solve: Hard faizanxmulla#8, faizanxmulla#9

Author: faizanxmulla

interview-query-solutions/Hard/08-cumulative-distribution.sql

@@ -0,0 +1,17 @@
+WITH hist as (
+    SELECT   u.id, COUNT(c.user_id) AS frequency
+    FROM     users u LEFT JOIN comments c ON u.id=c.user_id
+    GROUP BY 1
+),
+freq as (
+    SELECT   frequency, COUNT(*) AS num_users
+    FROM     hist
+    GROUP BY 1
+)
+SELECT   f1.frequency, SUM(f2.num_users) AS cum_total
+FROM     freq f1 LEFT JOIN freq f2 ON f1.frequency >= f2.frequency
+GROUP BY 1
+
+
+
+-- NOTE: was easy in hindsight but couldn't figure out on own.

interview-query-solutions/Hard/09-page-recommendations.sql

@@ -0,0 +1,23 @@
+SELECT   ps.page_id as page, 
+         ps.postal_code, 
+         1.0 * COUNT(CASE WHEN ps.postal_code = u.postal_code THEN r.user_id ELSE null END) / 
+               COUNT(r.user_id) as percentage
+FROM     page_sponsorships ps JOIN recommendations r ON ps.page_id=r.page_id 
+                              JOIN users u ON r.user_id=u.id
+GROUP BY 1, 2
+ORDER BY 1
+
+
+-- my attempt:
+
+SELECT   ps.page_id, ps.postal_code, ROUND(1.0 * COUNT(user_id) / COUNT(*) OVER(), 1) as percentage
+FROM     page_sponsorships ps JOIN recommendations r ON ps.page_id=r.page_id 
+                              JOIN users u ON r.user_id=u.id and ps.postal_code=u.postal_code
+GROUP BY 1, 2
+
+
+
+-- NOTE: 
+
+-- was using postal_code condition in the join itself, which was a mistake. 
+-- also had to drop ROUND() function to pass second test case.