new codes

Author: ankurbhambri

first_last.sql

@@ -1,5 +1,5 @@
 -- By default, the frame clause is set to 'RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW'.
--- It will access all records from the start to the end of the partition with 'RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING'.
+-- It will access/copy the final column value from the start to the end of that column with 'RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING'.
 -- If you use 'ROWS' instead of 'RANGE' and end with 'CURRENT ROW' like this - 'ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW' - it will consider only the current row.
 
 SELECT *,

lead_lag.sql

@@ -36,4 +36,13 @@ select *,
 	case when salary > lag(salary) over(partition by dept order by id) then 'Higher than previous employee'
 	when salary = lag(salary) over(partition by dept order by id) then 'Same as previous employee' 
 	else 'Less than previous employee' end
-from employees;
+from employees;
+
+
+
+-- Wthout LAG or LEAD
+
+with cte as (select row_number() over() id, amount from orders)
+select a.id, a.amount - b.amount as rv  from cte a left join cte b on a.id =  b.id + 1 -- for previous add
+union all
+select a.id, a.amount - b.amount  from cte a left join cte b on a.id =  b.id - 1 -- for next subract

median.sql

@@ -0,0 +1,20 @@
+-- median (odd set of numbers) = ((n+1)/2)th term
+-- median (even set of numbers) = ((n/2)th term + ((n/2)+1)th term)/2
+
+with cte as (
+    select student_id, sat_writing, rank() over(order by sat_writing) rn
+    from sat_scores
+),
+cte2 as (
+    select 
+    case 
+        when
+            (count(*) % 2) = 0 then (count(*) / 2) + (count(*) / 2) + 1 -- even median
+        else
+            (count(*) + 1 ) / 2 end as cn -- odd median
+    from cte
+)
+select cte.student_id, cte2.cn
+from cte, cte2
+where cte.rn=cte2.cn
+order by cte.student_id

recursive_cte.sql

@@ -1,36 +1,44 @@
-WITH RECURSIVE ManagerHierarchy AS (
+-- Counting 1 to N
+
+with recursive cte as (
+	select 1 as n
+	union all
+	select n + 1 from cte where n < 10
+)
+select * from cte;
+
+-- Pairs of number like [(0, 1), (1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]
+
+WITH RECURSIVE pairs AS (
+    SELECT 0 AS first_number, 1 AS second_number
+    UNION ALL
+    SELECT first_number + 2, second_number + 2
+    FROM pairs
+    WHERE first_number + 2 <= 10
+)
+SELECT * from pairs;
+
+-- Classical example of employee manager finding recursively
+
+WITH RECURSIVE cte AS (
     -- Anchor member: Select top-level managers (where manager_id is NULL)
     SELECT
-		id, name, manager_id, 0 AS level
+		employee_id, name, manager_id
     FROM
 		Employees
     WHERE
-		manager_id IS NULL
+		manager_id IS NULL -- starting point with root node
 
     UNION ALL
 
     -- Recursive member: Join with Employees table to find subsequent managers
     SELECT
-		e.id, e.name,
-		e.manager_id, 
-		mh.level + 1 AS level
+		e.employee_id, e.name, e.manager_id
     FROM
 		Employees e
     JOIN
-		ManagerHierarchy mh
+		cte c
 	ON
-		e.manager_id = mh.id
-)
--- Final query: Select from the CTE to retrieve all levels of managers
-SELECT * FROM ManagerHierarchy
-ORDER BY level, id;
-
-
-
-
-with recursive cte as (
-	select 1 as n
-	union all
-	select n + 1 from cte where n < 10
+		e.employee_id = c.manager_id
 )
-select * from cte;
+SELECT * FROM cte;

tricky_questions/Cookbook_Recipes.sql

@@ -0,0 +1,36 @@
+/*
+
+https://platform.stratascratch.com/coding/2089-cookbook-recipes/discussion?code_type=1
+
+pairs cte will give this
+
+First_number  Second_number
+    0	         1
+    2	         3
+    4	         5
+    6	         7
+    8	         9
+    10	         11
+    12	         13
+    14	         15
+
+*/
+
+WITH pairs as (
+    select 
+        (ROW_NUMBER() OVER (ORDER BY page_number) - 1 ) * 2  AS first_number,
+        (ROW_NUMBER() OVER (ORDER BY page_number) - 1 ) * 2 + 1 AS second_number
+    from cookbook_titles
+)
+SELECT
+    p.first_number AS left_page_number,
+    a.title AS left_title,
+    b.title AS right_title
+FROM
+    pairs p
+LEFT JOIN
+    cookbook_titles a ON a.page_number = p.first_number
+LEFT JOIN
+    cookbook_titles b ON b.page_number = p.second_number
+ORDER BY
+    p.first_number;

tricky_questions/Marketing_Campaign_Success.sql

@@ -0,0 +1,15 @@
+with cte as (
+    select *, 
+        rank() over(partition by user_id order by created_at) as rn 
+    from marketing_campaign
+),
+first_day as (
+    select * from cte where rn = 1
+),
+another_day as (
+    select * from cte where rn > 1
+)
+select count(distinct a.user_id) from another_day a 
+left join first_day b 
+on a.user_id = b.user_id and a.product_id = b.product_id
+where b.product_id is null

tricky_questions/Monthly_Percentage_Difference.sql

@@ -0,0 +1,35 @@
+-- https://platform.stratascratch.com/coding/10319-monthly-percentage-difference/discussion?code_type=1
+
+-- Using lag window function
+
+with cte as (
+    select
+        to_char(created_at, 'YYYY-MM') year_month,
+        sum(value) as curr_rv,
+        lag(sum(value)) over() prev_rv
+    from sf_transactions
+    group by to_char(created_at, 'YYYY-MM')
+)
+select
+    year_month,
+    (curr_rv - prev_rv) * 100 / prev_rv as revenue_diff_pct
+from cte
+order by year_month;
+
+
+-- Without lag and using left join
+
+with cte as (
+    select
+        to_char(created_at, 'YYYY-MM') as year_month,
+        sum(value) as curr_rv
+    from sf_transactions
+    group by to_char(created_at, 'YYYY-MM')
+)
+select
+    c1.year_month, c1.curr_rv, c2.curr_rv, c2.year_month,
+    (c1.curr_rv - c2.curr_rv) * 100 / c2.curr_rv as revenue_diff_pct
+from cte c1
+left join cte c2
+on c1.year_month = to_char(to_date(c2.year_month, 'YYYY-MM') + interval '1 month', 'YYYY-MM')
+order by c1.year_month;

word_count.sql

@@ -0,0 +1,11 @@
+-- https://platform.stratascratch.com/coding/9814-counting-instances-in-text/official-solution?code_type=1
+
+-- Counting Instances in Text
+
+SELECT 
+    word,nentry                                       
+FROM  
+    ts_stat('SELECT to_tsvector(contents) FROM google_file_store') 
+WHERE
+         ILIKE 'bull' or word ILIKE 'bear'
+