Combination Sum II LeetCode Solution | Easy Approach

Combination Sum II | 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
Output: 
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]

Example 2:

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

Constraints:

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

Combination Sum II Solutions

✅Time: 
✅Space: 

C++

 class Solution {
 public:
  vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
    vector<vector<int>> ans;

    sort(begin(candidates), end(candidates));
    dfs(candidates, 0, target, {}, ans);
    return ans;
  }

 private:
  void dfs(const vector<int>& A, int s, int target, vector<int>&& path,
           vector<vector<int>>& ans) {
    if (target < 0)
      return;
    if (target == 0) {
      ans.push_back(path);
      return;
    }

    for (int i = s; i < A.size(); ++i) {
      if (i > s && A[i] == A[i - 1])
        continue;
      path.push_back(A[i]);
      dfs(A, i + 1, target - A[i], move(path), ans);
      path.pop_back();
    }
  }
};

Java

 class Solution {
  public List<List<Integer>> combinationSum2(int[] candidates, int target) {
    List<List<Integer>> ans = new ArrayList<>();

    Arrays.sort(candidates);
    dfs(0, candidates, target, new ArrayList<>(), ans);
    return ans;
  }

  private void dfs(int s, int[] candidates, int target, List<Integer> path,
                   List<List<Integer>> ans) {
    if (target < 0)
      return;
    if (target == 0) {
      ans.add(new ArrayList<>(path));
      return;
    }

    for (int i = s; i < candidates.length; ++i) {
      if (i > s && candidates[i] == candidates[i - 1])
        continue;
      path.add(candidates[i]);
      dfs(i + 1, candidates, target - candidates[i], path, ans);
      path.remove(path.size() - 1);
    }
  }
}

Python

 class Solution:
  def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
    ans = []

    def dfs(s: int, target: int, path: List[int]) -> None:
      if target < 0:
        return
      if target == 0:
        ans.append(path.copy())
        return

      for i in range(s, len(candidates)):
        if i > s and candidates[i] == candidates[i - 1]:
          continue
        path.append(candidates[i])
        dfs(i + 1, target - candidates[i], path)
        path.pop()

    candidates.sort()
    dfs(0, target, [])
    return ans

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