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: 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.
product_id is a foreign key (reference column) to Product table.
Each row of this table shows a sale on the product product_id in a certain year.
Note that the price is per unit.
Table: Product
+--------------+---------+ | Column Name | Type | +--------------+---------+ | product_id | int | | product_name | varchar | +--------------+---------+ product_id is the primary key (column with unique values) of this table. Each row of this table indicates the product name of each product.
Write a solution to report the product_name, year, and price for each sale_id in the Sales table.
Return the resulting table in any order.
The result format is in the following example.
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 | +---------+------------+------+----------+-------+ Product table: +------------+--------------+ | product_id | product_name | +------------+--------------+ | 100 | Nokia | | 200 | Apple | | 300 | Samsung | +------------+--------------+ Output: +--------------+-------+-------+ | product_name | year | price | +--------------+-------+-------+ | Nokia | 2008 | 5000 | | Nokia | 2009 | 5000 | | Apple | 2011 | 9000 | +--------------+-------+-------+ Explanation: From sale_id = 1, we can conclude that Nokia was sold for 5000 in the year 2008. From sale_id = 2, we can conclude that Nokia was sold for 5000 in the year 2009. From sale_id = 7, we can conclude that Apple was sold for 9000 in the year 2011.
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. product_id is a foreign key (reference column) to Product table. Each row of this table shows a sale on the product product_id in a certain year. Note that the price is per unit. Table: Product +--------------+---------+ | Column Name | Type | +--------------+---------+ | product_id | int | | product_name | varchar | +--------------+---------+ product_id is the primary key (column with unique values) of this table. Each row of this table indicates the product name of each product. Write a solution to report the product_name, year, and price for each sale_id in the Sales table. Return the resulting
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"],"Product":["product_id","product_name"]},"rows":{"Sales":[[1,100,2008,10,5000],[2,100,2009,12,5000],[7,200,2011,15,9000]],"Product":[[100,"Nokia"],[200,"Apple"],[300,"Samsung"]]}}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 #1068: Product Sales Analysis I
// Auto-generated Java example from rust.
class Solution {
public void exampleSolution() {
}
}
// Reference (rust):
// // Accepted solution for LeetCode #1068: Product Sales Analysis I
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1068: Product Sales Analysis I
// # Write your MySQL query statement below
// SELECT product_name, year, price
// FROM
// Sales
// JOIN Product USING (product_id);
// "#
// }
// Accepted solution for LeetCode #1068: Product Sales Analysis I
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1068: Product Sales Analysis I
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1068: Product Sales Analysis I
// # Write your MySQL query statement below
// SELECT product_name, year, price
// FROM
// Sales
// JOIN Product USING (product_id);
// "#
// }
# Accepted solution for LeetCode #1068: Product Sales Analysis I
# Auto-generated Python example from rust.
def example_solution() -> None:
return
# Reference (rust):
# // Accepted solution for LeetCode #1068: Product Sales Analysis I
# pub fn sql_example() -> &'static str {
# r#"
# -- Accepted solution for LeetCode #1068: Product Sales Analysis I
# # Write your MySQL query statement below
# SELECT product_name, year, price
# FROM
# Sales
# JOIN Product USING (product_id);
# "#
# }
// Accepted solution for LeetCode #1068: Product Sales Analysis I
pub fn sql_example() -> &'static str {
r#"
-- Accepted solution for LeetCode #1068: Product Sales Analysis I
# Write your MySQL query statement below
SELECT product_name, year, price
FROM
Sales
JOIN Product USING (product_id);
"#
}
// Accepted solution for LeetCode #1068: Product Sales Analysis I
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1068: Product Sales Analysis I
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1068: Product Sales Analysis I
// # Write your MySQL query statement below
// SELECT product_name, year, price
// FROM
// Sales
// JOIN Product USING (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.