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Employee Importance
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Employee Importance
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/*
LeetCode 690:
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.
Solution:
Language: C++
Tag: BFS DFS
hint: since one employee has at most one direct leader and may have several subordinates.-- no need to consider duplicate
*/
/*
// 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:
int getImportance(vector<Employee*> employees, int id) {
/*
// 1. BFS - O(n) time + O(n) space
unordered_map<int, Employee*> hash;
for (Employee* item : employees) {
hash[item->id] = item;
}
int res = 0;
//One employee has at most one direct leader and may have several subordinates.-- no need hashID
queue<int> que;
que.push(id);
while (!que.empty()) {
int cur_ID = que.front();
que.pop();
res += hash[cur_ID]->importance;
for (int sub : hash[cur_ID]->subordinates) {
que.push(sub);
}
}
return res;
*/
// 2. DFS -- O(n) time + O(n) space
unordered_map<int, Employee*> hash;
for (Employee* item : employees) {
hash[item->id] = item;
}
return DFS_helper(hash, id);
}
int DFS_helper(unordered_map<int, Employee*> &hash, int id) {
int res = 0;
res += hash[id]->importance;
for (int sub : hash[id]->subordinates) {
res += DFS_helper(hash, sub);
}
return res;
}
};