LeetCode 2853 - Highest Salaries Difference

The problem provides a database table named Salaries with three columns: | Column | Meaning | | --- | --- | | empname | Employee name | | department | Employee department | | salary | Employee salary | The combination of (empname, department) is unique, meaning there are no…

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LeetCode Problem 2853

Difficulty: 🟢 Easy
Topics: Database

Solution

Problem Understanding

The problem provides a database table named Salaries with three columns:

Column Meaning
emp_name Employee name
department Employee department
salary Employee salary

The combination of (emp_name, department) is unique, meaning there are no duplicate employee entries within the same department.

The task is to compute the absolute difference between:

  1. The highest salary in the Engineering department
  2. The highest salary in the Marketing department

The final output should contain a single column named salary_difference.

In other words, we need to:

  • Find the maximum salary among all Engineering employees
  • Find the maximum salary among all Marketing employees
  • Compute:
ABS(max_engineering_salary - max_marketing_salary)

The problem guarantees that both Engineering and Marketing departments contain at least one employee. This guarantee is important because it means we never need to handle missing departments or NULL maximum values.

Even though this is labeled as an easy database problem, there are still a few implementation details worth understanding carefully.

A naive implementation might incorrectly:

  • Compare every Engineering employee with every Marketing employee
  • Forget to take the absolute value
  • Use averages or totals instead of maximum salaries
  • Fail when one department contains only a single employee

The input size is not explicitly stated, but SQL aggregate functions such as MAX() are designed to efficiently process large datasets. Since we only need two aggregate values, the optimal solution is straightforward and efficient.

Approaches

Brute Force Approach

A brute force approach would separate employees into two groups:

  • Engineering employees
  • Marketing employees

Then it could compare every Engineering salary against every Marketing salary to determine the largest salary in each department manually.

Conceptually, the algorithm would:

  1. Iterate through Engineering employees to determine the highest Engineering salary
  2. Iterate through Marketing employees to determine the highest Marketing salary
  3. Compute the absolute difference

In raw procedural terms, this works correctly because checking all employees guarantees we eventually discover the maximum salary for each department.

However, in SQL, manually simulating this kind of row-by-row comparison is unnecessary and inefficient. Database systems already provide optimized aggregate functions such as MAX() that solve this problem directly.

Optimal Approach

The key observation is that we only care about one value per department:

  • Maximum Engineering salary
  • Maximum Marketing salary

SQL aggregate functions are specifically designed for this type of computation.

We can use conditional aggregation:

  • Compute the maximum salary where department = 'Engineering'
  • Compute the maximum salary where department = 'Marketing'
  • Subtract the two values
  • Apply ABS() to ensure the result is non-negative

This approach is optimal because it scans the table once and computes both values simultaneously.

Approach Comparison

Approach Time Complexity Space Complexity Notes
Brute Force O(n) O(1) Manually scans rows and tracks department maximums
Optimal O(n) O(1) Uses SQL aggregate functions and conditional aggregation

Algorithm Walkthrough

Optimal SQL Algorithm

  1. Scan all rows in the Salaries table.
  2. For rows belonging to the Engineering department, track the maximum salary using:
MAX(CASE WHEN department = 'Engineering' THEN salary END)

This condition ensures only Engineering salaries participate in the aggregate.

  1. For rows belonging to the Marketing department, track the maximum salary using:
MAX(CASE WHEN department = 'Marketing' THEN salary END)
  1. Subtract the two maximum values.
  2. Apply the ABS() function so the result is always non-negative.
  3. Return the final value using the column alias salary_difference.

Why it works

The correctness comes from the behavior of the MAX() aggregate function.

For each department condition:

  • Only matching rows contribute salary values
  • Non-matching rows contribute NULL
  • MAX() ignores NULL values

Therefore:

  • The Engineering expression evaluates to the highest Engineering salary
  • The Marketing expression evaluates to the highest Marketing salary

Taking the absolute difference between these two values produces the required answer.

Python Solution

LeetCode database problems are solved using SQL rather than executable Python code. However, the platform still displays a language section, so the equivalent SQL query is shown below.

