39. Combination Sum

• Difficulty: Medium

• Topics: Array, Backtracking

• Similar Questions:

  • Letter Combinations of a Phone Number

  • Combination Sum II

  • Combinations

  • Combination Sum III

  • Factor Combinations

  • Combination Sum IV

Problem:

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]
]

Solutions:

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<int> path;
        vector<vector<int>> ret;
        helper(candidates, 0, target, path, ret);

        return ret;
    }

    void helper(vector<int>& candidates, int pos, int target, vector<int>& path, vector<vector<int>>& ret) {  
        if (target < 0) return;
        if (target == 0) {
            ret.push_back(path);
            return;
        }
        if (pos == candidates.size())   return;

        path.push_back(candidates[pos]);
        helper(candidates, pos, target - candidates[pos], path, ret);
        path.pop_back();
        helper(candidates, pos + 1, target, path, ret);
    }
};