Microblog - Create a new non-root user with sudo privileges on Ubuntu based DigitalOcean Droplet configured with SSH

Vatsal Mistry — Sat, 30 May 2020 18:44:22 +0000

This post will walk you through steps of creating a new non root user and configure sudo privileges with it. This post not just focuses on DigitalOcean Droplet but rather can be used for any linux based system.

Well, so you've spin up your Digital Ocean droplet and configured it with your SSH and have a root user to login from your terminal or bash.

But it is not a good option to use the root user, as the root user has the maximum level of access and rights, which might accidentally cause some problem. So a better choice is to create a new user.

Let's follow through the steps to create a new user and assign sudo privileges to it.

Login to your droplet using terminal or bash or PuTTY(Windows).
```
ssh root@<ip-address-of-your-droplet>
```
Create a new user, <new-user> (Replace <new-user> with your desired username).
```
adduser <new-user>
```
Congratulations, 😃 . You've created a new user. You can verify it using,
```
id <new-user>
```
Let us add the <new-user> to the sudo group to get the sudo privileges.
```
usermod -aG sudo <new-user>
```
Now let us add SSH key for the new user so that next time we can connect to droplet using <new-user> using SSH. Switch the user from root to using,
```
su - <new-user>
```
Create a directory .ssh into home dir. Change permissions for that dir. Follow the commands,
```
sudo mkdir ~/.ssh
sudo chmod 700 ~/.ssh
```

Create a file named authorized_keys and paste your SSH key in there. Also change the permission for it.

sudo vim ~/.ssh/authorized_keys
Paste the key and save the file.
sudo chmod 600 ~/.ssh/authorized_keys

Note: If you want to know how to generate SSH keys follow this doc from Git here.

Congratulations, 😃 . You've successfully created a non-root <new-user> with sudo privileges to work with. You can now login your droplet using,

```bash
ssh <new-user>@<ip-address-of-your-droplet>
```

Thanks for reading through.
Comment down if you have any doubts.
Like if you find this post useful. I welcome your opinions and constructive criticism. You can find me on twitter or my website.
Cheers!

How easy type conversion in Python helped me! A solution to problem from Code Jam 2019.

Vatsal Mistry — Thu, 18 Apr 2019 19:03:49 +0000

Foregone Solution — A problem from Google Code Jam's Qualifier Round this year felt like so easy to code.

You can checkout the full problem description at this LINK.

Problem(Source : Google Code Jam 2019 Website)

Someone just won the Code Jam lottery, and we owe them N jamcoins! However, when we tried to print out an over sized check, we encountered a problem. The value of N, which is an integer, includes at least one digit that is a 4... and the 4 key on the keyboard of our over sized check printer is broken.

Fortunately, we have a workaround: we will send our winner two checks for positive integer amounts A and B, such that neither A nor B contains any digit that is a 4, and A + B = N. Please help us find any pair of values A and B that satisfy these conditions.

Approach

While reading out the problem description, I read the input constraints and range which were around 10¹⁰⁰. So it was a bad idea to think of a very straight loop-based solution, that would strike the time complexity showing the nightmarish message “ Time limit Exceeded ! ”.

So I thought of dealing this with the help of strings. Python saved me!

Assuming you’ve read the problem description through the link above, there is a number N which may contain the digit 4 with any number of occurrences. The task was to find out two numbers A and B such that A and B doesn’t contain any occurrence of the digit 4 and summation of A and B should equal N(A+B=N).

So I thought of creating a new number let’s assume A for now, that would be same in length as N(length means number of digits in the number). The idea is while constructing the number A, we would analyze the number N digit by digit. For each occurrence of digit 4 in number N, we would add digit 1 in the number A. For rest of the digits in N, add digit 0 in the A. Following this, I got a number A.

I obtained the other number B by subtracting newly constructed A from N. And you see, I got two numbers A and B that do not contain digit 4 and the summation is equal to N.

For implementing this logic, I could not think of anything, but Python. The easy type conversions from integer to string or string to list and viz made it so easy to achieve this working.

Implementation

Code.

My View

This way it cracked all the test cases, no matter how large the input number was!

I believe the same logic can be implemented in various other languages, possibly the best in terms of time would be in C/C++, but Python just makes it super easy. You can switch from different data types with just few keywords and as you want.

I encourage any reader to try out this in some other language or in Python with less lines of code.

" The joy of coding Python should be in seeing short, concise, readable classes that express a lot of action in a small amount of clear code — not in reams of trivial code that bores the reader to death. "

-- Guido Van Rossum

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import antigravity

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Thanks for reading through. I welcome feedback and constructive criticism. I can be reached on Twitter @mistryvatsal11 or through my personal website mistryvatsal.github.io.

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