SELECT
    ABS(
        MAX(CASE WHEN department = 'Engineering' THEN salary END) -
        MAX(CASE WHEN department = 'Marketing' THEN salary END)
    ) AS salary_difference
FROM Salaries;

The query contains a single table scan and computes both department maximums simultaneously.

The CASE expressions selectively include salaries from only one department at a time. Every row that does not belong to the target department contributes NULL, which is ignored by MAX().

After computing both maximum salaries, the query subtracts them and wraps the result in ABS() to guarantee a non-negative difference.

Finally, the result is returned with the required column name salary_difference.

Go Solution

Database problems on LeetCode are solved in SQL, not Go. The equivalent SQL solution is still the correct submission regardless of the selected language.

SELECT
    ABS(
        MAX(CASE WHEN department = 'Engineering' THEN salary END) -
        MAX(CASE WHEN department = 'Marketing' THEN salary END)
    ) AS salary_difference
FROM Salaries;

There are no Go-specific implementation concerns because the execution happens entirely inside the SQL engine. Integer overflow is not an issue under normal LeetCode constraints, and aggregate functions handle NULL values automatically.

Worked Examples

Example 1

Input table:

emp_name department salary
Kathy Engineering 50000
Roy Marketing 30000
Charles Engineering 45000
Jack Engineering 85000
Benjamin Marketing 34000
Anthony Marketing 42000
Edward Engineering 102000
Terry Engineering 44000
Evelyn Marketing 53000
Arthur Engineering 32000

Step 1: Find Maximum Engineering Salary

Engineering Employee Salary Current Maximum
Kathy 50000 50000
Charles 45000 50000
Jack 85000 85000
Edward 102000 102000
Terry 44000 102000
Arthur 32000 102000

Final Engineering maximum:

102000

Step 2: Find Maximum Marketing Salary

Marketing Employee Salary Current Maximum
Roy 30000 30000
Benjamin 34000 34000
Anthony 42000 42000
Evelyn 53000 53000

Final Marketing maximum:

53000

Step 3: Compute Absolute Difference

ABS(102000 - 53000)
= ABS(49000)
= 49000

Final output:

salary_difference
49000

Complexity Analysis

Measure Complexity Explanation
Time O(n) The table is scanned once
Space O(1) Only aggregate values are stored

The database engine processes each row one time while maintaining two running maximum values. No additional data structures proportional to the input size are required.

Test Cases

# Example case from problem statement
# Engineering max = 102000
# Marketing max = 53000
# Difference = 49000

# Single employee per department
# Engineering max = 70000
# Marketing max = 50000
# Difference = 20000

# Marketing larger than Engineering
# Engineering max = 40000
# Marketing max = 90000
# Difference = 50000

# Equal maximum salaries
# Engineering max = 80000
# Marketing max = 80000
# Difference = 0

# Multiple employees with same maximum
# Engineering max = 95000
# Marketing max = 60000
# Difference = 35000

# Very small salaries
# Engineering max = 1
# Marketing max = 2
# Difference = 1

# Large salary values
# Engineering max = 1000000000
# Marketing max = 999999999
# Difference = 1

Test Case Summary

Test Why
Provided example Validates standard behavior
Single employee per department Ensures minimum valid input works
Marketing salary larger Verifies ABS() is necessary
Equal maximum salaries Ensures zero difference handled correctly
Duplicate maximum values Confirms repeated values do not affect correctness
Small salary values Tests lower numeric boundaries
Large salary values Tests large integer handling

Edge Cases

Departments With Only One Employee

A department may contain exactly one employee. A buggy implementation might incorrectly assume multiple rows exist when computing maximum salaries.

This solution handles the case naturally because MAX() works correctly even when only one value is present. The single salary becomes the maximum automatically.

Marketing Maximum Greater Than Engineering Maximum

Some implementations may subtract Engineering from Marketing directly without using ABS(). This would produce negative answers when Marketing has the larger salary.

Using:

ABS(engineering_max - marketing_max)

guarantees the result is always non-negative, exactly as required.

Multiple Employees Sharing the Maximum Salary

Several employees may have the same highest salary in a department. A naive implementation could mistakenly double count or incorrectly process duplicates.

The MAX() aggregate function only returns the largest value itself, regardless of how many times it appears. Therefore, duplicate maximum salaries do not affect correctness.