How should I practice Lowest Common Manager for Employee Groups?

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Lowest Common Manager for Employee Groups

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Problem statement

A company reporting hierarchy is a rooted binary tree. The tree is encoded by parallel arrays employeeIds and managerIds. For each index i, managerIds[i] is the direct manager of employeeIds[i]; the root has manager -1.

You also receive a batch of employee groups in queries. For each group, return the employee ID of its lowest common manager: the deepest employee whose subtree contains every employee in that group. An employee is in their own subtree, so a one-employee query returns that employee.

Return one manager ID per query in the original query order. The hierarchy is fixed across the entire batch, so preprocess it to support large or frequent groups efficiently.

Function

lowestCommonManagers(employeeIds: int[], managerIds: int[], queries: int[][]) → int[]

Examples

Example 1

employeeIds = [1,2,3,4,5,6,7]managerIds = [-1,1,1,2,2,3,3]queries = [[4,5],[4,6,7],[2,4,5]]return = [2,1,2]

Employees 4 and 5 report under 2. The group containing 4, 6, and 7 spans both children of root 1. Employee 2 is itself an ancestor of the final group.

Example 2

employeeIds = [10,20,30,40,50,60]managerIds = [-1,10,10,20,20,40]queries = [[60],[60,50],[30,60]]return = [60,20,10]

A singleton group returns 60. Employees 60 and 50 first meet at 20, while 30 and 60 first meet at root 10.

Constraints

1 <= employeeIds.length == managerIds.length <= 100000.
Employee IDs are unique integers from 1 through 1000000000.
Exactly one entry of managerIds is -1; every other entry names an employee in employeeIds.
The manager relationships form one connected acyclic rooted tree, and each manager has at most two direct reports.
1 <= queries.length <= 100000.
Each query is non-empty, contains unique employee IDs, and every queried ID exists in the hierarchy.
The total number of employee IDs across all queries is at most 100000.