216. Combination Sum III

Description of Problem

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

Solution

Tags: Backtracking Divide and Conquer

Explanation

For any candidate number, either use it or not to use.

Code (Rust)

impl Solution {
    pub fn combination_sum3(k: i32, n: i32) -> Vec<Vec<i32>> {
    let mut ans = vec![];
        let mut current_selection = vec![];
        Self::helper(&[1,2,3,4,5,6,7,8,9], n, k as usize, &mut current_selection, &mut ans);
        return ans;
    }

    fn helper(
        candidates : &[i32], 
        target: i32,
        k : usize,
        current_selection: &mut Vec<i32>,
        ans: &mut Vec<Vec<i32>>
    ) {
        // Meet the target
        if target == 0  && current_selection.len() == k{
            // Append Answer
            ans.push(current_selection.clone());
        // Do not have any candidate or current sum exceed the target
        }else if candidates.len() == 0 || target < 0 {
            return;
        }else {
            // Select the first candidate and then calculate the result
            current_selection.push(candidates[0]);
            Solution::helper(
                &candidates[1..], 
                target - candidates[0], 
                k,
                current_selection, 
                ans
            );
            current_selection.pop();

            // Skip the first candidate and then calculate the result
            Solution::helper(
                &candidates[1..], 
                target, 
                k,
                current_selection, 
                ans
            );
        }
    }
}