690. Employee Importance

690. Employee Importance (Easy)

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.

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.

Companies:
Amazon, Uber

Related Topics:
Hash Table, Depth-first Search, Breadth-first Search

Similar Questions:

• Nested List Weight Sum (Easy)

Solution 1. DFS

// OJ: https://leetcode.com/problems/employee-importance/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
private:
    unordered_map<int, Employee*> m;
    int collect(int id) {
        auto e = m[id];
        int ans = e->importance;
        for (auto &s : e->subordinates) ans += collect(s);
        return ans;
    }
public:
    int getImportance(vector<Employee*> employees, int id) {
        for (auto &e : employees) m[e->id] = e;
        return collect(id);
    }
};