LeetCode 615: Average Salary: Departments vs Company Solution in Python – A Step-by-Step Guide

Imagine you’re an HR analyst tasked with digging into payroll data—comparing how each department’s average salary stacks up against the company’s overall average, month by month. You’ve got employee records with salaries and department IDs, and you need to crunch the numbers to see who’s above or below the company line. That’s the core of LeetCode 615: Average Salary: Departments vs Company, a medium-level problem that’s all about aggregating and comparing data across tables. Using Python, we’ll explore two solutions: the Best Solution, a dictionary-based approach with monthly grouping for efficiency, and an Alternative Solution, a list-based method with explicit loops that’s simpler but less optimized. With detailed examples, beginner-friendly breakdowns—especially for the dictionary method—and clear code, this guide will help you analyze those salaries, whether you’re new to coding or leveling up. Let’s dive into the payroll and start calculating!

What Is LeetCode 615: Average Salary: Departments vs Company?

In LeetCode 615: Average Salary: Departments vs Company, you’re given two tables:

employee: id, department_id, salary, month.
department: id, department_name.

Your task is to compute, for each month and department, the average salary of employees in that department and the company-wide average salary, then compare them with labels like "higher," "lower," or "same." For example, if the sales team averages $5000 in January while the company averages $4500, sales is "higher." This problem tests your ability to group and aggregate data, typically posed as an SQL query with joins and GROUP BY, but we’ll solve it in Python for a coding twist.

Problem Statement

Input: Two lists (or tables):

employee: [id, department_id, salary, month].
department: [id, department_name].

Output: A list of [month, department_name, comparison] where comparison is "higher," "lower," or "same."
Rules:

Compute average salary per department and company-wide per month.
Compare department average to company average.
Return all month-department pairs present in data.

Constraints

employee has at least 1 row.
1 ≤ id, department_id ≤ 1000.
1 ≤ salary ≤ 10⁵.
month is a string (e.g., "Jan", "Feb").
department has at least 1 row.

Examples

Input:

Output:

Input:

Output:

These examples show the comparison—let’s solve it!

Understanding the Problem: Comparing Salaries

To solve LeetCode 615: Average Salary: Departments vs Company in Python, we need to group employee data by month and department, compute averages, compare them to the company-wide average per month, and label the results. A naive approach might process data multiple times, but we can optimize with aggregation. We’ll use:

Best Solution (Dictionary-Based Aggregation): O(n + m) time, O(n + m) space—fast and scalable (n = employees, m = departments).
Alternative Solution (List-Based Processing): O(n * m) time, O(n) space—simple but less efficient.

Let’s start with the dictionary-based solution, breaking it down for beginners.

Best Solution: Dictionary-Based Aggregation with Monthly Grouping

Why This Is the Best Solution

The dictionary-based approach is the top choice for LeetCode 615 because it’s efficient—O(n + m) time with O(n + m) space—and uses dictionaries to group data by month and department in one pass, mimicking SQL’s GROUP BY with minimal overhead. It’s scalable and intuitive once you grasp the structure. It’s like sorting payroll into monthly folders and comparing at a glance!

How It Works

Think of this as organizing a payroll ledger:

Step 1: Map Departments:

Create a dictionary from department mapping id to name.

Step 2: Aggregate Salaries:

Use nested dictionaries: month -> department_id -> [total_salary, count].
Track company totals: month -> [total_salary, count].

Step 3: Compute Averages:

For each month, calculate company average and department averages.

Step 4: Compare and Label:

For each department-month pair, compare to company average, label "higher," "lower," or "same."

Step 5: Return Result:

List of [month, department_name, comparison].

It’s like a payroll calculator—group and compare!

Step-by-Step Example

Example:

employee:
id | department_id | salary | month
1  | 1             | 5000   | Jan
2  | 2             | 4000   | Jan
3  | 1             | 6000   | Jan
department:
id | department_name
1  | Sales
2  | Engineering

Step 1: dept_map = {1: "Sales", 2: "Engineering"}.
Step 2: Aggregate:

month_data["Jan"] = {1: [11000, 2], 2: [4000, 1]}.
company_data["Jan"] = [15000, 3].

Step 3: Averages:

Company: 15000 / 3 = 5000.
Sales (1): 11000 / 2 = 5500.
Engineering (2): 4000 / 1 = 4000.

Step 4: Compare:

Sales: 5500 > 5000, "higher".
Engineering: 4000 < 5000, "lower".

