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Combination Sum II LeetCode Solution

Problem – Combination Sum II LeetCode Solution

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

Example 2:

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


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

Combination Sum II LeetCode Solution in Java

 public List<List<Integer>> combinationSum2(int[] cand, int target) {
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    List<Integer> path = new ArrayList<Integer>();
    dfs_com(cand, 0, target, path, res);
    return res;
void dfs_com(int[] cand, int cur, int target, List<Integer> path, List<List<Integer>> res) {
    if (target == 0) {
        res.add(new ArrayList(path));
        return ;
    if (target < 0) return;
    for (int i = cur; i < cand.length; i++){
        if (i > cur && cand[i] == cand[i-1]) continue;
        path.add(path.size(), cand[i]);
        dfs_com(cand, i+1, target - cand[i], path, res);

Combination Sum II LeetCode Solution in Python

def combinationSum2(self, candidates, target):
    # Sorting is really helpful, se we can avoid over counting easily
    result = []
    self.combine_sum_2(candidates, 0, [], result, target)
    return result
def combine_sum_2(self, nums, start, path, result, target):
    # Base case: if the sum of the path satisfies the target, we will consider 
    # it as a solution, and stop there
    if not target:
    for i in xrange(start, len(nums)):
        # Very important here! We don't use `i > 0` because we always want 
        # to count the first element in this recursive step even if it is the same 
        # as one before. To avoid overcounting, we just ignore the duplicates
        # after the first element.
        if i > start and nums[i] == nums[i - 1]:

        # If the current element is bigger than the assigned target, there is 
        # no need to keep searching, since all the numbers are positive
        if nums[i] > target:

        # We change the start to `i + 1` because one element only could
        # be used once
        self.combine_sum_2(nums, i + 1, path + [nums[i]], 
                           result, target - nums[i])

Combination Sum II LeetCode Solution in C++

class Solution {
    vector<vector<int> > combinationSum2(vector<int> &num, int target) 
        vector<vector<int>> res;
        vector<int> local;
        findCombination(res, 0, target, local, num);
        return res;
    void findCombination(vector<vector<int>>& res, const int order, const int target, vector<int>& local, const vector<int>& num)
            for(int i = order;i<num.size();i++) // iterative component
                if(num[i]>target) return;
                if(i&&num[i]==num[i-1]&&i>order) continue; // check duplicate combination
                findCombination(res,i+1,target-num[i],local,num); // recursive componenet
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