Solutions for SQL 50 Study Plan on LeetCode
1757 - Recyclable and Low Fat Products
SELECT product_id
FROM Products
WHERE low_fats = 'Y'
AND recyclable = 'Y'SELECT name
FROM Customer
WHERE referee_id != 2 OR referee_id IS nullSELECT name, population, area
FROM WORLD
WHERE area >= 3000000
OR population >= 25000000SELECT DISTINCT author_id as id
FROM Views
WHERE viewer_id >= 1
AND author_id = viewer_id
ORDER BY author_idSELECT tweet_id
FROM Tweets
WHERE length(content) > 151378 - Replace Employee ID With The Unique Identifier
SELECT unique_id, name
FROM Employees e
LEFT JOIN EmployeeUNI eu
ON e.id = eu.id1068 - Product Sales Analysis I
SELECT product_name, year, price
FROM Sales s
LEFT JOIN Product p
ON s.product_id = p.product_id1581 - Customer Who Visited but Did Not Make Any Transactions
SELECT customer_id, COUNT(*) as count_no_trans
FROM Visits
WHERE visit_id NOT IN (SELECT DISTINCT visit_id FROM Transactions)
GROUP BY customer_idSELECT w1.id
FROM Weather w1, Weather w2
WHERE DATEDIFF(w1.recordDate, w2.recordDate) = 1
AND w1.temperature > w2.temperature
-- OR
SELECT w1.id
FROM Weather w1, Weather w2
WHERE w1.temperature > w2.temperature
AND SUBDATE(w1.recordDate, 1) = w2.recordDate1661 - Average Time of Process per Machine
SELECT machine_id, ROUND(AVG(end - start), 3) AS processing_time
FROM
(SELECT machine_id, process_id,
MAX(CASE WHEN activity_type = 'start' THEN timestamp END) AS start,
MAX(CASE WHEN activity_type = 'end' THEN timestamp END) AS end
FROM Activity
GROUP BY machine_id, process_id) AS subq
GROUP BY machine_idSELECT name, bonus
FROM Employee e
LEFT JOIN Bonus b
ON e.empId = b.empId
WHERE bonus < 1000
OR bonus IS NULL1280 - Students and Examinations
SELECT a.student_id, a.student_name, b.subject_name, COUNT(c.subject_name) AS attended_exams
FROM Students a
JOIN Subjects b
LEFT JOIN Examinations c
ON a.student_id = c.student_id
AND b.subject_name = c.subject_name
GROUP BY 1, 3
ORDER BY 1, 3 570. Managers with at Least 5 Direct Reports
SELECT name
FROM Employee
WHERE id IN
(SELECT managerId
FROM Employee
GROUP BY managerId
HAVING COUNT(*) >= 5
)
-- OR
SELECT a.name
FROM Employee a
JOIN Employee b
WHERE a.id = b.managerId
GROUP BY b.managerId
HAVING COUNT(*) >= 5SELECT
s.user_id,
ROUND(
COALESCE(
SUM(
CASE WHEN ACTION = 'confirmed' THEN 1 END
) / COUNT(*), 0),2)
AS confirmation_rate
FROM Signups s
LEFT JOIN Confirmations c
ON s.user_id = c.user_id
GROUP BY s.user_id; -- odd id, "boring", rating desc
SELECT *
FROM Cinema
WHERE id % 2 <> 0
AND description <> "boring"
ORDER BY rating DESC-- avg(selling), round 2
SELECT p.product_id,
ROUND(SUM(price * units) / SUM(units), 2) AS average_price
FROM Prices p
LEFT JOIN UnitsSold s
ON p.product_id = s.product_id
AND purchase_date BETWEEN start_date AND end_date
GROUP BY p.product_id-- avg(exp_yr), round 2, by project
SELECT project_id, ROUND(AVG(experience_years), 2) average_years
FROM Project p
LEFT JOIN Employee e
ON p.employee_id = e.employee_id
