Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.
Table: Employee
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| employee_id | int |
| department_id | int |
| primary_flag | varchar |
+---------------+---------+
(employee_id, department_id) is the primary key (combination of columns with unique values) for this table.
employee_id is the id of the employee.
department_id is the id of the department to which the employee belongs.
primary_flag is an ENUM (category) of type ('Y', 'N'). If the flag is 'Y', the department is the primary department for the employee. If the flag is 'N', the department is not the primary.
Employees can belong to multiple departments. When the employee joins other departments, they need to decide which department is their primary department. Note that when an employee belongs to only one department, their primary column is 'N'.
Write a solution to report all the employees with their primary department. For employees who belong to one department, report their only department.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Employee table: +-------------+---------------+--------------+ | employee_id | department_id | primary_flag | +-------------+---------------+--------------+ | 1 | 1 | N | | 2 | 1 | Y | | 2 | 2 | N | | 3 | 3 | N | | 4 | 2 | N | | 4 | 3 | Y | | 4 | 4 | N | +-------------+---------------+--------------+ Output: +-------------+---------------+ | employee_id | department_id | +-------------+---------------+ | 1 | 1 | | 2 | 1 | | 3 | 3 | | 4 | 3 | +-------------+---------------+ Explanation: - The Primary department for employee 1 is 1. - The Primary department for employee 2 is 1. - The Primary department for employee 3 is 3. - The Primary department for employee 4 is 3.
Problem summary: Table: Employee +---------------+---------+ | Column Name | Type | +---------------+---------+ | employee_id | int | | department_id | int | | primary_flag | varchar | +---------------+---------+ (employee_id, department_id) is the primary key (combination of columns with unique values) for this table. employee_id is the id of the employee. department_id is the id of the department to which the employee belongs. primary_flag is an ENUM (category) of type ('Y', 'N'). If the flag is 'Y', the department is the primary department for the employee. If the flag is 'N', the department is not the primary. Employees can belong to multiple departments. When the employee joins other departments, they need to decide which department is their primary department. Note that when an employee belongs to only one department, their primary column is 'N'. Write a solution to report all the employees with
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers":{"Employee":["employee_id","department_id","primary_flag"]},"rows":{"Employee":[["1","1","N"],["2","1","Y"],["2","2","N"],["3","3","N"],["4","2","N"],["4","3","Y"],["4","4","N"]]}}Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1789: Primary Department for Each Employee
// Auto-generated Java example from rust.
class Solution {
public void exampleSolution() {
}
}
// Reference (rust):
// // Accepted solution for LeetCode #1789: Primary Department for Each Employee
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1789: Primary Department for Each Employee
// # Write your MySQL query statement below
// SELECT employee_id, department_id
// FROM Employee
// WHERE primary_flag = 'Y'
// UNION
// SELECT employee_id, department_id
// FROM Employee
// GROUP BY 1
// HAVING COUNT(1) = 1;
// "#
// }
// Accepted solution for LeetCode #1789: Primary Department for Each Employee
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1789: Primary Department for Each Employee
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1789: Primary Department for Each Employee
// # Write your MySQL query statement below
// SELECT employee_id, department_id
// FROM Employee
// WHERE primary_flag = 'Y'
// UNION
// SELECT employee_id, department_id
// FROM Employee
// GROUP BY 1
// HAVING COUNT(1) = 1;
// "#
// }
# Accepted solution for LeetCode #1789: Primary Department for Each Employee
# Auto-generated Python example from rust.
def example_solution() -> None:
return
# Reference (rust):
# // Accepted solution for LeetCode #1789: Primary Department for Each Employee
# pub fn sql_example() -> &'static str {
# r#"
# -- Accepted solution for LeetCode #1789: Primary Department for Each Employee
# # Write your MySQL query statement below
# SELECT employee_id, department_id
# FROM Employee
# WHERE primary_flag = 'Y'
# UNION
# SELECT employee_id, department_id
# FROM Employee
# GROUP BY 1
# HAVING COUNT(1) = 1;
# "#
# }
// Accepted solution for LeetCode #1789: Primary Department for Each Employee
pub fn sql_example() -> &'static str {
r#"
-- Accepted solution for LeetCode #1789: Primary Department for Each Employee
# Write your MySQL query statement below
SELECT employee_id, department_id
FROM Employee
WHERE primary_flag = 'Y'
UNION
SELECT employee_id, department_id
FROM Employee
GROUP BY 1
HAVING COUNT(1) = 1;
"#
}
// Accepted solution for LeetCode #1789: Primary Department for Each Employee
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1789: Primary Department for Each Employee
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1789: Primary Department for Each Employee
// # Write your MySQL query statement below
// SELECT employee_id, department_id
// FROM Employee
// WHERE primary_flag = 'Y'
// UNION
// SELECT employee_id, department_id
// FROM Employee
// GROUP BY 1
// HAVING COUNT(1) = 1;
// "#
// }
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.