[python/leetcode]employee importance/dfs

💬 문제

(leetcode.com/problems/employee-importance/)

 

You are given a data structure of employee information, which includes the employee's unique id, their importance value and their direct subordinates' id.

For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. They have importance value 15, 10 and 5, respectively. Then employee 1 has a data structure like [1, 15, [2]], and employee 2 has [2, 10, [3]], and employee 3 has [3, 5, []]. Note that although employee 3 is also a subordinate of employee 1, the relationship is not direct.

Now given the employee information of a company, and an employee id, you need to return the total importance value of this employee and all their subordinates.

 

💬 코드

"""
# Definition for Employee.
class Employee:
    def __init__(self, id: int, importance: int, subordinates: List[int]):
        self.id = id
        self.importance = importance
        self.subordinates = subordinates
"""

class Solution:
    def getImportance(self, employees: List['Employee'], id: int) -> int:
        graph = {e.id: e for e in employees}
        def dfs(i):
            employee = graph[i]
            return employee.importance + sum(dfs(j) for j in employee.subordinates)
        return dfs(id)

Employee는 class고 id, importance, subordinates를 속성으로 가지고 있다. 이를 무시하고 속성을 튜플에 저장해서 dfs를 사용하려 했다가 최대 재귀 횟수 초과라는 뼈아픈 실패를 맛봤다.