DEV Community: VARUN

CA 32 - Filter Assignments

VARUN — Sun, 29 Mar 2026 17:32:51 +0000

Get all movies (films) that have a rental rate greater than $3.

SELECT title, rental_rate FROM film WHERE rental_rate > 3;

Here, we need to fetch the title and rental_rate from film table and it must fetch the data which is having the rental_rate > 3 so we use the WHERE rental_rate > 30 in the condition.

Get all movies that have a rental rate greater than $3 and a replacement cost less than $20.

SELECT title, rental_rate, Replacement_cost FROM film WHERE rental_rate >3 and replacement_cost < 20;

It is same as the previous promblem in extra we fetch a replacement_cost and using the condition replacement_cost < 20.

Get all movies that are either rated as 'PG' or have a rental rate of $0.99.

SELECT rating, rental_rate FROM film WHERE rating = 'PG' or rental_rate = 0.99;

Here, we need to fetch the rating and rental_rate from film table and with the condition rating = 'PG' or rental_rate = 0.99.

Show the first 10 movies sorted by rental rate (highest first).

SELECT rental_rate FROM film ORDER BY rental_rate DESC LIMIT 10;

Here, we need to fetch rental_rate from film table and with the condition of top 10 highest value so we are using decsending order and LIMIT 10 to the top 10 highest value.

Skip the first 5 movies and fetch the next 3 sorted by rental rate in ascending order.

SELECT title,rental_rate FROM film ORDER BY rental_rate ASC OFFSET 5 LIMIT 3;

Here, we need to fetch the rental_rate data from film table and to skip the first five and to give the next 3 data sorted in ascending order.

Get all movies with a rental duration between 3 and 7 days.

SELECT title, rental_duration FROM film WHERE rental_duration BETWEEN 3 and 7;

Here, we are fetching the data from title and rental_duration FROM film and that rental_duration must be between 3 and 7.

Get all movies where the title starts with 'A' and ends with 'e'

SELECT title FROM film WHERE title LIKE 'A%e';

Here, we need to get the title of movie start with 'A' and ends with 'e' from the film table.

Find all movies where the title contains "Man", "Men", or "Woman".

SELECT title FROM film WHERE title LIKE '%Man%' OR title LIKE '%Men%' OR title LIKE '%Woman%';

Here, we need to get the title of the movie which contain "MAN","MEN"and "WOMEN" so we used the LIKE'%Man%', LIKE'%Men%' and LIKE'%Woman%'

Find all movies where the special features are not listed (i.e., special_features is NULL).

SELECT * FROM film WHERE special_features IS NULL;

Here we use IS NULL to find movies with no special feature.

Find the first 10 movies with a rental rate of $2.99 or $4.99, a rating of 'R', and a title containing the word "L".

SELECT * FROM film WHERE rental_rate IN (2.99, 4.99) AND rating = 'R' AND title LIKE '%L%' LIMIT 10;

CA 31 - Select Queries from DVD Rental database

VARUN — Sun, 29 Mar 2026 15:12:37 +0000

Retrieve film titles and their rental rates. Use column aliases to rename title as "Movie Title" and rental_rate as "Rate".

SELECT title AS "Movie Title",rental_rate AS "Rate"FROM film;

Here,title is renamed into Movie Title amd rental_rate is renamed into Rate from film table and to show the output.

List customer names and their email addresses. Alias first_name and last_name as "First Name" and "Last Name".

SELECT first_name AS "First Name",last_name AS "Last Name",email FROM customer;

Here, first_name is ranamed into First Name and last_name as Last Name and fetch the email from the customer table.

Get a list of films sorted by rental rate in descending order. If two films have the same rental rate, sort them alphabetically by title.

SELECT title,rental_rate FROM film ORDER BY rental_rate DESC,title ASC;

Here, title and rental_rate to be fetch from film table and to order the rental_rate to descending order and title as ascending order.

Retrieve actor names sorted by last name, then first name.

SELECT first_name,last_nameFROM actor ORDER BY last_name ASC,first_name ASC;

Here, the first_name and last_name to be fetch from actor table and to order the last name and first name in asccending order.

List all unique replacement costs from the film table.

SELECT DISTINCT replacement_cost FROM film;

Here, the we need to fetch the unique replacement_cost from the film table so we used DISTINCT in the query.

List all films' title and length in minutes. Alias length as "Duration (min)".

SELECT title,length AS "Duration (min)"FROM film;

Here, from film table the title and the length data to be fetch and to rename the length as Duration(min) so we use length AS "Duration (min)".

Retrieve customer first and last names along with their active status. Alias active as "Is Active".

SELECT first_name,last_name,active AS "Is Active" FROM customer;

Here,the first_name, last_name and active from customer table and to rename the active to Is Active so we use active AS "Is Active".

