Combination Sum LeetCode Solution | Easy Approach

Combination Sum | Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

It is guaranteed that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

Example 1:

Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.

Example 2:

Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:

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

Constraints:

• 1 <= candidates.length <= 30
• 1 <= candidates[i] <= 200
• All elements of candidates are distinct.
• 1 <= target <= 500

Combination Sum Solutions

✅Time: 

✅Space: O(target)

C++

 class Solution {
 public:
  vector<vector<int>> combinationSum(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) {
      path.push_back(A[i]);
      dfs(A, i, target - A[i], move(path), ans);
      path.pop_back();
    }
  }
};

Java

 class Solution {
  public List<List<Integer>> combinationSum(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) {
      path.add(candidates[i]);
      dfs(i, candidates, target - candidates[i], path, ans);
      path.remove(path.size() - 1);
    }
  }
}

Python

 class Solution:
  def combinationSum(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.clone())
        return

      for i in range(s, len(candidates)):
        path.append(candidates[i])
        dfs(i, target - candidates[i], path)
        path.pop()

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

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