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.
Move from brute-force thinking to an efficient approach using core interview patterns strategy.
Table: Sales
+-------------+-------+ | Column Name | Type | +-------------+-------+ | sale_id | int | | product_id | int | | year | int | | quantity | int | | price | int | +-------------+-------+ (sale_id, year) is the primary key (combination of columns with unique values) of this table. Each row records a sale of a product in a given year. A product may have multiple sales entries in the same year. Note that the per-unit price.
Write a solution to find all sales that occurred in the first year each product was sold.
For each product_id, identify the earliest year it appears in the Sales table.
Return all sales entries for that product in that year.
Return a table with the following columns: product_id, first_year, quantity, and price.
Return the result in any order.
Example 1:
Input: Sales table: +---------+------------+------+----------+-------+ | sale_id | product_id | year | quantity | price | +---------+------------+------+----------+-------+ | 1 | 100 | 2008 | 10 | 5000 | | 2 | 100 | 2009 | 12 | 5000 | | 7 | 200 | 2011 | 15 | 9000 | +---------+------------+------+----------+-------+ Output: +------------+------------+----------+-------+ | product_id | first_year | quantity | price | +------------+------------+----------+-------+ | 100 | 2008 | 10 | 5000 | | 200 | 2011 | 15 | 9000 | +------------+------------+----------+-------+
Problem summary: Table: Sales +-------------+-------+ | Column Name | Type | +-------------+-------+ | sale_id | int | | product_id | int | | year | int | | quantity | int | | price | int | +-------------+-------+ (sale_id, year) is the primary key (combination of columns with unique values) of this table. Each row records a sale of a product in a given year. A product may have multiple sales entries in the same year. Note that the per-unit price. Write a solution to find all sales that occurred in the first year each product was sold. For each product_id, identify the earliest year it appears in the Sales table. Return all sales entries for that product in that year. Return a table with the following columns: product_id, first_year, quantity, and price. Return the result in any order.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers":{"Sales":["sale_id","product_id","year","quantity","price"]},"rows":{"Sales":[[1,100,2008,10,5000],[2,100,2009,12,5000],[7,200,2011,15,9000]]}}product-sales-analysis-ii)product-sales-analysis-iv)product-sales-analysis-v)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1070: Product Sales Analysis III
// Auto-generated Java example from rust.
class Solution {
public void exampleSolution() {
}
}
// Reference (rust):
// // Accepted solution for LeetCode #1070: Product Sales Analysis III
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1070: Product Sales Analysis III
// # Write your MySQL query statement below
// SELECT
// product_id,
// year AS first_year,
// quantity,
// price
// FROM Sales
// WHERE
// (product_id, year) IN (
// SELECT
// product_id,
// MIN(year) AS year
// FROM Sales
// GROUP BY product_id
// );
// "#
// }
// Accepted solution for LeetCode #1070: Product Sales Analysis III
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1070: Product Sales Analysis III
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1070: Product Sales Analysis III
// # Write your MySQL query statement below
// SELECT
// product_id,
// year AS first_year,
// quantity,
// price
// FROM Sales
// WHERE
// (product_id, year) IN (
// SELECT
// product_id,
// MIN(year) AS year
// FROM Sales
// GROUP BY product_id
// );
// "#
// }
# Accepted solution for LeetCode #1070: Product Sales Analysis III
# Auto-generated Python example from rust.
def example_solution() -> None:
return
# Reference (rust):
# // Accepted solution for LeetCode #1070: Product Sales Analysis III
# pub fn sql_example() -> &'static str {
# r#"
# -- Accepted solution for LeetCode #1070: Product Sales Analysis III
# # Write your MySQL query statement below
# SELECT
# product_id,
# year AS first_year,
# quantity,
# price
# FROM Sales
# WHERE
# (product_id, year) IN (
# SELECT
# product_id,
# MIN(year) AS year
# FROM Sales
# GROUP BY product_id
# );
# "#
# }
// Accepted solution for LeetCode #1070: Product Sales Analysis III
pub fn sql_example() -> &'static str {
r#"
-- Accepted solution for LeetCode #1070: Product Sales Analysis III
# Write your MySQL query statement below
SELECT
product_id,
year AS first_year,
quantity,
price
FROM Sales
WHERE
(product_id, year) IN (
SELECT
product_id,
MIN(year) AS year
FROM Sales
GROUP BY product_id
);
"#
}
// Accepted solution for LeetCode #1070: Product Sales Analysis III
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1070: Product Sales Analysis III
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1070: Product Sales Analysis III
// # Write your MySQL query statement below
// SELECT
// product_id,
// year AS first_year,
// quantity,
// price
// FROM Sales
// WHERE
// (product_id, year) IN (
// SELECT
// product_id,
// MIN(year) AS year
// FROM Sales
// GROUP BY product_id
// );
// "#
// }
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.