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‎README.md

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| # | Title | Difficulty | Tag | Time, ms | Time, %
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|------|----------------|-------------|-------------|----------|--------
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| 3554 |[Find Category Recommendation Pairs](src/main/java/g3501_3600/s3554_find_category_recommendation_pairs)| Hard | Database | 623 | 82.76
20912092
| 3553 |[Minimum Weighted Subgraph With the Required Paths II](src/main/java/g3501_3600/s3553_minimum_weighted_subgraph_with_the_required_paths_ii)| Hard | Array, Depth_First_Search, Tree | 65 | 100.00
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| 3552 |[Grid Teleportation Traversal](src/main/java/g3501_3600/s3552_grid_teleportation_traversal)| Medium | Array, Hash_Table, Breadth_First_Search, Matrix | 146 | 98.62
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| 3551 |[Minimum Swaps to Sort by Digit Sum](src/main/java/g3501_3600/s3551_minimum_swaps_to_sort_by_digit_sum)| Medium | Array, Hash_Table, Sorting | 213 | 99.23

‎src/main/java/g0101_0200/s0175_combine_two_tables/readme.md

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@@ -77,6 +77,14 @@ There is no address in the address table for the personId = 1 so we return null
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```sql
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# Write your MySQL query statement below
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SELECT FirstName, LastName, City, State
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FROM Person LEFT JOIN Address USING (PersonId)
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SELECT
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FirstName,
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LastName,
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City,
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State
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FROM
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Person
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LEFT JOIN
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Address
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USING (PersonId);
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```

‎src/main/java/g0101_0200/s0176_second_highest_salary/readme.md

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```sql
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# Write your MySQL query statement below
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SELECT ifnull(
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(SELECT distinct(Salary)
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FROM Employee
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ORDER BY Salary DESC
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LIMIT 1
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OFFSET 1), NULL) SecondHighestSalary;
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SELECT
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IFNULL(
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(
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SELECT DISTINCT Salary
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FROM Employee
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ORDER BY Salary DESC
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LIMIT 1 OFFSET 1
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),
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NULL
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) AS SecondHighestSalary;
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```

‎src/main/java/g0101_0200/s0178_rank_scores/readme.md

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```sql
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# Write your MySQL query statement below
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select Score, DENSE_RANK() OVER(order by Score Desc) as "Rank" from Scores order by "Rank" Asc;
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SELECT
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Score,
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DENSE_RANK() OVER (ORDER BY Score DESC) AS Rank
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FROM
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Scores
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ORDER BY
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Rank ASC;
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```

‎src/main/java/g0101_0200/s0180_consecutive_numbers/readme.md

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```sql
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# Write your MySQL query statement below
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SELECT DISTINCT l1.num AS ConsecutiveNums
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FROM Logs l1
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JOIN Logs l2 ON l1.id = l2.id - 1
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JOIN Logs l3 ON l1.id = l3.id - 2
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WHERE l1.num = l2.num AND l2.num = l3.num;
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SELECT DISTINCT
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l1.num AS ConsecutiveNums
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FROM
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Logs l1
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JOIN Logs l2 ON l1.id = l2.id - 1
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JOIN Logs l3 ON l1.id = l3.id - 2
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WHERE
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l1.num = l2.num
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AND l2.num = l3.num;
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```

‎src/main/java/g0101_0200/s0181_employees_earning_more_than_their_managers/readme.md

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```sql
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# Write your MySQL query statement below
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select a.Name as Employee from Employee a left join Employee b on a.ManagerId=b.Id
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where a.Salary > b.Salary and a.ManagerId is not null
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SELECT
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a.Name AS Employee
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FROM
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Employee a
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LEFT JOIN Employee b ON a.ManagerId = b.Id
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WHERE
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a.Salary > b.Salary
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AND a.ManagerId IS NOT NULL;
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```

‎src/main/java/g0101_0200/s0182_duplicate_emails/readme.md

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```sql
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# Write your MySQL query statement below
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SELECT Email FROM Person GROUP BY Email HAVING COUNT(Email) > 1;
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SELECT
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Email
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FROM
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Person
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GROUP BY
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Email
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HAVING
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COUNT(Email) > 1;
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```

‎src/main/java/g0101_0200/s0183_customers_who_never_order/readme.md

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```sql
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# Write your MySQL query statement below
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SELECT c.Name as Customers
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FROM Customers as c
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LEFT JOIN Orders as o
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ON c.Id = o.CustomerId
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WHERE o.CustomerId is null
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SELECT
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c.Name AS Customers
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FROM
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Customers AS c
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LEFT JOIN Orders AS o ON c.Id = o.CustomerId
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WHERE
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o.CustomerId IS NULL;
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```

‎src/main/java/g0101_0200/s0184_department_highest_salary/readme.md

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Sel.Name AS Employee,
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Sel.Salary AS Salary
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FROM
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(
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SELECT
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Name,
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Salary,
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DepartmentId,
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DENSE_RANK() OVER (PARTITION BY DepartmentId ORDER BY Salary DESC) AS dr
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FROM Employee
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) AS Sel
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INNER JOIN Department d ON d.Id = Sel.DepartmentId
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WHERE Sel.dr = 1
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(
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SELECT
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Name,
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Salary,
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DepartmentId,
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DENSE_RANK() OVER (
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PARTITION BY DepartmentId
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ORDER BY Salary DESC
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) AS dr
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FROM
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Employee
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) AS Sel
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INNER JOIN Department d ON d.Id = Sel.DepartmentId
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WHERE
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Sel.dr = 1;
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```