Retrieve the list of film categories sorted alphabetically.

SELECT name from category order by name ASC;

Here, to take the list of film name from caregory and to sort the film name in ascending order.

List films by length, sorted in descending order. Include only the title and length.

SELECT title lenght FROM film ORDER BY title DESC,length DESC;

Here, list all the title and length from film and to order the title and lenght by descending order.

Retrieve all actor names, sorted by their first name in descending order.

SELECT first_name FROM actor ORDER BY first_name DESC;

Here,from actor table the first_name to be fetch and to order it by descending order.

List all unique ratings available in the film table.

SELECT DISTINCT rating FROM film;

Here,for the unique rating we are using DISTINCT to get unique rating from film table.

Find all unique rental durations from the film table.

SELECT DISTINCT rental_duration FROM film;

Here,for the unique rental_duation we are using DISTINCT to get unique rental_duration from film table.

Retrieve the first unique customer ID based on active status. Include the customer_id and active columns, and order by customer_id.

SELECT DISTINCT ON (active)customer_id,active FROM customer ORDER BY active,customer_id;

Here,we are fetching the data from customer which must give unique customer_id which must be active that means this query gives the customer_id which are in active.

List the earliest rental date for each customer. Include customer_id and rental_date, and order by customer_id.

SELECT customer_id,MIN(rental_date) AS rental_date FROM rental GROUP BY customer_id ORDER BY customer_id;

Here, we need to fetch the earliest rental data so we are using MIN(rental_date) and also including the customer_id and rental_date which must be in the order of customer_id from rental table.

List the 10 shortest films by length. Include the title and length.

SELECT title,length FROM film LIMIT 10;

Here, we need to fetch the title, length from the film table and it should be shortest of 10 films so using LIMIt 10.

Retrieve all unique values of store_id from the inventory table.

SELECT DISTINCT store_id FROM inventory;

Here, we need to fetch unique store_id from inventory table so we are using DISTINCT store_id.

Find all unique replacement_cost values in the film table. Sort the results in ascending order.

SELECT DISTINCT replacement_cost FROM film ORDER BY replacement_cost ASC;

Here, we need to select unique replacement_cost from film table and that cost should be in ascending order, for unique replacement_cost we are usihg DISTINCT replacement_cost in the query.

CA 30 - Basic Select SQL Queries

VARUN — Thu, 26 Mar 2026 18:21:25 +0000

The questions is to query all the attributes in city table so i us
[SELECT * from city]
*->denotes all the attribute(columns) for every row
from->from the table city.

The question is the query only the names of japanese city from the city by using the countrycode JPN.
[select name from city where countrycode = 'JPN';]
name->they only names of japanese city from city table
where->show the condition where the query to fetch the name(only japanese so they gave countrycode in the question) by using the country code we can take japanese name

This question is same as the previous one, here they ask american cities by using countrycode = 'USA' and also with the population > 100000
[select * from city where population>100000 and countrycode='USA';]

In this question they ask all city names from the table station but the condition is do not start with vowels and not to repeat duplicates.
for not repeat duplicate we use distinct and not include the words starting with vowels so we use NOT LIKE

It is same as 3rd problem in the blog.Here, the difference is population > 120000 and all other query are same.

This is same as the 2nd question in the blog there is no difference between these two questions.

In this question they ask the query all the city with ID=1661 from the CITY table.
[select * from city where ID=1661;]

Here, they ask to query the list of CITY and STATE from the table STATION.so the query is
[Select city,state from station;]

BACKEND-DAY:01-CLASS BLOG

VARUN — Wed, 25 Mar 2026 17:48:59 +0000

Write a blog on IP address and Subnet

VARUN — Mon, 23 Mar 2026 15:32:15 +0000

Internet: It means a small group networks connected together.
Example: city-A,city-B,city-C,....etc are connected by roads
working:
send request from your device and goes through router across the network and reaches the server and get response.

IP - Internet Protocol
it is unique for very device
eg: like a mobile number

types of ip:
1.public ip :
2.private ip : college network...(wifi)

eg: 192.168.1.10
__..._

Subnet: breaking a big network into small parts.
IP:192.168.1.10
Subnet: 255.255.255.0
network id:192.168.1
host id:10

subnets:192.168.1.1
192.168.1.2
192.168.1.3
.
.
.
.
.

CA 15 - Merge Two Linked List

VARUN — Sun, 22 Mar 2026 09:50:33 +0000

Problem
You are given two sorted linked lists:
list1
list2
Merge them into one sorted linked list.

