39, Combination Sum

I 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: 2and3are 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

Related Topics

• Array
• Backtracking

II Solution

Approach 1: Backtracking

pub fn combination_sum(candidates: Vec<i32>, target: i32) -> Vec<Vec<i32>> {
    //Self::backtracking_1(candidates, target)
    Self::backtracking_2(candidates, target)
}

fn backtracking_1(candidates: Vec<i32>, target: i32) -> Vec<Vec<i32>> {
    const BACKTRACK: fn(usize, &[i32], i32, &mut Vec<i32>, &mut Vec<Vec<i32>>) =
        |idx, candidates, target, combine, res| {
            if target < 0 {
                return;
            }

            if target == 0 {
                res.push(combine.clone());
                return;
            }

            for i in idx..candidates.len() {
                combine.push(candidates[i]);

                BACKTRACK(i, candidates, target - candidates[i], combine, res);

                combine.pop();
            }
        };
    let mut res = vec![];

    BACKTRACK(0, &candidates, target, &mut vec![], &mut res);

    res
}

fn backtracking_2(candidates: Vec<i32>, target: i32) -> Vec<Vec<i32>> {
    const BACKTRACK: fn(usize, &[i32], i32, &mut Vec<i32>, &mut Vec<Vec<i32>>) =
        |idx, candidates, target, combine, res| {
            if idx == candidates.len() {
                return;
            }

            if target == 0 {
                res.push(combine.clone());
                return;
            }

            BACKTRACK(idx + 1, candidates, target, combine, res);

            if target - candidates[idx] >= 0 {
                combine.push(candidates[idx]);

                BACKTRACK(idx, candidates, target - candidates[idx], combine, res);

                combine.pop();
            }
        };
    let mut res = vec![];

    BACKTRACK(0, &candidates, target, &mut vec![], &mut res);

    res
}

public List<List<Integer>> combinationSum(int[] candidates, int target) {
    //return this.backtracking1(candidates, target);
    return this.backtracking2(candidates, target);
}

@FunctionalInterface
interface QuintConsumer<A, B, C, D, E> {
    void accept(A a, B b, C c, D d, E e);
}

QuintConsumer<Integer, int[], Integer, List<Integer>, List<List<Integer>>> backtrack1 =
        (idx, candidates, target, combine, res) -> {
            if (target < 0) {
                return;
            }

            if (target == 0) {
                res.add(new ArrayList<>(combine));
                return;
            }

            for (int i = idx; i < candidates.length; i++) {
                combine.addLast(candidates[i]);

                this.backtrack1.accept(i, candidates, target - candidates[i], combine, res);

                combine.removeLast();
            }
        };

List<List<Integer>> backtracking1(int[] candidates, int target) {
    List<List<Integer>> res = new ArrayList<>();

    this.backtrack1.accept(0, candidates, target, new ArrayList<>(), res);

    return res;
}

QuintConsumer<Integer, int[], Integer, List<Integer>, List<List<Integer>>> backtrack2 =
        (idx, candidates, target, combine, res) -> {
            if (idx == candidates.length) {
                return;
            }

            if (target == 0) {
                res.add(new ArrayList<>(combine));
                return;
            }

            this.backtrack2.accept(idx + 1, candidates, target, combine, res);

            if (target - candidates[idx] >= 0) {
                combine.addLast(candidates[idx]);

                this.backtrack2.accept(idx, candidates, target - candidates[idx], combine, res);

                combine.removeLast();
            }
        };

List<List<Integer>> backtracking2(int[] candidates, int target) {
    List<List<Integer>> res = new ArrayList<>();

    this.backtrack2.accept(0, candidates, target, new ArrayList<>(), res);

    return res;
}