update 690.employee-importance.java

Author: marklcrns

690.employee-importance.java

@@ -0,0 +1,100 @@
+/*
+ * @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;
+};
+*/
+
+class Solution {
+	public HashMap<Integer, Employee> roster = new HashMap<Integer, Employee>();
+
+	public int getImportance(List<Employee> employees, int id) {
+		// 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?
+		// Nested loops?
+
+		// 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
+
+		for (Employee e : employees) {
+			this.roster.put(e.id, e);
+		}
+
+		return calculateImportance(id);
+	}
+
+	public int calculateImportance(int id) {
+		Employee lead =  this.roster.get(id);
+		int totalImp = lead.importance;
+
+		if (lead.subordinates.size() == 0) {
+			return totalImp;
+		} else {
+			for (int subId : lead.subordinates) {
+				totalImp += calculateImportance(subId);
+			}
+		}
+
+		return totalImp;
+	}
+}