Combination Sum II LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [๐Ÿ’ฏCorrect]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Combination Sum II in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemCombination Sum IIโ€“ LeetCode Problem

Combination Sum IIโ€“ LeetCode Problem

Problem:

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 IIโ€“ LeetCode Solutions
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();
    }
  }
};
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);
    }
  }
}
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)
        return

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

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

    return ans

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