HackerRank Set pop(), remove(), and discard() Python Solution | In 7 Step-by-Step Explanation

HackerRank Python Series: Working Through the Set discard(), remove(), and pop() Challenge

Hi there! Welcome back to our HackerRank Python Series! I’m looking forward to talking about the Set discard(), remove(), and pop() challenge today. I taught Python for a while, and to be honest, sets really confused me at first. They are useful, but they have some strange things about them. I’d like to walk you through this problem step by step, as if we were working on it together in class, until it makes sense to you. Does that sound good? Let’s get started.

What’s This Problem All About?

Okay, so you start with a set of numbers and then use these commands to change them. You have to use those commands and add up the things that are left. It’s a good way to practice with Python sets because they are great for things that are unique – there are no duplicates to worry about. The three main methods are pop(), remove(), and discard(). They all change the set in different ways, so it’s important to know when to choose which one.

Here’s a little table I’d scribble on a whiteboard if we were in class:

Each method’s got its own vibe, and we’ll see how they work in the code.

Let’s look at the code.

I’ll go over this step by step, as if we’re writing the code together, so that everything makes sense. Picture us working this out with a notebook nearby and writing down a few notes.

Step 1: Getting the Number of Elements

s = int(input())

This part is pretty basic – it asks for the count of numbers we’ll use. The input() takes what you enter as text, and int() changes it to a number.

Step 2: Making the Set

n = set(map(int, input().split()))

We’re getting the numbers now. What if you type “1 2 3 2”? The input(). split() breaks it up into pieces, like [‘1’, ‘2’, ‘3’, ‘2’]. Then, map(int, …) changes them into numbers, and set() puts them in a set, getting rid of any that are the same. It turns into {1, 2, 3}. It’s always cool how sets handle duplicates like that. It really helps.

Step 3: Looping Through Commands

for i in range(int(input())):

    cmd = input().split()

This part uses int(input()) to get the number of commands, which tells us how many loops to do. For each one, cmd = input(). split() takes a command, like “pop” or “discard 5,” and makes a list out of it, like [‘discard’, ‘5’]. It feels like going through a list of things to do one at a time.

Step 4: Using pop()

if cmd[0] == 'pop':

        n.pop()

To remove a random element, we execute n.pop() when the command is “pop.” You never know which one to pick from a shuffled bunch.

Step 5: Using remove()

  elif cmd[0] == 'remove':

        n.remove(int(cmd[1]))

To focus on a particular number from the command for “remove,” we utilize n.remove(int(cmd[1])). Although it is perfectly accurate, you will encounter a KeyError if that number is not in the set. Before I started looking at what was actually in the set, I used to make a lot of mistakes with that.

Step 6: Using discard()

elif cmd[0] == 'discard':

        n.discard(int(cmd[1]))

“Discard” is like the relaxed sibling of remove. It tries to delete the number with n.discard(int(cmd[1])), but if it’s not there, it just shrugs and keeps going. I wish I’d leaned on discard() more when I was starting out – it would’ve saved me some frustration.

Step 7: Adding Up the Leftovers

print(sum(n))

Lastly, print() shows the sum(n), which adds up all of the remaining items in the set. A straightforward sum of the remaining components is the outcome that HackerRank is aiming for.

The Complete Code All at Once
The complete code is provided here, along with some comments to help with comprehension:

s = int(input())

n = set(map(int, input().split()))

for i in range(int(input())):

    cmd = input().split()  

    if cmd[0] == 'pop':

        n.pop()  

    elif cmd[0] == 'remove':

        n.remove(int(cmd[1]))  

    elif cmd[0] == 'discard':

        n.discard(int(cmd[1]))  

print(sum(n))

Things to Keep in Mind

I’m going to wrap up a class before you leave, so let’s talk about the important things:

• pop(): Removes a random element by grabbing it. It’s fast, but if the set is empty, it will fail.
• remove(x): Aims for a particular element. Works perfectly, but if the element is absent, it fails.
• discard(x): Likewise aims for a particular element, but disregards its absence. My go-to method for avoiding trouble is this.

Quick Tip: Not sure if an element’s in the set? Stick with discard(). It’s like having a backup plan when your code’s acting up.

Wrapping It Up

Good job on finishing the HackerRank Set discard(), remove(), and pop() challenge! We have thoroughly examined how sets work, explained the distinctions between these approaches, and developed a solution that effectively handles the commands.

Sets might seem strange initially – I sure had trouble with them – but you’ve come a long way. This knowledge will be useful for other coding tasks and actual projects, too.

Related Post:

HackerRank Python Solution – Using itertools combinations_with_replacement | In 5 Easy Steps

Incorrect RegeX Challenge on HackerRank: 5-Step Python Regex Validation Tutorial

Python Official Documentation for Sets:

URL: https://docs.python.org/3/library/stdtypes.html#set

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