690. Employee Importance

https://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.

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.

----

Intuition

Explore the hierarchy starting from given id, and add all importance

Save the employee id, employee in hashmap for quick lookup of subordinate from starting employee id

DFS or BFS both work 

	/*
	// Definition for Employee.
	class Employee {
	public int id;
	public int importance;
	public List<Integer> subordinates;
	};
	*/
	class Solution {
	public int getImportance(List<Employee> employees, int id) {
	if (employees == null || employees.size() == 0) {
	return 0;
	}
	Map<Integer, Employee> map = new HashMap<Integer, Employee>();
	for (int i = 0; i < employees.size(); i++) {
	map.put(employees.get(i).id, employees.get(i));
	}
	return dfs(employees, map.get(id), map);
	}
	private int dfs(List<Employee> employees, Employee emp, Map<Integer, Employee> map) {
	int ans = emp.importance;
	for (int sub: emp.subordinates) {
	ans += dfs(employees, map.get(sub), map);
	}
	return ans;
	}
	}

view raw 690.EmployeeImportanceDFS.java hosted with ❤ by GitHub

	/*
	// Definition for Employee.
	class Employee {
	public int id;
	public int importance;
	public List<Integer> subordinates;
	};
	*/
	class Solution {
	public int getImportance(List<Employee> employees, int id) {
	if (employees == null || employees.size() == 0) {
	return 0;
	}
	Map<Integer, Employee> map = new HashMap<Integer, Employee>();
	for (int i = 0; i < employees.size(); i++) {
	map.put(employees.get(i).id, employees.get(i));
	}
	Queue<Employee> q = new LinkedList<Employee>();
	q.offer(map.get(id));
	int ans = 0;
	while (!q.isEmpty()) {
	Employee current = q.poll();
	ans += current.importance;
	for (Integer sub: current.subordinates) {
	q.offer(map.get(sub));
	}
	}
	return ans;
	}
	}

view raw 690.EmployeeImportanceBFS.java hosted with ❤ by GitHub