# LeetCode 1045: Customers Who Bought All Products

## Problem Restatement

We have two tables:

`Customer(customer_id, product_key)` — records each purchase.

`Product(product_key)` — all products in the catalog.

We need to return the `customer_id` of customers who bought **all** products in the catalog.

## Input and Output

| Table | Schema |
|---|---|
| `Customer` | `customer_id INT`, `product_key INT` |
| `Product` | `product_key INT` |

Return a table with one column: `customer_id`.

## Examples

```
Customer:               Product:
+-------------+----------+      +-------------+
| customer_id | product_key |    | product_key |
+-------------+----------+      +-------------+
| 1           | 5         |    | 5           |
| 2           | 6         |    | 6           |
| 3           | 5         |    +-------------+
| 3           | 6         |
| 1           | 6         |
+-------------+----------+
```

Customers who bought both `5` and `6`: customers `1` and `3`.

Answer:

```
+-------------+
| customer_id |
+-------------+
| 1           |
| 3           |
+-------------+
```

## Key Insight

For a customer to have bought all products, the count of distinct products they purchased must equal the total number of products.

Group by `customer_id` and filter with `HAVING COUNT(DISTINCT product_key) = (SELECT COUNT(*) FROM Product)`.

## Edge Cases

- Match the aggregate to the question: use raw `COUNT` or `SUM` for rows/amounts, and `COUNT(DISTINCT ...)` only for unique entities.
- Tie cases should return every qualifying row unless the statement asks for a single row.
- For outer joins, keep filters on the joined table in the `ON` clause when unmatched rows must remain visible.

## Common Pitfalls

- Do not put aggregate filters in `WHERE`; use `HAVING` after `GROUP BY`.
- When the answer needs all tied winners, avoid `LIMIT 1` unless the statement explicitly asks for one row.

## Implementation

```sql
SELECT customer_id
FROM Customer
GROUP BY customer_id
HAVING COUNT(DISTINCT product_key) = (SELECT COUNT(*) FROM Product);
```

## Code Explanation

`GROUP BY customer_id` aggregates each customer's purchases.

`COUNT(DISTINCT product_key)` counts unique products each customer bought.

`(SELECT COUNT(*) FROM Product)` gives the total number of products.

The `HAVING` clause keeps only customers who bought every product.