LeetCode-in-Java

40. Combination Sum II

Medium

Given a collection of candidate numbers (candidates) 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.

Note: The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8

Output:

[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
] 

Example 2:

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

Output:

[
[1,2,2],
[5]
] 

Constraints:

• 1 <= candidates.length <= 100
• 1 <= candidates[i] <= 50
• 1 <= target <= 30

Solution

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

public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> sums = new ArrayList<>();
        // optimize
        Arrays.sort(candidates);
        combinationSum(candidates, target, 0, sums, new LinkedList<>());
        return sums;
    }

    private void combinationSum(
            int[] candidates,
            int target,
            int start,
            List<List<Integer>> sums,
            LinkedList<Integer> sum) {
        if (target == 0) {
            // make a deep copy of the current combination
            sums.add(new ArrayList<>(sum));
            return;
        }
        for (int i = start; i < candidates.length && target >= candidates[i]; i++) {
            // If candidate[i] equals candidate[i-1], then solutions for i is subset of
            // solution of i-1
            if (i == start || (i > start && candidates[i] != candidates[i - 1])) {
                sum.addLast(candidates[i]);
                // call on 'i+1' (not i) to avoid duplicate usage of same element
                combinationSum(candidates, target - candidates[i], i + 1, sums, sum);
                sum.removeLast();
            }
        }
    }
}

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