GROUP BY project_id1633. Percentage of Users Attended a Contest
-- % desc, contest_id asc, round 2
SELECT r.contest_id,
ROUND(COUNT(DISTINCT r.user_id) * 100 / (SELECT COUNT(DISTINCT user_id) FROM Users), 2) AS percentage
FROM Register r
GROUP BY r.contest_id
ORDER BY percentage DESC, r.contest_id ASC;1211 Queries Quality and Percentage
--quality - avg(rating/position), poor query % - %(rating < 3), round 2
SELECT query_name,
ROUND(AVG(rating/position), 2) AS quality,
ROUND(SUM(IF(rating < 3, 1, 0)) * 100/ COUNT(rating), 2) AS poor_query_percentage
FROM Queries
GROUP BY query_name
-- OR
SELECT query_name,
ROUND(AVG(rating/position), 2) AS quality,
ROUND(SUM(
CASE WHEN rating < 3 THEN 1 ELSE 0 END
) * 100/ COUNT(rating), 2) AS poor_query_percentage
FROM Queries
GROUP BY query_name-- month, country, count(trans), total(amt), count(approved_trans), total(amt)
SELECT DATE_FORMAT(trans_date, '%Y-%m') month, country,
COUNT(state) trans_count,
SUM(IF(state = 'approved', 1, 0)) approved_count,
SUM(amount) trans_total_amount,
SUM(IF(state = 'approved', amount, 0)) approved_total_amount
FROM Transactions
GROUP BY 1, 2
-- OR
SELECT DATE_FORMAT(trans_date, '%Y-%m') month, country,
COUNT(state) trans_count,
SUM(CASE WHEN state = 'approved' THEN 1 ELSE 0 END) approved_count,
SUM(amount) trans_total_amount,
SUM(CASE WHEN state = 'approved' THEN amount ELSE 0 END) approved_total_amount
FROM Transactions
GROUP BY 1, 21174. Immediate Food Delivery II
SELECT
ROUND((COUNT(CASE WHEN d.order_date = d.customer_pref_delivery_date THEN 1 END) / COUNT(*)) * 100, 2) immediate_percentage
FROM Delivery d
WHERE d.order_date = (
SELECT
MIN(order_date)
FROM Delivery
WHERE customer_id = d.customer_id
);
-- OR
SELECT ROUND(AVG(temp.order_date=temp.customer_pref_delivery_date) * 100, 2) immediate_percentage
FROM (
SELECT *, RANK() OVER(partition by customer_id ORDER BY order_date) od
FROM Delivery) temp
WHERE temp.od = 1WITH login_date AS (SELECT player_id, MIN(event_date) AS first_login
FROM Activity
GROUP BY player_id),
recent_login AS (
SELECT *, DATE_ADD(first_login, INTERVAL 1 DAY) AS next_day
FROM login_date)
SELECT ROUND((SELECT COUNT(DISTINCT(player_id))
FROM Activity
WHERE (player_id, event_date) IN
(SELECT player_id, next_day FROM recent_login)) / (SELECT COUNT(DISTINCT player_id) FROM Activity), 2) AS fraction2356. Number of Unique Subjects Taught by Each Teacher
SELECT teacher_id, COUNT(DISTINCT subject_id) cnt
FROM Teacher
GROUP BY teacher_id1141. User Activity for the Past 30 Days I
SELECT activity_date as day, COUNT(DISTINCT user_id) AS active_users
FROM Activity
WHERE activity_date BETWEEN DATE_SUB('2019-07-27', INTERVAL 29 DAY) AND '2019-07-27'
GROUP BY activity_date1070. Product Sales Analysis III
SELECT s.product_id, s.year AS first_year, s.quantity, s.price
FROM Sales s
JOIN (
SELECT product_id, MIN(year) AS year
FROM sales
GROUP BY product_id
) p
ON s.product_id = p.product_id
AND s.year = p.year
-- OR
WITH first_year_sales AS (
