DEV Community

Solving a Job Application Code Challenge

Ben Greenberg on October 16, 2017

As I discussed in an earlier post, I recently graduated from the Flatiron School's online immersive full stack bootcamp. For the past few weeks I h...

Read full post

Ryan Palo • Oct 16 '17

Do you have the big long string of text? You've got me hooked and I want to solve it, but it seems like it would be more fun to actually get the secret word at the end :)

Ben Greenberg • Oct 16 '17 • Edited

Hey Ryan,

Not sure why, but my paste of the long string in this comment did not come in correctly as was mentioned below. You can find it on repl.it at repl.it/MivX.

Enjoy! :)

Ben

Ryan Palo • Oct 17 '17

Victory! I found an interesting performance quirk, at least for Ruby. Thanks for a great puzzle and write-up!

Ben Greenberg • Oct 17 '17 • Edited

Your solution in Ruby is awesome. I also love the dedication to benchmarking your two different variations and very interesting results on .count and it's optimization. I didn't know that before. Overall, Ruby is just plain awesome.

Ryan Palo • Oct 17 '17

Thanks! Yeah, I was super sure the way you did it would be faster. Ruby is one of my favorite things!

Richard Orelup • Oct 16 '17

Running against the blob above it seems the answer is a word followed by some misc character (the last 4 almost spelling out an f-bomb). Did I get the right answer or is something off? I did a codepen with your code and it seems to give the same answer.

Ben Greenberg • Oct 16 '17 • Edited

For those interested, I put this up on repl.it. I'm not sure why but my paste in the comment above of the very long string did not paste correctly. The repl.it has the correct long string: repl.it/MivX

Miguel Ángel García • Oct 18 '17

My solution. In python.

from collections import Counter

with open('test.input') as fd:  # the string is in the "test.input" file
    c = Counter(fd.read())

string = ''.join(sorted(c.keys(), key=lambda x: -1 * c[x]))
print(string[:string.index('_')])

Marco Alka • Oct 17 '17 • Edited

Probably not the most elegant way... I really need more practice with Rust :D

fn main() {
    let blob = "...";

    let mut result = blob
        .chars()
        .fold(vec![(None, 0); 27], |mut counter, c| {
            if c == '_' {
                if counter[26].0.is_none() { counter[26].0 = Some('_'); }
                counter[26].1 += 1;
            }
            else {
                let pos = (c as u8 - 'a' as u8) as usize;

                if counter[pos].0.is_none() { counter[pos].0 = Some(c); }
                counter[pos].1 += 1;
            }

            counter
        })
    ;

    result.sort_by(|a,b| b.1.cmp(&a.1));
    let result = result
        .iter()
        .fold(String::new(), |mut s, c| { s.push(c.0.unwrap_or(' ')); s })
    ;

    let result: Vec<&str> = result
        .split('_')
        .collect()
    ;

    println!("{}", result[0]);
}

Matt Weingaertner • Oct 17 '17 • Edited

Great Post! Thanks!

My C# approach would be something like this:

static void Main(string[] args) {
    var sortedString = longString
        .OrderBy(x => x.Equals('_') ? 2 : 1 )
        .ThenBy(x => x)
        .ToArray(); 
    Console.WriteLine(sortedString); 
}

Matt Weingaertner • Oct 17 '17

The problem with fast shooting, is always the same, mistakes! :/

static void Main(string[] args) {
    var keyString = new string(longString
        .OrderByDescending(x => longString.Count(c => c.Equals(x)))
        .Distinct()               
        .ToArray());             
    Console.WriteLine(keyString.Split('_')[0]); 
}

Peter Kim Frank • Oct 16 '17

@benhayehudi this is a great write-up. Really highlights the way you approached and then solved the problem. As a Javascript novice myself, I appreciated the step-by-step explanation of the decisions you made.

How did this coding challenge compare to others you've taken?

Ben Greenberg • Oct 16 '17

Thanks Peter!

This coding challenge ranked as a bit more involved for an initial challenge, i.e. as a requirement for inclusion with a resume submission and not something to do after an initial phone screen. I think it's actually a smart way for a place to be more upfront about their needs and skills required before taking up too much time of the people interested and their own time.

darkdoc • Oct 18 '17

Hey!
Can someone check my python solution? Any hints where/how can the code be better?

repl.it/Mnem/0

Ryan Palo • Oct 18 '17 • Edited

Hi! A couple of notes that you can take or leave.

You don't necessarily have to loop through and initialize each key for strngs manually. Here's a couple of alternatives.

# Your version
strngs = {}
for char in initchar:
    strngs[char] = 0

# you can create a dict out of a sequence of keys and starting value
# credit: @edh_developer
strngs = dict.fromkeys(initchar, 0)

# you can also use a defaultdict, which creates the key with a default value
# if it doesn't find the key there on access
from collections import defaultdict
strngs = defaultdict(int) # defaults to zero

You can use shorthand notation if you're modifying a value inplace.

# Your version
strngs[char] = strngs[char] + 1

# shortcut
strngs[char] += 1

You can use dict.items to get a list of pairs out of a dict that is much easier to sort. Then you don't have to manually sort and delete items out of your dict.

# Your version
while (strngs != {}):
    maximum = max(strngs, key=strngs.get)
    word = word + maximum
    del strngs[maximum]

# Could be
pairs = strngs.items()
desc = sorted(pairs, key=lambda x: x[1]) # the key= argument lets us
                                        # sort by the second item
desc_letters = ''.join(x[0] for x in desc) # Grab the first value (letter) from each pair

Some things to look up here for more information:

In general, when you're building a string, it's usually more performant, easier, and less error prone to build the whole thing as a list and join it at the end.

# Instead of this:
word = ""
for item in things:
    word = word + complicated_process(item)

# Since strings are immutable, this forces you to throw away
# your old string and create a new one *each time* which is expensive

# More efficient:
word  = []
for item in things:
    word.append(complicated_process(item))

letters = ''.join(word)

# Use list comprehensions (see link below) for even faster,
# cleaner, more pythonic code :)

word = [complicated_process(item) for item in things]
letters = ''.join(word)

# Use a generator expression (like above) for *even better* results!
letters = ''.join(complicated_process(item) for item in things)

Here's a link on list comprehensions

Hopefully this helps. I'm not sure how comfortable you are with Python, so I tried to cover everything. Let me know if you have any questions 😁

Dimitar Venkov • Oct 18 '17 • Edited

Here's a py3 version using the builtin dict and iterators:

from operator import itemgetter
long_string = "..."

counter = dict()
for c in iter(long_string):
counter[c] = counter.get(c, 0) + 1

sorted_chars = [k for (k, v) in sorted(counter.items(), key=itemgetter(1), reverse=True)]
word = "".join(sorted_chars[:sorted_chars.index('_')] )

print(word)