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: Customer
+-------------+---------+ | Column Name | Type | +-------------+---------+ | customer_id | int | | product_key | int | +-------------+---------+ This table may contain duplicates rows.customer_idis not NULL.product_key is a foreign key (reference column) toProducttable.
Table: Product
+-------------+---------+ | Column Name | Type | +-------------+---------+ | product_key | int | +-------------+---------+ product_key is the primary key (column with unique values) for this table.
Write a solution to report the customer ids from the Customer table that bought all the products in the Product table.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Customer table: +-------------+-------------+ | customer_id | product_key | +-------------+-------------+ | 1 | 5 | | 2 | 6 | | 3 | 5 | | 3 | 6 | | 1 | 6 | +-------------+-------------+ Product table: +-------------+ | product_key | +-------------+ | 5 | | 6 | +-------------+ Output: +-------------+ | customer_id | +-------------+ | 1 | | 3 | +-------------+ Explanation: The customers who bought all the products (5 and 6) are customers with IDs 1 and 3.
Problem summary: Table: Customer +-------------+---------+ | Column Name | Type | +-------------+---------+ | customer_id | int | | product_key | int | +-------------+---------+ This table may contain duplicates rows. customer_id is not NULL. product_key is a foreign key (reference column) to Product table. Table: Product +-------------+---------+ | Column Name | Type | +-------------+---------+ | product_key | int | +-------------+---------+ product_key is the primary key (column with unique values) for this table. Write a solution to report the customer ids from the Customer table that bought all the products in the Product table. Return the result table in any order. The result format is in the following example.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers":{"Customer":["customer_id","product_key"],"Product":["product_key"]},"rows":{"Customer":[[1,5],[2,6],[3,5],[3,6],[1,6]],"Product":[[5],[6]]}}Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1045: Customers Who Bought All Products
// Auto-generated Java example from rust.
class Solution {
public void exampleSolution() {
}
}
// Reference (rust):
// // Accepted solution for LeetCode #1045: Customers Who Bought All Products
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1045: Customers Who Bought All Products
// # Write your MySQL query statement below
// SELECT customer_id
// FROM Customer
// GROUP BY 1
// HAVING COUNT(DISTINCT product_key) = (SELECT COUNT(1) FROM Product);
// "#
// }
// Accepted solution for LeetCode #1045: Customers Who Bought All Products
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1045: Customers Who Bought All Products
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1045: Customers Who Bought All Products
// # Write your MySQL query statement below
// SELECT customer_id
// FROM Customer
// GROUP BY 1
// HAVING COUNT(DISTINCT product_key) = (SELECT COUNT(1) FROM Product);
// "#
// }
# Accepted solution for LeetCode #1045: Customers Who Bought All Products
# Auto-generated Python example from rust.
def example_solution() -> None:
return
# Reference (rust):
# // Accepted solution for LeetCode #1045: Customers Who Bought All Products
# pub fn sql_example() -> &'static str {
# r#"
# -- Accepted solution for LeetCode #1045: Customers Who Bought All Products
# # Write your MySQL query statement below
# SELECT customer_id
# FROM Customer
# GROUP BY 1
# HAVING COUNT(DISTINCT product_key) = (SELECT COUNT(1) FROM Product);
# "#
# }
// Accepted solution for LeetCode #1045: Customers Who Bought All Products
pub fn sql_example() -> &'static str {
r#"
-- Accepted solution for LeetCode #1045: Customers Who Bought All Products
# Write your MySQL query statement below
SELECT customer_id
FROM Customer
GROUP BY 1
HAVING COUNT(DISTINCT product_key) = (SELECT COUNT(1) FROM Product);
"#
}
// Accepted solution for LeetCode #1045: Customers Who Bought All Products
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1045: Customers Who Bought All Products
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1045: Customers Who Bought All Products
// # Write your MySQL query statement below
// SELECT customer_id
// FROM Customer
// GROUP BY 1
// HAVING COUNT(DISTINCT product_key) = (SELECT COUNT(1) FROM Product);
// "#
// }
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.