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

Solution

The logics for the template are:

  • start_index: the start index of the next potential candidate in candidates.
  • is_leaf: if remaining == 0 we have reached our target using elements in path.
  • get_edges: all candidates in candidates[start_index:].
  • is_valid: candidate[i] is valid if it does not exceed remaining and is not a duplicate.

We will follow the same steps as in Deduplication to validate and avoid a duplicate candidate.

Implementation

def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
    def dfs(start_index, path, remaining):
        if remaining == 0:
            ans.append(path[:])
        for i in range(start_index, len(candidates)):
            if remaining - candidates[i] < 0:   # avoid results exceeding target
                break
            elif i != start_index and candidates[i] == candidates[i-1]: # avoid duplicates
                continue
            path.append(candidates[i])
            dfs(i+1, path, remaining - candidates[i])
            path.pop()
    candidates.sort()
    ans = []
    dfs(0, [], target)
    return ans

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Question 1 out of 10

Which of the two traversal algorithms (BFS and DFS) can be used to find whether two nodes are connected?


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