Backtracking - Combination Sum II

Given a collection of candidate numbers candidates (which may contain duplicates) and a target number target, find all unique combinations in candidates where the candidate numbers sum to target. Each number in candidates may only be used once in the combination.

Input: candidates = [10,1,2,7,6,1,5], target = 8
Output: [[1,1,6],[1,2,5],[1,7],[2,6]]
Explanation: 1+1+6=8, 1+2+5=8, 1+7=8, 2+6=8.

Input: candidates = [2,5,2,1,2], target = 5
Output: [[1,2,2],[5]]

Sort the array first. The key difference from Combination Sum I is that each element can only be used once (recurse with i+1) and we must skip duplicates at the same tree level to avoid duplicate combinations.

Sort candidates to bring duplicates adjacent and enable pruning.
In the loop, skip if i > start && candidates[i] == candidates[i-1].
Recurse with i+1 (not i) since each element is used at most once.
Prune when candidates[i] > remaining.

import java.util.*; class Solution { public List<List<Integer>> combinationSum2(int[] candidates, int target) { Arrays.sort(candidates); List<List<Integer>> result = new ArrayList<>(); backtrack(candidates, target, 0, new ArrayList<>(), result); return result; } private void backtrack(int[] candidates, int remaining, int start, List<Integer> current, List<List<Integer>> result) { if (remaining == 0) { result.add(new ArrayList<>(current)); return; } for (int i = start; i < candidates.length; i++) { if (candidates[i] > remaining) break; // Skip duplicates at the same depth level if (i > start && candidates[i] == candidates[i - 1]) continue; current.add(candidates[i]); backtrack(candidates, remaining - candidates[i], i + 1, current, result); current.remove(current.size() - 1); } } }

Complexity Analysis:

Time complexity: O(2^n) in the worst case since each element is either included or excluded.
Space complexity: O(n) for the recursion stack.