39. Combination Sum

Description of the Problem

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.

The test cases are generated such 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
• 2 <= candidates[i] <= 40
• All elements of candidates are distinct.
• 1 <= target <= 40

Solution

Explantion

Hints: For every target value, we either select first candidate or not to select first candidate. (See the below code)

Code (Rust)

impl Solution {
    pub fn combination_sum(candidates: Vec<i32>, target: i32) -> Vec<Vec<i32>> {
        let mut ans = vec![];
        let mut current_selection = vec![];
        Solution::helper(&candidates, target, &mut current_selection, &mut ans);
        return ans;
    }

    fn helper(
        candidates : &[i32], 
        target: i32,
        current_selection: &mut Vec<i32>,
        ans: &mut Vec<Vec<i32>>
    ) {
        // Meet the target
        if target == 0 {
            // 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, target - candidates[0], current_selection, ans);
            current_selection.pop();

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