Employee Importance LeetCode Solution

Last updated on October 5th, 2024 at 04:34 pm

Here, We see Employee Importance LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

Breadth First Search, Depth First Search, Hash Table

Companies

Uber

Level of Question

Medium

Employee Importance LeetCode Solution

Employee Importance LeetCode Solution

Problem Statement

You have a data structure of employee information, including the employee’s unique ID, importance value, and direct subordinates’ IDs.

You are given an array of employees employees where:

  • employees[i].id is the ID of the ith employee.
  • employees[i].importance is the importance value of the ith employee.
  • employees[i].subordinates is a list of the IDs of the direct subordinates of the ith employee.

Given an integer id that represents an employee’s ID, return the total importance value of this employee and all their direct and indirect subordinates.

Example 1:

emp1 tree

Input: employees = [[1,5,[2,3]],[2,3,[]],[3,3,[]]], id = 1
Output: 11
Explanation: Employee 1 has an importance value of 5 and has two direct subordinates: employee 2 and employee 3. They both have an importance value of 3. Thus, the total importance value of employee 1 is 5 + 3 + 3 = 11.

Example 2:

emp2 tree

Input: employees = [[1,2,[5]],[5,-3,[]]], id = 5
Output: -3
Explanation: Employee 5 has an importance value of -3 and has no direct subordinates. Thus, the total importance value of employee 5 is -3.

1. Employee Importance LeetCode Solution C++

class Solution {
public:
    int getImportance(vector<Employee*> employees, int id) {
        unordered_map<int, Employee*>m;
        for(auto x: employees) m[x->id] = x;
        int sum = 0;
        deque<Employee*>q;
        q.push_back(m[id]);
        while(!q.empty()){
            auto p = q.front();
            q.pop_front();
            for(auto x: p->subordinates) q.push_back(m[x]);
            sum += p->importance;
        }
        return sum;
    }
};

2. Employee Importance LeetCode Solution Java

class Solution {
    public int getImportance(List<Employee> employees, int id) {
        Map<Integer, Employee> inputMap = new HashMap<>();
		for(Employee e : employees) {
			inputMap.put(e.id, e);
		}
		return helper(inputMap, id);
	}
	private static int helper(Map<Integer, Employee> inputMap, int id) {
		int imp = inputMap.get(id).importance;
		for(int subId : inputMap.get(id).subordinates) {
			imp += helper(inputMap, subId);
		}
		return imp;
	}
}

3. Employee Importance LeetCode Solution JavaScript

var GetImportance = function(employees, id) {
    let employeeMap = new Map();
    for (employee of employees) {
        employeeMap.set(employee.id, {importance : employee.importance, sub : employee.subordinates})
    }
    let totalImportance = 0;
    let queue = [id];
    while(queue.length > 0) {
        let currentEmployee = employeeMap.get(queue.shift());
        totalImportance += currentEmployee.importance;
        queue.push(...currentEmployee.sub)
    }
    return totalImportance
};

4. Employee Importance LeetCode Solution Python

class Solution(object):
    def getImportance(self, employees, id):
        def dfs(emp):
            imp = emps[emp].importance            
            for s in emps[emp].subordinates:
                imp += dfs(s)
            return imp
        emps= {emp.id: emp for emp in employees}
        return dfs(id)
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