diff --git a/README.md b/README.md
index 608c120..e07ef7f 100644
--- a/README.md
+++ b/README.md
@@ -60,28 +60,43 @@ ON s.product_id = p.product_id
 
 [1581 - Customer Who Visited but Did Not Make Any Transactions](https://leetcode.com/problems/customer-who-visited-but-did-not-make-any-transactions/)
 ```sql
-SELECT customer_id, COUNT(*) as count_no_trans
-FROM Visits 
-WHERE visit_id NOT IN (SELECT DISTINCT visit_id FROM Transactions)
-GROUP BY customer_id
+select v.customer_id, count(v.visit_id) as count_no_trans
+from Visits v
+left join Transactions t
+On v.visit_id = t.visit_id
+Where t.transaction_id is Null
+group by v.customer_id;
+
+--2nd approach: using Subquery (This approach is not ideal as this problem can be solved using joins)
+SELECT customer_id, COUNT(visit_id) as count_no_trans 
+FROM Visits
+WHERE visit_id NOT IN (
+	SELECT visit_id FROM Transactions
+	)
+GROUP BY customer_id;
 ```
 
 [197 - Rising Temperature](https://leetcode.com/problems/rising-temperature/) 
 ```sql
-SELECT w1.id 
-FROM Weather w1, Weather w2
-WHERE DATEDIFF(w1.recordDate, w2.recordDate) = 1
-AND w1.temperature > w2.temperature
+SELECT 
+    w1.id
+FROM 
+    Weather w1
+JOIN 
+    Weather w2
+ON 
+    DATEDIFF(w1.recordDate, w2.recordDate) = 1
+WHERE 
+    w1.temperature > w2.temperature;
 
--- OR
-SELECT w1.id
-FROM Weather w1, Weather w2
-WHERE w1.temperature > w2.temperature
-AND SUBDATE(w1.recordDate, 1) = w2.recordDate
+-- OR We can use lag() window function to get previous_temperature and record_date columns beside our table and filter it
 ```
 
+
 [1661 - Average Time of Process per Machine](https://leetcode.com/problems/average-time-of-process-per-machine/)
 ```sql
+
+-- 1ST approach is by writing subquery to make temporary table first and then we apply math function
 SELECT machine_id, ROUND(AVG(end - start), 3) AS processing_time
 FROM 
 (SELECT machine_id, process_id, 
@@ -90,28 +105,41 @@ FROM
  FROM Activity 
   GROUP BY machine_id, process_id) AS subq
 GROUP BY machine_id
+
+-- OR this is another approach but this put math expession in one go 
+SELECT 
+    machine_id,
+    round(SUM(CASE WHEN activity_type='start' THEN timestamp*-1 ELSE timestamp END)
+    / (SELECT COUNT(DISTINCT process_id)),3) AS processing_time
+FROM 
+    Activity
+GROUP BY machine_id;
+
 ```
 
 [577 - Employee Bonus](https://leetcode.com/problems/employee-bonus/solutions/)
 ```sql
+-- Note that e.name b.bonus can also be written in select statement
 SELECT name, bonus
 FROM Employee e
 LEFT JOIN Bonus b
 ON e.empId = b.empId
-WHERE bonus < 1000
-OR bonus IS NULL
+WHERE bonus < 1000 OR bonus IS NULL
 ```
 
 [1280 - Students and Examinations](https://leetcode.com/problems/students-and-examinations/)
 ```sql
-SELECT a.student_id, a.student_name, b.subject_name, COUNT(c.subject_name) AS attended_exams
-FROM Students a
-JOIN Subjects b
-LEFT JOIN Examinations c
-ON a.student_id = c.student_id
-AND b.subject_name = c.subject_name
-GROUP BY 1, 3
-ORDER BY 1, 3 
+-- Beautiful and brain storming one cross join is must
+
+select st.student_id, st.student_name, s.subject_name, 
+count(e.student_id) as attended_exams
+
+from Students st
+cross join Subjects s
+left join Examinations e
+On st.student_id = e.student_id and s.subject_name = e.subject_name
+group by 1,2,3
+order by 1,2,3 
 ```
 [570. Managers with at Least 5 Direct Reports](https://leetcode.com/problems/managers-with-at-least-5-direct-reports)
 ```sql
@@ -124,13 +152,6 @@ WHERE id IN
    HAVING COUNT(*) >= 5
   )
 
--- OR
-SELECT a.name
-FROM Employee a
-JOIN Employee b
-WHERE a.id = b.managerId
-GROUP BY b.managerId
-HAVING COUNT(*) >= 5
 ```
 
 [1934. Confirmation Rate](https://leetcode.com/problems/confirmation-rate/)