You are given a table,
Projects, containing three columns:
End_Date. It is guaranteed that the difference between the
End_Date and the
Start_Date is equal to 1 day for each row in the table.
End_Date of the tasks are consecutive, then they are part of the same project. Samantha is interested in finding the total number of different projects completed.
Write a query to output the start and end dates of projects listed by the number of days it took to complete the project in ascending order. If there is more than one project that have the same number of completion days, then order by the start date of the project.
Here, we'll discuss two different SQL approaches that make use of window functions and date arithmetic to solve this problem. Both approaches have different methods of grouping tasks into projects.
This approach groups tasks into projects based on their
Start_Date. If a task's
Start_Date is consecutive to another task's
End_Date, they belong to the same project. The
ROW_NUMBER window function generates a sequence of numbers, which is subtracted from the
Start_Date to create a grouping identifier.
WITH CTE AS ( SELECT Start_Date, End_Date, DATEADD(day, - ROW_NUMBER() OVER(ORDER BY Start_Date), Start_Date) [Grouping] FROM Projects ) SELECT MIN(Start_Date) [Project_Start_Date], MAX(End_Date) [Project_End_Date] FROM CTE GROUP BY Grouping ORDER BY DATEDIFF(day, MIN(Start_Date), MAX(End_Date)), MIN(Start_Date)
This approach, on the other hand, groups tasks into projects based on their
End_Date. Similar to the first approach,
ROW_NUMBER is used to generate a sequence of numbers, which is subtracted from the
End_Date to create a grouping identifier.
WITH CTE AS ( SELECT Start_Date, End_Date, DATEADD(day, - ROW_NUMBER() OVER(ORDER BY End_Date), End_Date) [Grouping] FROM Projects ) SELECT MIN(Start_Date) [Project_Start_Date], MAX(End_Date) [Project_End_Date] FROM CTE GROUP BY Grouping ORDER BY DATEDIFF(day, MIN(Start_Date), MAX(End_Date)), MIN(Start_Date)
Both of these solutions offer an efficient way to tackle the problem using window functions and date arithmetic. They provide an excellent showcase of some of the more advanced features of SQL and how they can be used to solve complex problems.
While these solutions have been shown to be effective on HackerRank, it's important to keep in mind that performance can vary based on different factors like the specific RDBMS used, the database structure, and the size of the data.
You can find the original problem at HackerRank.
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