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feat: add sql solution to lc problem: No.3156 #2859

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May 20, 2024
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feat: add sql solution to lc problem: No.3156 #2859

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9799fe5
feat: update lc problems
yanglbme May 20, 2024
f1c100a
feat: add sql solution to lc problem: No.3156
yanglbme May 20, 2024
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feat: add sql solution to lc problem: No.3156 
No.3156.Employee Task Duration and Concurrent Tasks
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@yanglbme
yanglbme committed May 20, 2024
commit f1c100ab7b4a46f9c77417f1530884fce40bcc2c
36 changes: 35 additions & 1 deletion solution/3100-3199/3156.Employee Task Duration and Concurrent Tasks/README.md
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Expand Up @@ -114,7 +114,41 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3156.Em
#### MySQL

```sql

# Write your MySQL query statement below
WITH
T AS (
SELECT DISTINCT employee_id, start_time AS st
FROM Tasks
UNION DISTINCT
SELECT DISTINCT employee_id, end_time AS st
FROM Tasks
),
P AS (
SELECT
*,
LEAD(st) OVER (
PARTITION BY employee_id
ORDER BY st
) AS ed
FROM T
),
S AS (
SELECT
P.*,
COUNT(1) AS concurrent_count
FROM
P
INNER JOIN Tasks USING (employee_id)
WHERE P.st >= Tasks.start_time AND P.ed <= Tasks.end_time
GROUP BY 1, 2, 3
)
SELECT
employee_id,
FLOOR(SUM(TIME_TO_SEC(TIMEDIFF(ed, st)) / 3600)) AS total_task_hours,
MAX(concurrent_count) AS max_concurrent_tasks
FROM S
GROUP BY 1
ORDER BY 1;
```

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36 changes: 35 additions & 1 deletion solution/3100-3199/3156.Employee Task Duration and Concurrent Tasks/README_EN.md
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Expand Up @@ -113,7 +113,41 @@ Each row in this table contains the task identifier, the employee identifier, an
#### MySQL

```sql

# Write your MySQL query statement below
WITH
T AS (
SELECT DISTINCT employee_id, start_time AS st
FROM Tasks
UNION DISTINCT
SELECT DISTINCT employee_id, end_time AS st
FROM Tasks
),
P AS (
SELECT
*,
LEAD(st) OVER (
PARTITION BY employee_id
ORDER BY st
) AS ed
FROM T
),
S AS (
SELECT
P.*,
COUNT(1) AS concurrent_count
FROM
P
INNER JOIN Tasks USING (employee_id)
WHERE P.st >= Tasks.start_time AND P.ed <= Tasks.end_time
GROUP BY 1, 2, 3
)
SELECT
employee_id,
FLOOR(SUM(TIME_TO_SEC(TIMEDIFF(ed, st)) / 3600)) AS total_task_hours,
MAX(concurrent_count) AS max_concurrent_tasks
FROM S
GROUP BY 1
ORDER BY 1;
```

<!-- tabs:end -->
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35 changes: 35 additions & 0 deletions solution/3100-3199/3156.Employee Task Duration and Concurrent Tasks/Solution.sql
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Original file line number Diff line number Diff line change
@@ -0,0 +1,35 @@
# Write your MySQL query statement below
WITH
T AS (
SELECT DISTINCT employee_id, start_time AS st
FROM Tasks
UNION DISTINCT
SELECT DISTINCT employee_id, end_time AS st
FROM Tasks
),
P AS (
SELECT
*,
LEAD(st) OVER (
PARTITION BY employee_id
ORDER BY st
) AS ed
FROM T
),
S AS (
SELECT
P.*,
COUNT(1) AS concurrent_count
FROM
P
INNER JOIN Tasks USING (employee_id)
WHERE P.st >= Tasks.start_time AND P.ed <= Tasks.end_time
GROUP BY 1, 2, 3
)
SELECT
employee_id,
FLOOR(SUM(TIME_TO_SEC(TIMEDIFF(ed, st)) / 3600)) AS total_task_hours,
MAX(concurrent_count) AS max_concurrent_tasks
FROM S
GROUP BY 1
ORDER BY 1;
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