LeetCode 39. Combination Sum

Question

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.

It is guaranteed 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
• 1 <= candidates[i] <= 200
• All elements of candidates are distinct.
• 1 <= target <= 500

Source: https://leetcode.com/problems/combination-sum/

Solution

We are given a distinct candidates array, so we do not need to use a set to avoid duplicates like in LeetCode 47. Permutations II. Just by making elements of every combination in a non-decreasing order, we can get unique combinations.

Pre-sorting helps us find valid candidates faster. When the candidates array is sorted, we can pass start position rather than start value to the deeper layer of DFS tree.

private void dfs(int[] candidates, int start, int target,
                 LinkedList<Integer> current, List<List<Integer>> result) {
    if (target < 0) {
        return;
    }
    if (target == 0) {
        List<Integer> copy = new ArrayList<>(current);
        result.add(copy);
        return;
    }

    for (int candi : candidates) {
        // combinations in non-decreasing order
        if (candi >= start) {
            // backtrace
            current.addLast(candi);
            dfs(candidates, candi, target - candi, current, result);
            current.removeLast();
        }
    }
}

public List<List<Integer>> combinationSum(int[] candidates, int target) {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> comb = new LinkedList<>();
    dfs(candidates, 0, target, comb, result);
    return result;
}

// improve performance under the same time complexity
// through pre-sorting candidates so that we can find valid candidates easily
private void dfs2(int[] sortedCandidates, int startPos, int target,
                 LinkedList<Integer> current, List<List<Integer>> result) {
    if (target < 0) {
        return;
    }
    if (target == 0) {
        List<Integer> copy = new ArrayList<>(current);
        result.add(copy);
        return;
    }

    // combinations in non-decreasing order
    for (int i = startPos; i < sortedCandidates.length; i++) {
        int candi = sortedCandidates[i];
        // backtrace
        current.addLast(candi);
        dfs2(sortedCandidates, i, target - candi, current, result);
        current.removeLast();
    }
}

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> comb = new LinkedList<>();
    Arrays.sort(candidates);
    dfs2(candidates, 0, target, comb, result);
    return result;
}