leetcode 690. Employee Importance

Posted on |

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:

One employee has at most one direct leader and may have several subordinates.

The maximum number of employees won’t exceed 2000.

dfs

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
/*
// Employee info
class Employee {
public:
    // It's the unique ID of each node.
    // unique id of this employee
    int id;
    // the importance value of this employee
    int importance;
    // the id of direct subordinates
    vector<int> subordinates;
};
*/
class Solution {
public:
      bool vis[2000];
    map<int,Employee*> emap;
    int getImportance(vector<Employee*> employees, int id) {
        for(int i=0;i<employees.size();i++)
            for(Employee* e:employees){
            emap[e->id]=e;
        }
      
       int cost=0;
        dfs(id,cost);
        return cost;
    }
    void dfs(int id,int &cost){
        cost+=emap[id]->importance;
        vis[id]=true;
        for(auto ch:emap[id]->subordinates){
            dfs(ch,cost);
        }
        
    }
};

bfs

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
/*
// Employee info
class Employee {
public:
    // It's the unique ID of each node.
    // unique id of this employee
    int id;
    // the importance value of this employee
    int importance;
    // the id of direct subordinates
    vector<int> subordinates;
};
*/
class Solution {
public:
    map<int,Employee*> emap;
    int getImportance(vector<Employee*> employees, int id) {
        for(int i=0;i<employees.size();i++)
            for(Employee* e:employees){
            emap[e->id]=e;
        }
        queue<Employee*> q;
        q.push(emap[id]);
        int cost=0;
        while(!q.empty()){
           Employee* a=q.front();
            q.pop();
            cost+=a->importance;
            for(auto ch:a->subordinates){
                q.push(emap[ch]);
                
            }
            
        }
        return cost;
        
    }
};