## Given 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.

Note:

• 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.

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]
``````

## Using backtracking algorithm

The solution of this problem is as same as the Combination Sum problem. But to remove all duplicate combinations, we need to sort our array at first.

``````public static List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<>();
Arrays.sort(candidates);
combinationSum2(candidates, target, res, new ArrayList<>(), 0);

return res;
}

public static void combinationSum2(int[] candidates, int target, List<List<Integer>> res, List<Integer> values, int idx) {
if (target < 0) {
return;
}

if (target == 0) {
return;
}

for (int i = idx; i < candidates.length; ++i) {
if (idx != i && candidates[i] == candidates[i - 1]) {
continue;
}

combinationSum2(candidates, target - candidates[i], res, values, i + 1);
values.remove(values.size() - 1);
}
}
``````

The complexity of this solution:

• Time complexity: O(n * C(k, n)).

## Wrapping up

• Understanding about the structure of the backtracking algorithm.

Refer:

https://en.wikipedia.org/wiki/External_sorting