From 399e9931a8f9a888c79746861a6717028c1825de Mon Sep 17 00:00:00 2001
From: SukeshSrinivas <123338848+SukeshSrinivas@users.noreply.github.com>
Date: Wed, 8 Jan 2025 07:37:36 -0600
Subject: [PATCH 1/2] Done

---
 Problem 1 Report Contiguous Dates.sql | 22 ++++++++++++++++++++++
 1 file changed, 22 insertions(+)
 create mode 100644 Problem 1 Report Contiguous Dates.sql

diff --git a/Problem 1 Report Contiguous Dates.sql b/Problem 1 Report Contiguous Dates.sql
new file mode 100644
index 0000000..7392da4
--- /dev/null
+++ b/Problem 1 Report Contiguous Dates.sql	
@@ -0,0 +1,22 @@
+# Write your MySQL query statement below
+-- Step 1: Create a Common Table Expression (CTE) to compare the current row with the next two rows
+WITH ConsecutiveNumsCTE AS (
+    SELECT 
+        l1.id AS CurrentID,        -- ID of the current row
+        l1.num AS CurrentNum,      -- Number of the current row
+        l2.num AS NextNum,         -- Number from the next row
+        l3.num AS SecondNextNum    -- Number from the second next row
+    FROM Logs l1
+    -- Join the table with itself to get the next row (ID = CurrentID + 1)
+    LEFT JOIN Logs l2
+        ON l1.id = l2.id - 1
+    -- Join the table again to get the second next row (ID = CurrentID + 2)
+    LEFT JOIN Logs l3
+        ON l1.id = l3.id - 2
+)
+
+-- Step 2: Select numbers that appear three times consecutively
+SELECT DISTINCT CurrentNum AS ConsecutiveNums
+FROM ConsecutiveNumsCTE
+-- Filter rows where the current number is equal to both the next and the second next number
+WHERE CurrentNum = NextNum AND NextNum = SecondNextNum;
\ No newline at end of file

From ba8d1a24d01a16ece97a8af055a08105719cd20f Mon Sep 17 00:00:00 2001
From: SukeshSrinivas <123338848+SukeshSrinivas@users.noreply.github.com>
Date: Thu, 9 Jan 2025 14:21:07 -0600
Subject: [PATCH 2/2] Done

---
 Problem2StudentReportbyGeography.sql          | 11 ++++++++++
 ...3 Average Salary Department vs Company.sql | 22 +++++++++++++++++++
 Problem4 Game Play Analysis1.sql              |  2 ++
 3 files changed, 35 insertions(+)
 create mode 100644 Problem2StudentReportbyGeography.sql
 create mode 100644 Problem3 Average Salary Department vs Company.sql
 create mode 100644 Problem4 Game Play Analysis1.sql

diff --git a/Problem2StudentReportbyGeography.sql b/Problem2StudentReportbyGeography.sql
new file mode 100644
index 0000000..1c35320
--- /dev/null
+++ b/Problem2StudentReportbyGeography.sql
@@ -0,0 +1,11 @@
+# Write your MySQL query statement below
+WITH RankedStudents AS (
+    SELECT name, continent, ROW_NUMBER() OVER (PARTITION BY continent ORDER BY name) AS 'rnk' FROM student
+)
+
+SELECT MAX(CASE WHEN continent = 'America' THEN name END) AS America,
+    MAX(CASE WHEN continent = 'Asia' THEN name END) AS Asia,
+    MAX(CASE WHEN continent = 'Europe' THEN name END) AS Europe
+       FROM RankedStudents
+       GROUP BY rnk
+       ORDER BY rnk
\ No newline at end of file
diff --git a/Problem3 Average Salary Department vs Company.sql b/Problem3 Average Salary Department vs Company.sql
new file mode 100644
index 0000000..ed1997a
--- /dev/null
+++ b/Problem3 Average Salary Department vs Company.sql	
@@ -0,0 +1,22 @@
+# Write your MySQL query statement below
+WITH DepAndMonth AS(
+    SELECT DATE_FORMAT(pay_date, '%Y-%m') AS pay_month,
+s.employee_id, s.amount, s.pay_date, e.department_id
+FROM salary s
+LEFT JOIN employee e
+ON s.employee_id = e. employee_id
+),
+AverageSalaries AS (
+    
+    SELECT pay_month, amount, department_id,
+    AVG(amount) OVER (PARTITION BY pay_month) AS 'CompanyAverageSalary',
+    AVG(amount) OVER (PARTITION BY pay_month, department_id) AS 'DepartmentAverage' 
+    FROM DepAndMonth     
+)
+
+SELECT DISTINCTpay_month, department_id, CASE 
+WHEN DepartmentAverage > CompanyAverageSalary THEN 'higher'
+WHEN DepartmentAverage < CompanyAverageSalary THEN 'lower'
+ELSE 'same' END AS comparison
+FROM AverageSalaries
+ORDER BY pay_month
\ No newline at end of file
diff --git a/Problem4 Game Play Analysis1.sql b/Problem4 Game Play Analysis1.sql
new file mode 100644
index 0000000..b3f663b
--- /dev/null
+++ b/Problem4 Game Play Analysis1.sql	
@@ -0,0 +1,2 @@
+SELECT DISTINCT player_id, FIRST_VALUE(event_date) OVER (PARTITION BY player_id ORDER BY event_date) AS 'first_login' FROM ACTIVITY 
+