LeetCode. 39. Combination Sum

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Given distinct positive integers candidates and a target, return every unique combination whose values sum to the target. A candidate may be used any number of times.

Sort the candidates, then use depth-first backtracking. The helper carries a start index and the remaining sum. Each recursive choice begins at the chosen candidate’s index, so that candidate can be reused; starting there also prevents choices from moving back to a smaller value. Thus each combination is built in nondecreasing order and has exactly one representation, avoiding duplicate permutations.

After choosing a value, subtract it from the remainder, recurse, then remove the value to restore the current path. Since the candidates are positive and sorted, stop the loop as soon as a candidate exceeds the remainder. When the remainder reaches zero, copy the path into the result.

Let V be the number of search states visited and S the number of returned combinations. The search takes O(V + S·L) time, where L is the maximum combination length (copying each result costs up to L); sorting costs O(n log n). The recursion and current path use O(L) auxiliary space, excluding the returned results.

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        Arrays.sort(candidates);
        List<List<Integer>> result = new ArrayList<>();
        backtrack(candidates, 0, target, new ArrayList<>(), result);
        return result;
    }

    private void backtrack(
            int[] candidates,
            int start,
            int remaining,
            List<Integer> path,
            List<List<Integer>> result) {
        if (remaining == 0) {
            result.add(new ArrayList<>(path));
            return;
        }

        for (int i = start; i < candidates.length; i++) {
            int candidate = candidates[i];
            if (candidate > remaining) {
                break;
            }

            path.add(candidate);
            backtrack(candidates, i, remaining - candidate, path, result);
            path.remove(path.size() - 1);
        }
    }
}