39. 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

Approach 1: DFS.

Code (Python)

Approach 1:

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        if not candidates or min(candidates) > target:
            return []
        result = []
        current = []
        #self._combination_sum(candidates, current, 0, target, result)
        self._combination_sum2(candidates, current, 0, target, result)
        return result
    
    # method 1
    def _combination_sum(self, candidates, current, index, target, result):
        if target == 0:
            result.append([_ for _ in current])
            return
        if target < 0 or index >= len(candidates):
            return
        # do not use candidate[index]
        self._combination_sum(candidates, current, index + 1, target, result)
        # or use candidate[index], and can use again
        current.append(candidates[index])
        self._combination_sum(candidates, current, index, target - candidates[index], result)
        current.pop()
    
    # method 2
    def _combination_sum2(self, candidates, current, index, target, result):
        if target == 0:
            result.append([_ for _ in current])
            return
        if target < 0 or index >= len(candidates):
            return
            # # use i to limit duplicates -- limit the rest of the combination to using only the numbers on or after i
        for i in range(index, len(candidates)):
            self._combination_sum2(candidates, current + [candidates[i]], i, target - candidates[i], result)

Code (C++)

Approach 1:

class Solution {
private:
    vector<vector<int>> solutionSet;
    void combinationSum(vector<int>& candidates, int head, int target, vector<int>& solution) {
        if (target == 0)
        {
            solutionSet.push_back(solution);
            return;
        }
        for (int i = head; i < candidates.size(); ++i)
        {
            if (candidates[i] > target)
                break;
            solution.push_back(candidates[i]);
            combinationSum(candidates, i, target - candidates[i], solution);
            solution.pop_back();
        }
    }
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<int> solution;
        std::sort(candidates.begin(), candidates.end());
        combinationSum(candidates, 0, target, solution);
        return solutionSet;
    }
};