‎src/main/java/g3501_3600/s3554_find_category_recommendation_pairs/readme.md

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[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Java?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Java)
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[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Java?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Java/fork)
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## 3554\. Find Category Recommendation Pairs
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Table: `ProductPurchases`
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+-------------+------+
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| Column Name | Type |
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+-------------+------+
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| user_id | int |
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| product_id | int |
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| quantity | int |
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+-------------+------+
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(user_id, product_id) is the unique identifier for this table.
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Each row represents a purchase of a product by a user in a specific quantity.
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Table: `ProductInfo`
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+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| product_id | int |
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| category | varchar |
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| price | decimal |
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+-------------+---------+
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product_id is the unique identifier for this table.
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Each row assigns a category and price to a product.
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Amazon wants to understand shopping patterns across product categories. Write a solution to:
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1. Find all **category pairs** (where `category1` < `category2`)
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2. For **each category pair**, determine the number of **unique** **customers** who purchased products from **both** categories
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A category pair is considered **reportable** if at least `3` different customers have purchased products from both categories.
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Return _the result table of reportable category pairs ordered by **customer\_count** in **descending** order, and in case of a tie, by **category1** in **ascending** order lexicographically, and then by **category2** in **ascending** order._
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The result format is in the following example.
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**Example:**
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**Input:**
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ProductPurchases table:
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+---------+------------+----------+
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| user_id | product_id | quantity |
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+---------+------------+----------+
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| 1 | 101 | 2 |
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| 1 | 102 | 1 |
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| 1 | 201 | 3 |
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| 1 | 301 | 1 |
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| 2 | 101 | 1 |
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| 2 | 102 | 2 |
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| 2 | 103 | 1 |
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| 2 | 201 | 5 |
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| 3 | 101 | 2 |
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| 3 | 103 | 1 |
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| 3 | 301 | 4 |
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| 3 | 401 | 2 |
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| 4 | 101 | 1 |
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| 4 | 201 | 3 |
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| 4 | 301 | 1 |
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| 4 | 401 | 2 |
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| 5 | 102 | 2 |
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| 5 | 103 | 1 |
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| 5 | 201 | 2 |
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| 5 | 202 | 3 |
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+---------+------------+----------+
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ProductInfo table:
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+------------+-------------+-------+
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| product_id | category | price |
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+------------+-------------+-------+
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| 101 | Electronics | 100 |
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| 102 | Books | 20 |
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| 103 | Books | 35 |
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| 201 | Clothing | 45 |
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| 202 | Clothing | 60 |
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| 301 | Sports | 75 |
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| 401 | Kitchen | 50 |
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+------------+-------------+-------+
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**Output:**
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+-------------+-------------+----------------+
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| category1 | category2 | customer_count |
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+-------------+-------------+----------------+
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| Books | Clothing | 3 |
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| Books | Electronics | 3 |
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| Clothing | Electronics | 3 |
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| Electronics | Sports | 3 |
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+-------------+-------------+----------------+
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**Explanation:**
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* **Books-Clothing**:
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* User 1 purchased products from Books (102) and Clothing (201)
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* User 2 purchased products from Books (102, 103) and Clothing (201)
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* User 5 purchased products from Books (102, 103) and Clothing (201, 202)
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* Total: 3 customers purchased from both categories
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* **Books-Electronics**:
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* User 1 purchased products from Books (102) and Electronics (101)
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* User 2 purchased products from Books (102, 103) and Electronics (101)
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* User 3 purchased products from Books (103) and Electronics (101)
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* Total: 3 customers purchased from both categories
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* **Clothing-Electronics**:
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* User 1 purchased products from Clothing (201) and Electronics (101)
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* User 2 purchased products from Clothing (201) and Electronics (101)
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* User 4 purchased products from Clothing (201) and Electronics (101)
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* Total: 3 customers purchased from both categories
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* **Electronics-Sports**:
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* User 1 purchased products from Electronics (101) and Sports (301)
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* User 3 purchased products from Electronics (101) and Sports (301)
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* User 4 purchased products from Electronics (101) and Sports (301)
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* Total: 3 customers purchased from both categories
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* Other category pairs like Clothing-Sports (only 2 customers: Users 1 and 4) and Books-Kitchen (only 1 customer: User 3) have fewer than 3 shared customers and are not included in the result.
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The result is ordered by customer\_count in descending order. Since all pairs have the same customer\_count of 3, they are ordered by category1 (then category2) in ascending order.
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## Solution
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```sql
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# Write your MySQL query statement below
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SELECT
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pi1.category AS category1,
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pi2.category AS category2,
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COUNT(DISTINCT pp1.user_id) AS customer_count
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FROM
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ProductPurchases pp1,
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ProductPurchases pp2,
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ProductInfo pi1,
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ProductInfo pi2
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WHERE
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pp1.user_id = pp2.user_id
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AND pi1.category < pi2.category
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AND pp1.product_id = pi1.product_id
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AND pp2.product_id = pi2.product_id
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GROUP BY
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pi1.category,
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pi2.category
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HAVING
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COUNT(DISTINCT pp1.user_id) >= 3
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ORDER BY
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customer_count DESC,
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category1 ASC,
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category2 ASC;
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```

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