diff --git a/01-TheNumberOfSeniorsAndJuniorsToJoinTheCompany.sql b/01-TheNumberOfSeniorsAndJuniorsToJoinTheCompany.sql
new file mode 100644
index 0000000..0c8673f
--- /dev/null
+++ b/01-TheNumberOfSeniorsAndJuniorsToJoinTheCompany.sql
@@ -0,0 +1,12 @@
+-- Problem 1 : The Number of Seniors and Juniors to Join the Company	(https://leetcode.com/problems/the-number-of-seniors-and-juniors-to-join-the-company/)
+
+with cte as(
+select *, sum(salary) over (partition by experience order by salary, employee_id) as 'rsum'
+from candidates)
+select 'Senior' as experience, count(*) as 'accepted_candidates'
+from cte
+where experience='Senior' and rsum <= 70000
+union
+select 'Junior' as experince, count(*) as 'accepted_candidates'
+from cte 
+where experience = 'Junior' and rsum <= (select 70000 - ifnull(max(rsum),0) from cte where experience='Senior' and rsum <= 70000)
\ No newline at end of file
diff --git a/02-LeagueStatistics.sql b/02-LeagueStatistics.sql
new file mode 100644
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--- /dev/null
+++ b/02-LeagueStatistics.sql
@@ -0,0 +1,58 @@
+-- Problem 2 : League Statistics		(https://leetcode.com/problems/league-statistics/ )
+
+with cte as (
+select home_team_id as team_id, home_team_goals as goal_for, away_team_goals as goal_against
+from matches 
+union all 
+select away_team_id as team_id, away_team_goals as goal_for, home_team_goals as goal_against
+from matches)
+select team_name, 
+count(cte.team_id) as matches_played, 
+sum(
+    case 
+        when goal_for > goal_against then 3
+        when goal_for = goal_against then 1
+        else 0
+    end
+)
+ as points, 
+sum(goal_for) as goal_for, sum(goal_against) as goal_against, 
+sum(goal_for) - sum(goal_against) as goal_diff
+from teams
+join cte on teams.team_id=cte.team_id
+group by cte.team_id
+order by points desc, goal_diff desc, team_name
+
+
+/*
+just for my future reference! i could come up with the following solution at 1st which works but is too verbose and 
+unreadable. BUT I looked at the following solution and generated the above. so, it is a good reference for future
+*/
+with matches_mod as(
+select *, 
+(case 
+when home_team_goals > away_team_goals then 3
+when home_team_goals = away_team_goals then 1
+else 0
+end) as
+home_team_points, 
+(case 
+when home_team_goals < away_team_goals then 3
+when home_team_goals = away_team_goals then 1
+else 0
+end)
+away_team_points
+from matches),
+cte as (
+select home_team_id as team_id, home_team_points as points, home_team_goals as goal_for, away_team_goals as goal_against
+from matches_mod 
+union all 
+select away_team_id as team_id, away_team_points as points, away_team_goals as goal_for, home_team_goals as goal_against
+from matches_mod )
+
+select team_name, count(cte.team_id) as matches_played, sum(points) as points, sum(goal_for) as goal_for, sum(goal_against) as goal_against, 
+sum(goal_for) - sum(goal_against) as goal_diff
+from teams
+join cte on teams.team_id=cte.team_id
+group by cte.team_id
+order by points desc, goal_diff desc, team_name
diff --git a/03-SalesPerson.sql b/03-SalesPerson.sql
new file mode 100644
index 0000000..f18c849
--- /dev/null
+++ b/03-SalesPerson.sql
@@ -0,0 +1,21 @@
+-- Problem 3 : Sales Person (https://leetcode.com/problems/sales-person/ )
+
+-- using cte
+with cte as (
+select order_id, o.com_id, name as c_name, sales_id
+from company c
+join orders o on c.com_id = o.com_id
+)
+select sp.name
+from salesperson sp 
+left join cte c on sp.sales_id=c.sales_id
+group by sp.sales_id
+having sum(case when c.c_name='RED' then 1 else 0 end) = 0 
+
+-- more straight forward
+select name from salesperson 
+where sales_id not in (
+select sales_id
+from company c
+join orders o on c.com_id = o.com_id
+where c.name = 'RED')
\ No newline at end of file
diff --git a/04-FriendRequest-II.sql b/04-FriendRequest-II.sql
new file mode 100644
index 0000000..ccfe7d2
--- /dev/null
+++ b/04-FriendRequest-II.sql
@@ -0,0 +1,27 @@
+-- Friend Requests II (https://leetcode.com/problems/friend-requests-ii-who-has-the-most-friends/ )
+
+-- using cte 
+with cte as (
+    select requester_id as id from RequestAccepted
+    union all
+    select accepter_id as id from RequestAccepted
+),
+friend_count as (
+    select id, count(id) as num 
+    from cte 
+    group by id
+)
+select id, num
+from friend_count
+where num = (select max(num) from friend_count) -- it will also work if there are multiple people with max friends
+
+-- another appraoch
+select id, count(id) as num 
+from (
+    select requester_id as id from RequestAccepted
+    union all 
+    select accepter_id as id from RequestAccepted
+) as temp
+group by id
+order by num desc 
+limit 1 -- works because only one person has the most friends       
\ No newline at end of file