sql6: solution to problems 1, 2, 4

01-GamePlayAnalysis-II.sql

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+-- Problem 1 : Game Play Analysis II	(https://leetcode.com/problems/game-play-analysis-ii/)
+
+select player_id, device_id 
+from 
+(select player_id, device_id, min(event_date) from activity group by player_id) as a

02-GamePlayAnalysis-III.sql

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+select player_id, event_date, sum(games_played) 
+over (PARTITION BY player_id ORDER BY event_date) 
+as games_played_so_far from activity
+
+-- this will not work if we do not have order by event_date because the sum will be 
+-- calculated for all rows in the partition, it will return each row with full sum (12 for player_id 1)
+
+
+-- using subquery
+select a1.player_id, a1.event_date, 
+       (select sum(a2.games_played) 
+        from activity a2 
+        where a2.player_id = a1.player_id 
+          and a2.event_date <= a1.event_date) as games_played_so_far
+from activity a1
+
+-- subquery is not efficient, it will be executed for each row in the outer query
+-- so it takes O(n^2) time complexity, it will take 1st row then goes to the subquery 
+-- and match the player_id and event_date and run the sum

03-ShortestDistanceInAPlane.sql

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+-- Problem 3 : Shortest Distance in a Plane		(https://leetcode.com/problems/shortest-distance-in-a-plane/)
+
+select round(min(sqrt(pow(p1.x-p2.x,2) + pow(p1.y-p2.y,2))),2) as 'shortest' 
+from point2D p1
+inner join point2D p2
+on (p1.x < p2.x OR (p1.x = p2.x AND p1.y < p2.y))
+-- to my understanding: forward comparing, make p1x < p2x then include this
+-- p1x = p2x, then compare y — include only if p1y < p2y
+
+select round(min(sqrt(pow(p1.x-p2.x,2) + pow(p1.y-p2.y,2))),2) as 'shortest' 
+from point2D p1
+inner join point2D p2
+on (p1.x, p1.y) < (p2.x, p2.y)
+-- doesn't run on MySQL?!

04-CombineTwoTables.sql

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+-- Problem 4 : Combine Two Tables	(https://leetcode.com/problems/combine-two-tables/)
+
+select firstName, lastName, city, state
+from person p
+left join address a 
+on p.personId = a.personId

05-CustomerWithStrictlyIncreasingPurchases.sql

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+-- Problem 5 : Customers with Strictly Increasing Purchases	(https://leetcode.com/problems/customers-with-strictly-increasing-purchases/)
+
+with cte as (
+    select customer_id, year(order_date) as 'year', sum(price) as price 
+    from orders
+    group by year, customer_id
+)
+select customer_id
+from cte c1
+left join cte c2 on c1.customer_id=c2.customer_id 
+and c1.year+1 = c2.year 
+and c1.price < c2.price
+group by c1.customer_id
+having count(*) - count(c2.customer_id) = 1