Employee Importance

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.

Solution: Using DFS

    int getImportance(vector<Employee*> employees, int id) {
        unordered_set<int> s;
        unordered_map<int, Employee*> m;
        for (auto e : employees) m[e->id] = e;
        return helper(id, m, s);
    }

    int helper(int id, unordered_map<int, Employee*>& m, unordered_set<int>& s) {
        if (s.count(id)) return 0;
        s.insert(id);
        int res = m[id]->importance;
        for (int num : m[id]->subordinates) {
            res += helper(num, m, s);
        }
        return res;
    }