How should I practice Organization Reputation After Employee Departures?

Start by reading the full statement, solve the visible examples, then compare the pattern against related HSBC online assessment problems in FastPrep.

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Organization Reputation After Employee Departures

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HSBCINTERNOA

Problem statement

An organization has n employees with IDs from 1 through n. Employee i has efficiency efficiency[i - 1] and belongs to team teamId[i - 1]. The organization's reputation is the sum of the efficiencies of all employees who are still active.

Process the rows of queries in order. A row [employeeId, k] describes one day:

Fire the currently active employee with ID employeeId.
From that employee's team, up to k other active employees resign. Choose employees by lower efficiency first; when efficiencies are equal, choose the lower employee ID first.
If fewer than k active teammates remain, all of them resign.

Return an array containing the organization's reputation at the end of each day.

Every queried employeeId is active at the start of its day. Once an employee is fired or resigns, that employee never returns.

Function

getOrganizationReputation(efficiency: int[], teamId: int[], queries: int[][]) → long[]

Examples

Example 1

efficiency = [10,5,8,3]teamId = [1,1,2,1]queries = [[1,1],[3,0]]return = [13,5]

The initial reputation is 26. On the first day, employee 1 is fired and employee 4, the least-efficient remaining member of team 1, resigns. The reputation becomes 13. On the second day, employee 3 is fired and no teammate is requested, so the reputation becomes 5.

Example 2

efficiency = [4,4,7,1,9]teamId = [2,2,2,3,3]queries = [[2,2],[5,3]]return = [10,0]

After employee 2 is fired, the two remaining members of team 2 resign, leaving reputation 10. On the next day, employee 5 is fired. Team 3 has only employee 4 remaining, so that employee also resigns and the final reputation is 0.

Example 3

efficiency = [7]teamId = [9]queries = [[1,5]]return = [0]

The only employee is fired. No teammate remains, so the reputation is 0 even though k is larger than the team.

Constraints

1 <= efficiency.length = teamId.length <= 2 * 10^5
-10^9 <= efficiency[i] <= 10^9
1 <= teamId[i] <= 10^9
1 <= queries.length <= efficiency.length
Every row of queries contains exactly two integers [employeeId, k].
Each queried employeeId is active at the start of its day.
0 <= k <= efficiency.length
Reputation calculations use signed 64-bit integers.