Combination Sum Problem

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. Each number in C may only be used once in the combination.

All numbers (including target) will be positive integers.
Elements in a combination (a1, a2,…, ak) must be in non-descending order.(ie, a1 ≤ a2 ≤.. ≤ ak).
The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8,

A solution set is: [1, 7] [1, 2, 5] [2, 6] [1, 1, 6]

Solution: The solution uses backtracking technique, which uses the constraint of sum of the total numbers in set to be the target number. Since it is backtracking algorithm the complexity is exponential.


public class CrazyForCode {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        List<Integer> list = new ArrayList<Integer>();

        backTrack(result, list, candidates, target, 0);

        return result;
    private void backTrack(List<List<Integer>> result, 
        List<Integer> list, int[] candidates, int target, 
        int position) {

            int sum = 0;
            for (int x: list) {
                    sum += x;

            if (sum == target) {
                    result.add(new ArrayList<Integer>(list));

            if (sum < target) 
                for (int i = position; i < candidates.length; i++) 
                    if(position != i 
                        && candidates[i] == candidates[i-1])
                    backTrack(result, list, candidates, target, i+1);
                    list.remove(list.size() - 1);

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