Create employee-importance.py

Author: kamyu104

Python/employee-importance.py

@@ -0,0 +1,72 @@
+# Time:  O(n)
+# Space: O(h)
+
+# 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.
+
+"""
+# Employee info
+class Employee(object):
+    def __init__(self, id, importance, subordinates):
+        # It's the unique id of each node.
+        # unique id of this employee
+        self.id = id
+        # the importance value of this employee
+        self.importance = importance
+        # the id of direct subordinates
+        self.subordinates = subordinates
+"""
+class Solution(object):
+    def getImportance(self, employees, id):
+        """
+        :type employees: Employee
+        :type id: int
+        :rtype: int
+        """
+        if employees[id-1] is None:
+            return 0
+        result = employees[id-1].importance
+        for id in employees[id-1].subordinates:
+            result += self.getImportance(employees, id)
+        return result
+
+
+# Time:  O(n)
+# Space: O(w), w is the max number of nodes in the levels of the tree
+class Solution2(object):
+    def getImportance(self, employees, id):
+        """
+        :type employees: Employee
+        :type id: int
+        :rtype: int
+        """
+        result, q = 0, collections.deque([id])
+        while q:
+            curr = q.popleft()
+            employee = employees[curr-1]
+            result += employee.importance;
+            for id in employee.subordinates:
+                q.append(id)
+        return result