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216. Combination Sum III

Description

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

Solutions

Solution 1: Pruning + Backtracking (Two Approaches)

We design a function $dfs(i, s)$, which represents that we are currently enumerating the number $i$, and there are still numbers with a sum of $s$ to be enumerated. The current search path is $t$, and the answer is $ans$.

The execution logic of the function $dfs(i, s)$ is as follows:

Approach One:

  • If $s = 0$ and the length of the current search path $t$ is $k$, it means that a group of answers has been found. Add $t$ to $ans$ and then return.
  • If $i \gt 9$ or $i \gt s$ or the length of the current search path $t$ is greater than $k$, it means that the current search path cannot be the answer, so return directly.
  • Otherwise, we can choose to add the number $i$ to the search path $t$, and then continue to search, i.e., execute $dfs(i + 1, s - i)$. After the search is completed, remove $i$ from the search path $t$; we can also choose not to add the number $i$ to the search path $t$, and directly execute $dfs(i + 1, s)$.
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class Solution:
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        def dfs(i: int, s: int):
            if s == 0:
                if len(t) == k:
                    ans.append(t[:])
                return
            if i > 9 or i > s or len(t) >= k:
                return
            t.append(i)
            dfs(i + 1, s - i)
            t.pop()
            dfs(i + 1, s)

        ans = []
        t = []
        dfs(1, n)
        return ans

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class Solution {
    private List<List<Integer>> ans = new ArrayList<>();
    private List<Integer> t = new ArrayList<>();
    private int k;

    public List<List<Integer>> combinationSum3(int k, int n) {
        this.k = k;
        dfs(1, n);
        return ans;
    }

    private void dfs(int i, int s) {
        if (s == 0) {
            if (t.size() == k) {
                ans.add(new ArrayList<>(t));
            }
            return;
        }
        if (i > 9 || i > s || t.size() >= k) {
            return;
        }
        t.add(i);
        dfs(i + 1, s - i);
        t.remove(t.size() - 1);
        dfs(i + 1, s);
    }
}

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class Solution {
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int>> ans;
        vector<int> t;
        function<void(int, int)> dfs = [&](int i, int s) {
            if (s == 0) {
                if (t.size() == k) {
                    ans.emplace_back(t);
                }
                return;
            }
            if (i > 9 || i > s || t.size() >= k) {
                return;
            }
            t.emplace_back(i);
            dfs(i + 1, s - i);
            t.pop_back();
            dfs(i + 1, s);
        };
        dfs(1, n);
        return ans;
    }
};

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func combinationSum3(k int, n int) (ans [][]int) {
    t := []int{}
    var dfs func(i, s int)
    dfs = func(i, s int) {
        if s == 0 {
            if len(t) == k {
                ans = append(ans, slices.Clone(t))
            }
            return
        }
        if i > 9 || i > s || len(t) >= k {
            return
        }
        t = append(t, i)
        dfs(i+1