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: Project
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| project_id | int |
| employee_id | int |
+-------------+---------+
(project_id, employee_id) is the primary key of this table.
employee_id is a foreign key to Employee table.
Each row of this table indicates that the employee with employee_id is working on the project with project_id.
Table: Employee
+------------------+---------+ | Column Name | Type | +------------------+---------+ | employee_id | int | | name | varchar | | experience_years | int | +------------------+---------+ employee_id is the primary key of this table. It's guaranteed that experience_years is not NULL. Each row of this table contains information about one employee.
Write an SQL query that reports the average experience years of all the employees for each project, rounded to 2 digits.
Return the result table in any order.
The query result format is in the following example.
Example 1:
Input: Project table: +-------------+-------------+ | project_id | employee_id | +-------------+-------------+ | 1 | 1 | | 1 | 2 | | 1 | 3 | | 2 | 1 | | 2 | 4 | +-------------+-------------+ Employee table: +-------------+--------+------------------+ | employee_id | name | experience_years | +-------------+--------+------------------+ | 1 | Khaled | 3 | | 2 | Ali | 2 | | 3 | John | 1 | | 4 | Doe | 2 | +-------------+--------+------------------+ Output: +-------------+---------------+ | project_id | average_years | +-------------+---------------+ | 1 | 2.00 | | 2 | 2.50 | +-------------+---------------+ Explanation: The average experience years for the first project is (3 + 2 + 1) / 3 = 2.00 and for the second project is (3 + 2) / 2 = 2.50
Problem summary: Table: Project +-------------+---------+ | Column Name | Type | +-------------+---------+ | project_id | int | | employee_id | int | +-------------+---------+ (project_id, employee_id) is the primary key of this table. employee_id is a foreign key to Employee table. Each row of this table indicates that the employee with employee_id is working on the project with project_id. Table: Employee +------------------+---------+ | Column Name | Type | +------------------+---------+ | employee_id | int | | name | varchar | | experience_years | int | +------------------+---------+ employee_id is the primary key of this table. It's guaranteed that experience_years is not NULL. Each row of this table contains information about one employee. Write an SQL query that reports the average experience years of all the employees for each project, rounded to 2 digits. Return the result table in any order.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers":{"Project":["project_id","employee_id"],"Employee":["employee_id","name","experience_years"]},"rows":{"Project":[[1,1],[1,2],[1,3],[2,1],[2,4]],"Employee":[[1,"Khaled",3],[2,"Ali",2],[3,"John",1],[4,"Doe",2]]}}project-employees-ii)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1075: Project Employees I
// Auto-generated Java example from rust.
class Solution {
public void exampleSolution() {
}
}
// Reference (rust):
// // Accepted solution for LeetCode #1075: Project Employees I
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1075: Project Employees I
// # Write your MySQL query statement
// SELECT project_id, ROUND(AVG(experience_years), 2) AS average_years
// FROM
// Project
// JOIN Employee USING (employee_id)
// GROUP BY 1;
// "#
// }
// Accepted solution for LeetCode #1075: Project Employees I
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1075: Project Employees I
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1075: Project Employees I
// # Write your MySQL query statement
// SELECT project_id, ROUND(AVG(experience_years), 2) AS average_years
// FROM
// Project
// JOIN Employee USING (employee_id)
// GROUP BY 1;
// "#
// }
# Accepted solution for LeetCode #1075: Project Employees I
# Auto-generated Python example from rust.
def example_solution() -> None:
return
# Reference (rust):
# // Accepted solution for LeetCode #1075: Project Employees I
# pub fn sql_example() -> &'static str {
# r#"
# -- Accepted solution for LeetCode #1075: Project Employees I
# # Write your MySQL query statement
# SELECT project_id, ROUND(AVG(experience_years), 2) AS average_years
# FROM
# Project
# JOIN Employee USING (employee_id)
# GROUP BY 1;
# "#
# }
// Accepted solution for LeetCode #1075: Project Employees I
pub fn sql_example() -> &'static str {
r#"
-- Accepted solution for LeetCode #1075: Project Employees I
# Write your MySQL query statement
SELECT project_id, ROUND(AVG(experience_years), 2) AS average_years
FROM
Project
JOIN Employee USING (employee_id)
GROUP BY 1;
"#
}
// Accepted solution for LeetCode #1075: Project Employees I
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1075: Project Employees I
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1075: Project Employees I
// # Write your MySQL query statement
// SELECT project_id, ROUND(AVG(experience_years), 2) AS average_years
// FROM
// Project
// JOIN Employee USING (employee_id)
// GROUP BY 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.