SELECT s.product_id, MIN(s.year) as first_year
FROM Sales s
INNER JOIN Product p
ON s.product_id = p.product_id
GROUP BY s.product_id)
SELECT f.product_id, f.first_year, s.quantity, s.price
FROM first_year_sales f
JOIN Sales s
ON f.product_id = s.product_id
AND f.first_year = s.year596. Classes More Than 5 Students
SELECT class
FROM Courses
GROUP BY class
HAVING COUNT(student) >= 5SELECT user_id, COUNT(DISTINCT follower_id) AS followers_count
FROM Followers
GROUP BY user_id
ORDER BY user_id ASCSELECT COALESCE(
(SELECT num
FROM MyNumbers
GROUP BY num
HAVING COUNT(num) = 1
ORDER BY num DESC
LIMIT 1), null)
AS num1045. Customers Who Bought All Products
SELECT customer_id
FROM Customer
GROUP BY customer_id
HAVING COUNT(DISTINCT product_key) = (
SELECT COUNT(product_key)
FROM Product
)1731. The Number of Employees Which Report to Each Employee
SELECT e1.employee_id, e1.name, COUNT(e2.employee_id) reports_count, ROUND(AVG(e2.age)) average_age
FROM Employees e1, Employees e2
WHERE e1.employee_id = e2.reports_to
GROUP BY e1.employee_id
HAVING reports_count > 0
ORDER BY e1.employee_id1789. Primary Department for Each Employee
SELECT employee_id, department_id
FROM Employee
WHERE primary_flag = 'Y'
UNION
SELECT employee_id, department_id
FROM Employee
GROUP BY employee_id
HAVING COUNT(employee_id)=1
-- OR
SELECT employee_id,department_id
FROM Employee
WHERE primary_flag = 'Y' OR employee_id IN
(SELECT employee_id
FROM employee
GROUP BY employee_id
HAVING COUNT(department_id) = 1
)SELECT x, y, z,
CASE WHEN x + y > z AND x + z > y AND y + z > x THEN 'Yes'
ELSE 'No' END AS triangle
FROM TriangleWITH cte AS (
SELECT id, num,
LEAD(num) OVER (ORDER BY id) AS next,
LAG(num) OVER (ORDER BY id) AS prev
FROM Logs
)
SELECT DISTINCT(num) AS ConsecutiveNums
FROM cte
WHERE num = next AND num = prev1164. Product Price at a Given Date
SELECT product_id, new_price AS price
FROM products
WHERE (product_id, change_date) IN
(
SELECT product_id, MAX(change_date)
FROM products
WHERE change_date <= '2019-08-16'
GROUP BY product_id
)
UNION
SELECT product_id, 10 AS price
FROM products
WHEN product_id NOT IN
(
SELECT product_id
FROM products
WHERE change_date <= '2019-08-16'
)1978. Employees Whose Manager Left the Company
SELECT employee_id
FROM Employees
WHERE manager_id NOT IN (
SELECT employee_id
FROM Employees
)
AND salary < 30000
ORDER BY employee_id185. Department Top Three Salaries
WITH RankedSalaries AS
(SELECT
e.Id AS employee_id,
e.name AS employee,
e.salary,
e.departmentId,
DENSE_RANK() OVER (PARTITION BY e.departmentId ORDER BY e.salary DESC) AS salary_rank
FROM Employee e)
SELECT d.name AS Department,
r.employee,
r.salary
FROM Department d
JOIN RankedSalaries r ON r.departmentId = d.id
WHERE r.salary_rank <=3;SELECT user_id, CONCAT(UPPER(LEFT(name, 1)), LOWER(RIGHT(name, LENGTH(name)-1))) AS name
FROM Users
ORDER BY user_id1527. Patients With a Condition
SELECT patient_id, patient_name, conditions
FROM patients
