/*
    Given array & a target, find all unique combos that sum to target, nums can only be used once
    Ex. candidates = [10,1,2,7,6,1,5], target = 8 -> [[1,1,6],[1,2,5],[1,7],[2,6]]

    Backtracking, generate all combo sums, push/pop + index checking to explore new combos

    Time: O(2^n)
    Space: O(n)
*/

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        
        vector<int> curr;
        vector<vector<int>> result;
        
        dfs(candidates, target, 0, 0, curr, result);
        return result;
    }
private:
    void dfs(vector<int>& candidates, int target, int sum, int start, vector<int>& curr, vector<vector<int>>& result) {
        if (sum > target) {
            return;
        }
        if (sum == target) {
            result.push_back(curr);
            return;
        }
        for (int i = start; i < candidates.size(); i++) {
            if (i > start && candidates[i] == candidates[i - 1]) {
                continue;
            }
            curr.push_back(candidates[i]);
            dfs(candidates, target, sum + candidates[i], i + 1, curr, result);
            curr.pop_back();
        }
    }
};