Output: [["Jan", "Sales", "higher"], ["Jan", "Engineering", "lower"]].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

from collections import defaultdict

class Solution:
    def averageSalary(self, employee: List[List], department: List[List]) -> List[List]:
        # Step 1: Map department id to name
        dept_map = {d[0]: d[1] for d in department}

        # Step 2: Aggregate salaries by month and department
        month_data = defaultdict(lambda: defaultdict(lambda: [0, 0]))  # [total, count]
        company_data = defaultdict(lambda: [0, 0])  # [total, count]

        for _, dept_id, salary, month in employee:
            month_data[month][dept_id][0] += salary
            month_data[month][dept_id][1] += 1
            company_data[month][0] += salary
            company_data[month][1] += 1

        # Step 3: Compute averages and compare
        result = []
        for month in month_data:
            company_avg = company_data[month][0] / company_data[month][1]
            for dept_id in month_data[month]:
                dept_total, dept_count = month_data[month][dept_id]
                dept_avg = dept_total / dept_count
                comparison = ("same" if dept_avg == company_avg else
                             "higher" if dept_avg > company_avg else "lower")
                result.append([month, dept_map[dept_id], comparison])

        # Step 4: Return result
        return result

Line 1: Import defaultdict for ease.
Line 6: Map id to department_name.
Lines 9-15: Aggregate:

month_data: Nested dict for dept totals/counts.
company_data: Company totals/counts per month.

Lines 18-25: Compute and compare:

Calculate averages, label based on comparison.

Time Complexity: O(n + m)—one pass through each list.
Space Complexity: O(n + m)—store aggregated data.

This is like a payroll sorter—fast and neat!

Alternative Solution: List-Based Processing with Explicit Loops

Why an Alternative Approach?

The list-based approach processes data with explicit loops—O(n * m) time, O(n) space. It’s simpler to follow, avoiding nested dictionaries, but it’s less efficient due to repeated traversals. It’s like manually tallying salaries month by month!

How It Works

Picture this as a payroll tally:

Step 1: Map departments.
Step 2: Group employees by month.
Step 3: For each month:

Compute company average.
Compute each department’s average, compare.

Step 4: Build result list.

It’s like a step-by-step payroll check!

Step-by-Step Example

Example:

employee: [[1,1,5000,"Jan"], [2,2,4000,"Jan"], [3,1,6000,"Jan"]]
department: [[1,"Sales"], [2,"Engineering"]]

Step 1: dept_map = {1: "Sales", 2: "Engineering"}.
Step 2: Group: Jan = [[1,1,5000], [2,2,4000], [3,1,6000]].
Step 3: Process "Jan":

Company: (5000+4000+6000)/3 = 5000.
Sales: (5000+6000)/2 = 5500 > 5000, "higher".
Engineering: 4000/1 = 4000 < 5000, "lower".

Output: [["Jan", "Sales", "higher"], ["Jan", "Engineering", "lower"]].

Code for List Approach

class Solution:
    def averageSalary(self, employee: List[List], department: List[List]) -> List[List]:
        # Step 1: Map departments
        dept_map = {d[0]: d[1] for d in department}

        # Step 2: Group by month
        month_groups = {}
        for emp in employee:
            month = emp[3]
            if month not in month_groups:
                month_groups[month] = []
            month_groups[month].append(emp)

        # Step 3: Process each month
        result = []
        for month, emp_list in month_groups.items():
            # Company average
            company_total = sum(emp[2] for emp in emp_list)
            company_count = len(emp_list)
            company_avg = company_total / company_count

            # Department averages
            dept_totals = {}
            dept_counts = {}
            for _, dept_id, salary, _ in emp_list:
                dept_totals[dept_id] = dept_totals.get(dept_id, 0) + salary
                dept_counts[dept_id] = dept_counts.get(dept_id, 0) + 1

            # Compare
            for dept_id in dept_totals:
                dept_avg = dept_totals[dept_id] / dept_counts[dept_id]
                comparison = ("same" if dept_avg == company_avg else
                             "higher" if dept_avg > company_avg else "lower")
                result.append([month, dept_map[dept_id], comparison])

        return result

Lines 4-11: Group by month into lists.
Lines 14-28: Process:

Compute company avg, then dept avgs, compare.

Time Complexity: O(n * m)—multiple loops per month.
Space Complexity: O(n)—store grouped data.

It’s a manual payroll tally—clear but slower!

Comparing the Two Solutions

Dictionary (Best):

Pros: O(n + m) time, O(n + m) space, fast and clean.
Cons: Nested dicts less intuitive.

List (Alternative):

Pros: O(n * m) time, O(n) space, simpler flow.
Cons: Less efficient.

Dictionary wins for speed.

Additional Examples and Edge Cases

Input:

employee = [[1,1,5000,"Jan"]], department = [[1,"Sales"]]
Output: [["Jan", "Sales", "same"]].

Input: Empty employee

Output: [] (handled by constraints).

Both work well.

Complexity Breakdown

Dictionary: Time O(n + m), Space O(n + m).
List: Time O(n * m), Space O(n).

Dictionary is optimal.

Key Takeaways

Dictionary: Quick grouping—smart!
List: Step-by-step—clear!
Data: Aggregation is fun.
Python Tip: Dicts rock—see Python Basics.

Final Thoughts: Crunch Those Numbers

LeetCode 615: Average Salary: Departments vs Company in Python is a neat data analysis challenge. Dictionary-based aggregation offers efficiency, while list-based processing keeps it simple. Want more? Try LeetCode 177: Nth Highest Salary or LeetCode 184: Department Highest Salary. Ready to analyze? Head to Solve LeetCode 615 on LeetCode and compare those salaries today!