Algorithm Practice

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690 Employee Importance

690. Employee Importance

1. Question

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

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Input:

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[[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1

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Output:

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11

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Explanation:

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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.

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

2. Implementation

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class Solution {

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public int getImportance(List<Employee> employees, int id) {

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Map<Integer, Employee> map = new HashMap<>();

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for (Employee employee : employees) {

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map.put(employee.id, employee);

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}

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Queue<Employee> queue = new LinkedList<>();

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queue.add(map.get(id));

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int res = 0;

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while (!queue.isEmpty()) {

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Employee curEmployee = queue.remove();

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res += curEmployee.importance;

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for (Integer employeeId : curEmployee.subordinates) {

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queue.add(map.get(employeeId));

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}

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}

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return res;

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}

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}

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class Solution {

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public int getImportance(List<Employee> employees, int id) {

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Map<Integer, Employee> map = new HashMap<>();

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for (Employee employee : employees) {

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map.put(employee.id, employee);

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}

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int[] res = new int[1];

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getImportanceByDFS(map, id, res);

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return res[0];

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}

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public void getImportanceByDFS(Map<Integer, Employee> map, int id, int[] res) {

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res[0] += map.get(id).importance;

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for (Integer subId : map.get(id).subordinates) {

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getImportanceByDFS(map, subId, res);

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}

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}

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}

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3. Time & Space Complexity

BFS: 时间复杂度O(n), 空间复杂度O(n),因为用了HashMap

DFS: 时间复杂度O(n), 空间复杂度O(n),因为用了HashMap

Last modified 2yr ago