690.employee-importance.java

/*
 * @lc app=leetcode id=690 lang=java
 *
 * [690] Employee Importance
 *
 * https://leetcode.com/problems/employee-importance/description/
 *
 * algorithms
 * Easy (58.49%)
 * Total Accepted:    98.8K
 * Total Submissions: 168.9K
 * Testcase Example:  '[[1,2,[2]], [2,3,[]]]\n2'
 *
 * You are given a data structure of employee information, which includes the
 * employee's unique id, their importance value and their 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 their
 * 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.
 *
 *
 */
/*
// Definition for Employee.
class Employee {
public int id;
public int importance;
public List<Integer> subordinates;
};
*/


// Given information:
//		List if Employee class unique id, importance, List of subordinates' id
//		Max Employee 2000
//		At least 1 employee has direct leader and more than 1 subordinates
//		Return all importance value of current employee and the subordinates
//
// Recursion?
// Needs to have the indexes correlated to the employee id
// Convert List into HashMap
// Nested loops?
// Needs to have a way of skipping employees already been searched


// RECURSION - Depth-First Search
// Convert List into HashMap with id as the key
// Store total importance value
// Get root employee
	// if root employee has no subordinates
		// return importance value
	// else, loop over subordinates and recurse down
class Solution {
	public Map<Integer, Employee> roster;

	public int getImportance(List<Employee> employees, int id) {

		if (employees == null) return 0;
		roster = new HashMap<Integer, Employee>();

		for (Employee e : employees) {
			roster.put(e.id, e);
		}

		return calculateImportance(id);
	}

	public int calculateImportance(int id) {
		Employee lead =  roster.get(id);
		int totalImp = lead.importance;

		if (lead.subordinates.isEmpty()) {
			return totalImp;
		} else {
			for (int subId : lead.subordinates) {
				totalImp += calculateImportance(subId);
			}
		}

		return totalImp;
	}
}

// // NESTED LOOPS
// class Solution {
// 	HashMap<Integer, Employee> roster;
// 	public int getImportance(List<Employee> employees, int id) {
// 		if (employees == null) return 0;
// 		roster = new HashMap<>();
//
// 		for (Employee e : employees) {
// 			roster.put(e.id, e);
// 		}
//
// 		int totalImp = 0;
// 		Queue<Employee> q = new LinkedList<>();
// 		q.add(roster.get(id));
//
// 		while (!q.isEmpty()) {
// 			Employee curr = q.poll();
// 			totalImp += curr.importance;
// 			for (int subID : curr.subordinates) {
// 				q.add(roster.get(subID));
// 			}
// 		}
//
// 		return totalImp;
// 	}
// }