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This approach utilizes SQL's JOIN operations to identify missing information. We can use a FULL OUTER JOIN between the Employees and Salaries tables to ensure we capture all employee IDs, whether they appear in only one of the tables (missing from the other) or in both.
After joining, we will filter the results for rows where either the name or salary is NULL.
Time Complexity: O(N + M), where N is the number of employees and M is the number of salaries, because the join operation requires a scan of both tables.
Space Complexity: O(N + M) due to the storage of the join results.
1SELECT E.employee_id FROM Employees E FULL OUTER JOIN Salaries S ON E.employee_id = S.employee_id WHERE E.name IS NULL OR S.salary IS NULL ORDER BY employee_id;This SQL query uses a FULL OUTER JOIN to combine the data from the Employees and Salaries tables on the employee_id. The WHERE clause is used to filter out rows where either the name is NULL or the salary is NULL, indicating missing information from one of the tables.
This solution employs two separate queries combined with UNION to find missing records. The first query selects employee_ids from the Employees table that do not exist in the Salaries table (indicating missing salary records), and the second query selects employee_ids from the Salaries table that do not exist in the Employees table (indicating missing employee records).
Time Complexity: O(N * M) since each subquery potentially involves a scan of the Employees table for each entry in Salaries, and vice versa.
Space Complexity: O(N + M), primarily for the UNION operation result storage.
1SELECT employee_id FROM Employees WHERE employee_id NOT IN (SELECT employee_id FROM Salaries) UNION SELECTThis approach involves using a SQL full outer join to combine the Employees and Salaries tables on the employee_id column. By checking for null values in the resulting join, you can determine which employees have missing information (either name or salary).
Time Complexity: O(n + m), where n is the number of records in the Employees table and m is the number of records in the Salaries table as we need to check all employees and salaries.
Space Complexity: O(k), where k is the number of employee_ids with missing information, primarily due to the result storage.
1from sqlalchemy import create_engine, MetaData, Table, select, or_
2
3# Suppose we have an engine and the tables are already available
4def find_missing_information(engine):
metadata = MetaData(bind=engine)
employees = Table('Employees', metadata, autoload_with=engine)
salaries = Table('Salaries', metadata, autoload_with=engine)
# Full outer join equivalent using UNION of left and right joins
stmt = select([employees.c.employee_id.label('eid')]) \
.select_from(employees.join(salaries, employees.c.employee_id == salaries.c.employee_id, isouter=True)) \
.where(employees.c.name == None) \
.union(
select([salaries.c.employee_id])
.select_from(salaries.join(employees, employees.c.employee_id == salaries.c.employee_id, isouter=True))
.where(salaries.c.salary == None)
)
result = engine.execute(stmt).fetchall()
return sorted([row.eid for row in result])
This approach involves using two hash maps (dictionaries) to store information about employees and salaries. By comparing the keys of these two maps, we can determine the missing information by finding the employee IDs that are not present in both maps.
Time Complexity: O(n + m), where n is the size of the employees list and m is the size of the salaries list.
Space Complexity: O(n + m) due to the space used for storing the dictionaries of employees and salaries.
1function findMissingInformation(employees, salaries) {
2
This SQL script combines two subqueries using UNION: The first subquery lists employee_ids from the Employees table that do not appear in Salaries (missing salaries), and the second lists employee_ids from Salaries that do not appear in Employees (missing names). The result is returned in ascending order by employee_id.
In this approach, we establish a connection to the database and use SQLAlchemy to perform a full outer join between the Employees and Salaries tables using their employee IDs. The key here is checking for nulls in both the employee name and salary after the join. We achieve this using the left outer join and union of two conditions: when the name is missing and when the salary is missing. The union ensures all relevant IDs with missing information are captured.
In this JavaScript implementation, we utilize Map objects to store employees and salaries based on employee IDs. The function iterates over the maps and checks for any missing opposite pair for each map entry. It records missing IDs (those that are not common in both maps), converts them to a sorted array before returning.