Correct Idea (Two Pointers)
create new nodes.
relink existing nodes.
Core Logic:
Compare current nodes of both lists
Attach the smaller one
Move forward
Repeat

CA 07 - Kth Smallest

VARUN — Sun, 22 Mar 2026 09:44:27 +0000

Problem Statement
Given:
An array arr[]
An integer k
Find the k-th smallest element.
Example:
Input: arr = [10, 5, 4, 3, 48, 6], k = 4
Output: 5

Approach Max Heap (Priority Queue)
Idea:
Keep only k smallest elements
Use a max heap of size kHow It Works

Example:

arr = [10, 5, 4, 3, 48, 6]
k = 4

Steps:

Keep pushing into heap
Maintain only 4 smallest elements
Final heap contains smallest 4
Top = answer

If a smaller element comes replace the largest in heap

CA 06 - Find the Maximum and Minimum Element in the Array

VARUN — Sun, 22 Mar 2026 09:38:28 +0000

Problem Statement
Given an array, find:

The minimum element
The maximum element

Example:
Input: [1, 4, 3, 8, 6]
Output: [1, 8]

Approach
You only need one loop.

Idea:
Assume first element is both min and max
Traverse the array
Update values when needed

Step-by-Step Example
Input:
[1, 4, 3, 8, 6]
Process:
Start → min = 1, max = 1
4 → max = 4
3 → no change
8 → max = 8
6 → no change
Final:
[1, 8]

CA 05 - Reverse the array

VARUN — Sun, 22 Mar 2026 09:32:37 +0000

Reverse an Array — The Simplest Problem You’re Still Overthinking

Problem:
Given an array, reverse it.

That means:

First → Last
Second → Second Last
and so on ([GeeksforGeeks][1])

Example:
Input → [1, 4, 3, 2, 6, 5]
Output → [5, 6, 2, 3, 4, 1] ([GeeksforGeeks][2])

Code (Clean & Correct)

class Solution:
def reverseArray(self, arr):
left = 0
right = len(arr) - 1

    while left < right:
        arr[left], arr[right] = arr[right], arr[left]
        left += 1
        right -= 1

CA 08 - Sort 0s, 1s, and 2s

VARUN — Sun, 22 Mar 2026 08:43:09 +0000

Problem:
Given an array containing only 0s, 1s, and 2s, sort it without using any built-in sort function.

Solution: Dutch National Flag Algorithm
We divide the array into three regions:

Left → all 0s
Middle → all 1s
Right → all 2s We maintain 3 pointers:
low → next position for 0
mid → current element
high → next position for 2

💡 Algorithm:

If current is 0 → swap with low, move both forward
If current is 1 → move mid
If current is 2 → swap with high, move high backward

Code:
python
class Solution:
def sort012(self, arr):
low, mid = 0, 0
high = len(arr) - 1

    while mid <= high:
        if arr[mid] == 0:
            arr[low], arr[mid] = arr[mid], arr[low]
            low += 1
            mid += 1
        elif arr[mid] == 1:
            mid += 1
        else:
            arr[mid], arr[high] = arr[high], arr[mid]
            high -= 1

CA 09 - Move All Negative Elements to end

VARUN — Sun, 22 Mar 2026 08:35:31 +0000

Problem:
Keep order of positives SAME
Keep order of negatives SAME
Move all negatives to the end

This means:

You need a stable partition
NOT swapping randomly like quicksort

Correct Approach (Simple & Clean)
Idea:
First collect all positive numbers
Then collect all negative numbers
Put back into array

Example

Input:
[1, -1, 3, 2, -7, -5, 11, 6]
Step:
positives → [1, 3, 2, 11, 6]
negatives → [-1, -7, -5]
Final:
[1, 3, 2, 11, 6, -1, -7, -5]

CA 10 - Kadanes Algorithm

VARUN — Sun, 22 Mar 2026 08:23:04 +0000

The Problem
Given an array of integers.
The Task is to find the maximum sum of a contiguous subarray.

Example:
Input: [2, 3, -8, 7, -1, 2, 3]
Output: 11

Idea
If your current sum becomes negative:
Drop it immediately
Start fresh from next element

Algorithm Logic
At each index:

current_sum = max(arr[i], current_sum + arr[i])
max_sum = max(max_sum, current_sum)

Example

Array:
[2, 3, -8, 7, -1, 2, 3]
Step-by-step:

Start with 2 → current = 2
Add 3 → current = 5
Add -8 → current = -3 ❌ (bad → drop)
Start fresh at 7 → current = 7
Add -1 → current = 6
Add 2 → current = 8
Add 3 → current = 11 ✅

answer = 11

Python Code
class Solution:
def maxSubarraySum(self, arr):

    max_sum = arr[0]
    current_sum = arr[0]

    for i in range(1, len(arr)):
        current_sum = max(arr[i], current_sum + arr[i])
        max_sum = max(max_sum, current_sum)

    return max_sum

All numbers are negative:
[-2, -4]
Answer = -2 (not 0)
max_sum = arr[0]