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: Product
+--------------+---------+ | Column Name | Type | +--------------+---------+ | product_id | int | | product_name | varchar | | unit_price | int | +--------------+---------+ product_id is the primary key (column with unique values) of this table. Each row of this table indicates the name and the price of each product.
Table: Sales
+-------------+---------+ | Column Name | Type | +-------------+---------+ | seller_id | int | | product_id | int | | buyer_id | int | | sale_date | date | | quantity | int | | price | int | +-------------+---------+ This table can have duplicate rows. product_id is a foreign key (reference column) to the Product table. Each row of this table contains some information about one sale.
Write a solution to report the products that were only sold in the first quarter of 2019. That is, between 2019-01-01 and 2019-03-31 inclusive.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Product table: +------------+--------------+------------+ | product_id | product_name | unit_price | +------------+--------------+------------+ | 1 | S8 | 1000 | | 2 | G4 | 800 | | 3 | iPhone | 1400 | +------------+--------------+------------+ Sales table: +-----------+------------+----------+------------+----------+-------+ | seller_id | product_id | buyer_id | sale_date | quantity | price | +-----------+------------+----------+------------+----------+-------+ | 1 | 1 | 1 | 2019-01-21 | 2 | 2000 | | 1 | 2 | 2 | 2019-02-17 | 1 | 800 | | 2 | 2 | 3 | 2019-06-02 | 1 | 800 | | 3 | 3 | 4 | 2019-05-13 | 2 | 2800 | +-----------+------------+----------+------------+----------+-------+ Output: +-------------+--------------+ | product_id | product_name | +-------------+--------------+ | 1 | S8 | +-------------+--------------+ Explanation: The product with id 1 was only sold in the spring of 2019. The product with id 2 was sold in the spring of 2019 but was also sold after the spring of 2019. The product with id 3 was sold after spring 2019. We return only product 1 as it is the product that was only sold in the spring of 2019.
Problem summary: Table: Product +--------------+---------+ | Column Name | Type | +--------------+---------+ | product_id | int | | product_name | varchar | | unit_price | int | +--------------+---------+ product_id is the primary key (column with unique values) of this table. Each row of this table indicates the name and the price of each product. Table: Sales +-------------+---------+ | Column Name | Type | +-------------+---------+ | seller_id | int | | product_id | int | | buyer_id | int | | sale_date | date | | quantity | int | | price | int | +-------------+---------+ This table can have duplicate rows. product_id is a foreign key (reference column) to the Product table. Each row of this table contains some information about one sale. Write a solution to report the products that were only sold in the first quarter of 2019. That is, between 2019-01-01 and 2019-03-31 inclusive. Return the result
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers":{"Product":["product_id","product_name","unit_price"],"Sales":["seller_id","product_id","buyer_id","sale_date","quantity","price"]},"rows":{"Product":[[1,"S8",1000],[2,"G4",800],[3,"iPhone",1400]],"Sales":[[1,1,1,"2019-01-21",2,2000],[1,2,2,"2019-02-17",1,800],[2,2,3,"2019-06-02",1,800],[3,3,4,"2019-05-13",2,2800]]}}sales-analysis-ii)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1084: Sales Analysis III
// Auto-generated Java example from rust.
class Solution {
public void exampleSolution() {
}
}
// Reference (rust):
// // Accepted solution for LeetCode #1084: Sales Analysis III
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1084: Sales Analysis III
// # Write your MySQL query statement below
// SELECT product_id, product_name
// FROM
// Sales
// JOIN Product USING (product_id)
// GROUP BY 1
// HAVING COUNT(1) = SUM(sale_date BETWEEN '2019-01-01' AND '2019-03-31');
// "#
// }
// Accepted solution for LeetCode #1084: Sales Analysis III
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1084: Sales Analysis III
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1084: Sales Analysis III
// # Write your MySQL query statement below
// SELECT product_id, product_name
// FROM
// Sales
// JOIN Product USING (product_id)
// GROUP BY 1
// HAVING COUNT(1) = SUM(sale_date BETWEEN '2019-01-01' AND '2019-03-31');
// "#
// }
# Accepted solution for LeetCode #1084: Sales Analysis III
# Auto-generated Python example from rust.
def example_solution() -> None:
return
# Reference (rust):
# // Accepted solution for LeetCode #1084: Sales Analysis III
# pub fn sql_example() -> &'static str {
# r#"
# -- Accepted solution for LeetCode #1084: Sales Analysis III
# # Write your MySQL query statement below
# SELECT product_id, product_name
# FROM
# Sales
# JOIN Product USING (product_id)
# GROUP BY 1
# HAVING COUNT(1) = SUM(sale_date BETWEEN '2019-01-01' AND '2019-03-31');
# "#
# }
// Accepted solution for LeetCode #1084: Sales Analysis III
pub fn sql_example() -> &'static str {
r#"
-- Accepted solution for LeetCode #1084: Sales Analysis III
# Write your MySQL query statement below
SELECT product_id, product_name
FROM
Sales
JOIN Product USING (product_id)
GROUP BY 1
HAVING COUNT(1) = SUM(sale_date BETWEEN '2019-01-01' AND '2019-03-31');
"#
}
// Accepted solution for LeetCode #1084: Sales Analysis III
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1084: Sales Analysis III
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1084: Sales Analysis III
// # Write your MySQL query statement below
// SELECT product_id, product_name
// FROM
// Sales
// JOIN Product USING (product_id)
// GROUP BY 1
// HAVING COUNT(1) = SUM(sale_date BETWEEN '2019-01-01' AND '2019-03-31');
// "#
// }
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.