diff --git a/LeetcodeProblems/Databases/Human_Traffic_of_Stadium.sql b/LeetcodeProblems/Databases/Human_Traffic_of_Stadium.sql new file mode 100644 index 0000000..4719696 --- /dev/null +++ b/LeetcodeProblems/Databases/Human_Traffic_of_Stadium.sql @@ -0,0 +1,57 @@ +/* +Human Traffic of Stadium +https://leetcode.com/problems/human-traffic-of-stadium/ + +Table: Stadium + ++---------------+---------+ +| Column Name | Type | ++---------------+---------+ +| id | int | +| visit_date | date | +| people | int | ++---------------+---------+ +visit_date is the primary key for this table. +Each row of this table contains the visit date and visit id to the stadium with the number of people during the visit. +No two rows will have the same visit_date, and as the id increases, the dates increase as well. + +Write an SQL query to display the records with three or more rows with consecutive id's, and the number of people is greater than or equal to 100 for each. + +Return the result table ordered by visit_date in ascending order. + +The query result format is in the following example. + +Stadium table: ++------+------------+-----------+ +| id | visit_date | people | ++------+------------+-----------+ +| 1 | 2017-01-01 | 10 | +| 2 | 2017-01-02 | 109 | +| 3 | 2017-01-03 | 150 | +| 4 | 2017-01-04 | 99 | +| 5 | 2017-01-05 | 145 | +| 6 | 2017-01-06 | 1455 | +| 7 | 2017-01-07 | 199 | +| 8 | 2017-01-09 | 188 | ++------+------------+-----------+ + +Result table: ++------+------------+-----------+ +| id | visit_date | people | ++------+------------+-----------+ +| 5 | 2017-01-05 | 145 | +| 6 | 2017-01-06 | 1455 | +| 7 | 2017-01-07 | 199 | +| 8 | 2017-01-09 | 188 | ++------+------------+-----------+ +The four rows with ids 5, 6, 7, and 8 have consecutive ids and each of them has >= 100 people attended. Note that row 8 was included even though the visit_date was not the next day after row 7. +The rows with ids 2 and 3 are not included because we need at least three consecutive ids. +*/ + +SELECT DISTINCT(s1.id) as id, s1.visit_date, s1.people +FROM Stadium AS s1, Stadium AS s2, Stadium AS s3 +WHERE + (s1.people >= 100 AND s2.people >= 100 AND s3.people >= 100 AND s1.id = s2.id - 1 AND s2.id = s3.id - 1 AND s1.id = s3.id - 2) OR -- S1, S2, S3 + (s1.people >= 100 AND s2.people >= 100 AND s3.people >= 100 AND s1.id = s2.id + 1 AND s2.id = s3.id - 2 AND s1.id = s3.id - 1) OR -- S2. S1. S3 + (s1.people >= 100 AND s2.people >= 100 AND s3.people >= 100 AND s1.id = s2.id + 2 AND s2.id = s3.id - 1 AND s1.id = s3.id + 1) -- S2 S3 S1 +ORDER BY s1.id ASC diff --git a/LeetcodeProblems/Databases/Rank_Scores.sql b/LeetcodeProblems/Databases/Rank_Scores.sql new file mode 100644 index 0000000..4a59063 --- /dev/null +++ b/LeetcodeProblems/Databases/Rank_Scores.sql @@ -0,0 +1,56 @@ +/* +Rank Scores +https://leetcode.com/problems/rank-scores/ + +Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" between ranks. + ++----+-------+ +| Id | Score | ++----+-------+ +| 1 | 3.50 | +| 2 | 3.65 | +| 3 | 4.00 | +| 4 | 3.85 | +| 5 | 4.00 | +| 6 | 3.65 | ++----+-------+ +For example, given the above Scores table, your query should generate the following report (order by highest score): + ++-------+---------+ +| score | Rank | ++-------+---------+ +| 4.00 | 1 | +| 4.00 | 1 | +| 3.85 | 2 | +| 3.65 | 3 | +| 3.65 | 3 | +| 3.50 | 4 | ++-------+---------+ +Important Note: For MySQL solutions, to escape reserved words used as column names, you can use an apostrophe before and after the keyword. For example `Rank`. +*/ + +-- Solution 1 +SELECT Scores.Score, t3.Rank +FROM Scores JOIN ( + select Score, (@inc := @inc + 1) as "Rank" + FROM ( + SELECT Score as Score + FROM Scores + GROUP BY Score + ORDER BY Score DESC + ) AS t1 CROSS JOIN (SELECT @inc := 0 ) as t2 +) as t3 ON Scores.Score = t3.Score +ORDER BY t3.Rank ASC + + +-- Solution 2 +-- Set two variables, one to keep the previous Score and other to keep the current Rank +SELECT Score, + @curr := CASE WHEN @previous = (@previous := Score) THEN + CAST(@curr AS SIGNED INTEGER) + ELSE + CAST(@curr + 1 AS SIGNED INTEGER) + END AS "Rank" +FROM Scores, (SELECT @curr := 0, @previous := -1) as t1 +ORDER BY SCORE DESC + diff --git a/LeetcodeProblems/Databases/Trips_and_Users.sql b/LeetcodeProblems/Databases/Trips_and_Users.sql new file mode 100644 index 0000000..b8b3d75 --- /dev/null +++ b/LeetcodeProblems/Databases/Trips_and_Users.sql @@ -0,0 +1,71 @@ +/* +Trips and Users +https://leetcode.com/problems/trips-and-users/ + +SQL Schema +The Trips table holds all taxi trips. Each trip has a unique Id, while Client_Id and Driver_Id are both foreign keys to the Users_Id at the Users table. Status is an ENUM type of (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’). + ++----+-----------+-----------+---------+--------------------+----------+ +| Id | Client_Id | Driver_Id | City_Id | Status |Request_at| ++----+-----------+-----------+---------+--------------------+----------+ +| 1 | 1 | 10 | 1 | completed |2013-10-01| +| 2 | 2 | 11 | 1 | cancelled_by_driver|2013-10-01| +| 3 | 3 | 12 | 6 | completed |2013-10-01| +| 4 | 4 | 13 | 6 | cancelled_by_client|2013-10-01| +| 5 | 1 | 10 | 1 | completed |2013-10-02| +| 6 | 2 | 11 | 6 | completed |2013-10-02| +| 7 | 3 | 12 | 6 | completed |2013-10-02| +| 8 | 2 | 12 | 12 | completed |2013-10-03| +| 9 | 3 | 10 | 12 | completed |2013-10-03| +| 10 | 4 | 13 | 12 | cancelled_by_driver|2013-10-03| ++----+-----------+-----------+---------+--------------------+----------+ +The Users table holds all users. Each user has an unique Users_Id, and Role is an ENUM type of (‘client’, ‘driver’, ‘partner’). + ++----------+--------+--------+ +| Users_Id | Banned | Role | ++----------+--------+--------+ +| 1 | No | client | +| 2 | Yes | client | +| 3 | No | client | +| 4 | No | client | +| 10 | No | driver | +| 11 | No | driver | +| 12 | No | driver | +| 13 | No | driver | ++----------+--------+--------+ +Write a SQL query to find the cancellation rate of requests made by unbanned users (both client and driver must be unbanned) between Oct 1, 2013 and Oct 3, 2013. The cancellation rate is computed by dividing the number of canceled (by client or driver) requests made by unbanned users by the total number of requests made by unbanned users. + +For the above tables, your SQL query should return the following rows with the cancellation rate being rounded to two decimal places. + ++------------+-------------------+ +| Day | Cancellation Rate | ++------------+-------------------+ +| 2013-10-01 | 0.33 | +| 2013-10-02 | 0.00 | +| 2013-10-03 | 0.50 | ++------------+-------------------+ +*/ + +-- solution 1 +SELECT trips_per_date.Request_at AS "Day", Round(coalesce(cancelled_trips.count_of_cancelled_trips, 0) / trips_per_date.count_of_trips, 2) as "Cancellation Rate" +FROM ( + SELECT Trips.Request_at, count(*) as count_of_cancelled_trips + FROM Trips JOIN Users as users_drivers ON Trips.Driver_Id = users_drivers.Users_Id JOIN Users as users_clients ON Trips.Client_Id = users_clients.Users_Id + WHERE users_drivers.Banned = "No" AND users_clients.Banned = "No" AND (Trips.Status = "cancelled_by_driver" OR Trips.Status = "cancelled_by_client") + GROUP BY Trips.Request_at + ) as cancelled_trips RIGHT JOIN ( + SELECT Request_at, count(*) as count_of_trips + FROM Trips JOIN Users as users_drivers ON Trips.Driver_Id = users_drivers.Users_Id JOIN Users as users_clients ON Trips.Client_Id = users_clients.Users_Id + WHERE users_drivers.Banned = "No" AND users_clients.Banned = "No" + Group by Request_at + ) as trips_per_date ON cancelled_trips.Request_at = trips_per_date.Request_at +WHERE trips_per_date.Request_at >= Date("2013-10-01") AND trips_per_date.Request_at <= Date("2013-10-03") +ORDER BY trips_per_date.Request_at + +-- solution 2 +SELECT Trips.Request_at AS "Day", ROUND(count(IF(Trips.Status = "cancelled_by_driver" OR Trips.Status = "cancelled_by_client",TRUE,null)) / count(*), 2) as "Cancellation Rate" +FROM Trips JOIN Users as users_drivers ON Trips.Driver_Id = users_drivers.Users_Id JOIN Users as users_clients ON Trips.Client_Id = users_clients.Users_Id +WHERE users_drivers.Banned = "No" AND users_clients.Banned = "No" -- AND () + AND Trips.Request_at >= Date("2013-10-01") AND Trips.Request_at <= Date("2013-10-03") +GROUP BY Trips.Request_at +ORDER BY Trips.Request_at \ No newline at end of file diff --git a/README.md b/README.md index d9b2e8a..a95b292 100644 --- a/README.md +++ b/README.md @@ -77,10 +77,14 @@ To run a specific problem in your console, go to the file test, add the call to | Algoritmhs | | - | | [Heap Sort](/SortingAlgorithms/heapSort.js) | +| [Quick Sort](/SortingAlgorithms/QuickSort.js) | ### Databases | Problems | Level | Link | |-|-|-| +| [Trips and Users](/LeetcodeProblems/Databases/Trips_and_Users.sql) | Hard | https://leetcode.com/problems/trips-and-users/ | +| [Human Traffic of Stadium](/LeetcodeProblems/Databases/Human_Traffic_of_Stadium.sql) | Hard | https://leetcode.com/problems/human-traffic-of-stadium | +| [Rank Scores](/LeetcodeProblems/Databases/Rank_Scores.sql) | Medium | https://leetcode.com/problems/rank-scores | | [Consecutive Numbers](/LeetcodeProblems/Databases/Consecutive_Numbers.sql) | Medium | https://leetcode.com/problems/consecutive-numbers | | [Department Highest Salary](/LeetcodeProblems/Databases/Department_Highest_Salary.sql) | Medium | https://leetcode.com/problems/department-highest-salary | | [Exchange Seats](/LeetcodeProblems/Databases/Exchange_Seats.sql) | Medium | https://leetcode.com/problems/exchange-seats |