WHERE conditions LIKE '% DIAB1%'
OR conditions LIKE 'DIAB1%'DELETE p
FROM Person p, Person q
WHERE p.id > q.id
AND q.Email = p.EmailSELECT
(SELECT DISTINCT Salary
FROM Employee
ORDER BY Salary DESC
LIMIT 1 OFFSET 1)
AS SecondHighestSalary
-- HINT: subquery is used to return null if there is no SecondHighestSalary1517. Find Users With Valid E-Mails
SELECT *
FROM Users
WHERE mail REGEXP '^[A-Za-z][A-Za-z0-9_\.\-]*@leetcode\\.com$'1204. Last Person to Fit in the Bus
-- 1000 kg limit
-- name of last person
WITH CTE AS (
SELECT person_name, weight, turn, SUM(weight)
OVER(ORDER BY turn) AS total_weight
FROM Queue
)
SELECT person_name
FROM cte
WHERE total_weight <=1000
ORDER BY total_weight DESC
LIMIT 1;SELECT 'Low Salary' AS category, SUM(IF(income<20000,1,0)) AS accounts_count
FROM Accounts
UNION
SELECT 'Average Salary' AS category, SUM(IF(income>=20000 AND income<=50000,1,0)) AS accounts_count
FROM Accounts
UNION
SELECT 'High Salary' AS category, SUM(IF(income>50000,1,0)) AS accounts_count
FROM Accounts-- id, student
-- swap every two consecutives
-- num(students): odd? no swap for last one
SELECT id,
CASE WHEN MOD(id,2)=0 THEN (LAG(student) OVER (ORDER BY id))
ELSE (LEAD(student, 1, student) OVER (ORDER BY id))
END AS 'Student'
FROM Seat1327. List the Products Ordered in a Period
-- name, amt
-- >= 100 units, feb 2020
SELECT p.product_name, SUM(o.unit) AS unit
FROM Products p
LEFT JOIN Orders o
ON p.product_id = o.product_id
WHERE DATE_FORMAT(order_date, '%Y-%m') = '2020-02'
GROUP BY p.product_name
HAVING SUM(o.unit) >= 1001484. Group Sold Products By The Date
SELECT sell_date,
COUNT(DISTINCT product) AS num_sold,
GROUP_CONCAT(DISTINCT product) AS 'products'
FROM Activities
GROUP BY sell_date
ORDER BY sell_date(SELECT name AS results
FROM Users u
LEFT JOIN MovieRating mr
ON u.user_id = mr.user_id
GROUP BY name
ORDER BY COUNT(rating) DESC, name ASC
LIMIT 1)
UNION ALL
(SELECT title
FROM Movies m
LEFT JOIN MovieRating mr
ON m.movie_id = mr.movie_id
WHERE DATE_FORMAT(created_at, '%Y-%m') = '2020-02'
GROUP BY title
ORDER BY AVG(rating) DESC, title ASC
LIMIT 1
)-- pay: last 7 days (today inclusive) - avg.amt (round, 2)
SELECT visited_on, amount, ROUND(amount/7, 2) AS average_amount
FROM (
SELECT DISTINCT visited_on,
SUM(amount) OVER(ORDER BY visited_on RANGE BETWEEN INTERVAL 6 DAY PRECEDING AND CURRENT ROW) AS amount,
MIN(visited_on) OVER() day_1
FROM Customer
) t
WHERE visited_on >= day_1+6;602. Friend Requests II: Who Has the Most Friends
-- `union` selects only unique vals, so we use `union all` here
WITH CTE AS (
SELECT requester_id AS id FROM RequestAccepted
UNION ALL
SELECT accepter_id AS id FROM RequestAccepted
)
SELECT id, COUNT(id) AS num
FROM CTE
GROUP BY id
ORDER BY num DESC
LIMIT 1SELECT
ROUND(SUM(tiv_2016),2) AS tiv_2016
FROM insurance
WHERE tiv_2015 IN (SELECT tiv_2015 FROM insurance GROUP BY tiv_2015 HAVING COUNT(*) > 1)
AND (lat,lon) IN (SELECT lat,lon FROM insurance GROUP BY lat,lon HAVING COUNT(*) = 1)