Combination Sum

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

All numbers (including target) will be positive integers. The solution set must not contain duplicate combinations. Example 1:

Input: candidates = [2,3,6,7], target = 7, A solution set is: [ [7], [2,2,3] ]

Example 2:

Input: candidates = [2,3,5], target = 8, A solution set is: [ [2,2,2,2], [2,3,3], [3,5] ]

Solution: Using Backtracking

• First we need to sort the input array
• For backtracking, we need to pass the candidates and target
• We also pass the start index and previous choice vector to represent the current state
• The termination case is when the target is less than 0
• When target is equal to zero, we push the vector to result vector
• When target is larger, we try all the combinations start from the start index.

    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        vector<vector<int> > res;
        vector<int> out;
        sort(candidates.begin(), candidates.end());
        combinationSumDFS(candidates, target, 0, out, res);
        return res;
    }

    void combinationSumDFS(vector<int> &candidates, int target, int start, vector<int> &out, vector<vector<int> > &res) {
        if (target < 0) return;
        else if (target == 0) res.push_back(out);
        else {
            for (int i = start; i < candidates.size(); ++i) {
                out.push_back(candidates[i]);
                combinationSumDFS(candidates, target - candidates[i], i, out, res);
                out.pop_back();
            }
        }
    }