k, int n) {
List<List<Integer>> ans = new ArrayList<>();
dfs(k, n, 1, new ArrayList<>(), ans);
return ans;
}
private void dfs(int k, int n, int s, List<Integer> path, List<List<Integer>> ans) {
if (k == 0 && n == 0) {
ans.add(new ArrayList<>(path));
return;
}
if (k == 0 || n < 0)
return;
for (int i = s; i <= 9; ++i) {
path.add(i);
dfs(k - 1, n - i, i + 1, path, ans);
path.remove(path.size() - 1);
}
}
} // code provided by PROGIEZ
216. Combination Sum III LeetCode Solution in Python
class Solution:
def combinationSum3(self, k: int, n: int) -> list[list[int]]:
ans = []
def dfs(k: int, n: int, s: int, path: list[int]) -> None:
if k == 0 and n == 0:
ans.append(path)
return
if k == 0 or n < 0:
return
for i in range(s, 10):
dfs(k - 1, n - i, i + 1, path + [i])
dfs(k, n, 1, [])
return ans # code by PROGIEZ
