690. Employee Importance #

Problem #

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.

Example 1:

Input: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
Output: 11
Explanation:
Employee 1 has importance value 5, and he has two direct subordinates: employee 2 and employee 3. They both have importance value 3. So the total importance value of employee 1 is 5 + 3 + 3 = 11.

Note:

1. One employee has at most one direct leader and may have several subordinates.
2. The maximum number of employees won’t exceed 2000.

Problem Summary #

Given a data structure that stores employee information, it includes the employee’s unique id, importance, and the ids of direct subordinates. For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. Their corresponding importance values are 15, 10, and 5. Then the data structure for employee 1 is [1, 15, [2]], the data structure for employee 2 is [2, 10, [3]], and the data structure for employee 3 is [3, 5, []]. Note that although employee 3 is also a subordinate of employee 1, since they are not a direct subordinate, this is not reflected in employee 1’s data structure. Now, given all employee information for a company and a single employee id, return the sum of the importance values of this employee and all their subordinates.

Solution Approach #

• Easy problem. According to the problem statement, use DFS or BFS to find all employees subordinate to the given id, accumulate the importance of subordinate employees, and finally add the importance of the employee themself; that is the required result.

Code #

package leetcode

type Employee struct {
	Id           int
	Importance   int
	Subordinates []int
}

func getImportance(employees []*Employee, id int) int {
	m, queue, res := map[int]*Employee{}, []int{id}, 0
	for _, e := range employees {
		m[e.Id] = e
	}
	for len(queue) > 0 {
		e := m[queue[0]]
		queue = queue[1:]
		if e == nil {
			continue
		}
		res += e.Importance
		for _, i := range e.Subordinates {
			queue = append(queue, i)
		}
	}
	return res
}