How do you solve Primary Department for Each Employee on LeetCode?

Primary Department for Each Employee (LeetCode #1789) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to differentiate between primary and non-primary departments is crucial.

What is the brute force solution for Primary Department for Each Employee?

The brute force approach for Primary Department for Each Employee is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check each employee's departments one by one to determine which one is marked as primary. This is straightforward but inefficient, especially as the number of employees and departments grows. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Primary Department for Each Employee?

Primary Department for Each Employee uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Primary Department for Each Employee?

Always clarify the requirements before jumping into coding. Think about edge cases, such as employees with only one department. Discuss your thought process with the interviewer to demonstrate your understanding.

What are common mistakes when solving Primary Department for Each Employee?

Not handling cases where an employee has only one department correctly. Overlooking the need to check if a primary department has already been assigned.

#1789

Primary Department for Each Employee

Easy

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

In this approach, we will use a single pass through the employee data to create a mapping of employee IDs to their primary departments. This is efficient and avoids unnecessary checks.

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Algorithm

5 steps

1Step 1: Initialize a dictionary to store the primary department for each employee.
2Step 2: Loop through the employee records.
3Step 3: If the primary_flag is 'Y', store the department_id in the dictionary.
4Step 4: If the employee has no primary department, store their only department.
5Step 5: Convert the dictionary to a list of results.

solution.py9 lines

1def primary_department(employees):
2    primary_depts = {}
3    for emp in employees:
4        emp_id = emp['employee_id']
5        if emp['primary_flag'] == 'Y':
6            primary_depts[emp_id] = emp['department_id']
7        elif emp_id not in primary_depts:
8            primary_depts[emp_id] = emp['department_id']
9    return [{'employee_id': emp_id, 'department_id': dept} for emp_id, dept in primary_depts.items()]

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Complexity note: The time complexity is O(n) because we only loop through the employee list once. The space complexity is O(n) due to the dictionary storing the primary departments.

1Understanding how to differentiate between primary and non-primary departments is crucial.
2Using a dictionary to track primary departments can significantly reduce time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.