Ali Spittel

Posted on Nov 17, 2018 • Edited on Oct 28, 2019

Daily Coding Puzzles - Nov 11th - Nov 16th

#codingpuzzle #programming #challenge

Every day on Twitter, I post coding puzzles. These are quick coding challenges that increase in difficulty across the span of the week -- with Monday being the most beginner friendly and Friday being super tough. I love seeing other people's solutions as well, and so people post their solution to the problem in any programming language.

Here's more about them.

I wanted to try posting these here. I'm going to post each question from this week as a comment below, and then we will thread answers under those questions. Please feel free to post your solutions to the ones you are interested in below! Then, you can comment with insights into people's solutions below that! I will also add a meta thread if you have advice on how to format this in the future!

Excited to see your solutions!

Top comments (52)

Ali Spittel • Nov 17 '18

Tuesday - Don't give me five! (7 KYU)

In this kata you get the start number and the end number of a region and should return the count of all numbers except numbers with a 5 in it. The start and the end number are both inclusive!

codewars.com/kata/dont-give-me-five

Abubakar Nur Khalil • Nov 17 '18 • Edited

Quick JS solution

const dgmf = (init, lim) => {
    let count = 0

    for (var i = init-1;i < lim+1;i++) {
        count = String(i).includes('5') ? count : count + 1
        // count = i % 5 != 0 ? count + 1 : count
    }

    return count
}

Lukas Klinzing • Nov 18 '18

This removes more than 5s

Abubakar Nur Khalil • Nov 19 '18

True, just realized 😂

Changing the conditional expression should fix that

// ...
// Loop code
count = String(i).includes('5') ? count : count + 1

E. Choroba • Nov 17 '18

Perl solution, test included:

#! /usr/bin/perl
use warnings;
use strict;

sub count_without_5 {
    my ($from, $to) = @_;
    return scalar grep $_ !~ /5/, $from .. $to
}

use Test::More tests => 2;
is count_without_5(1,9), 8;
is count_without_5(4,17), 12;

Josh Pullen • Nov 17 '18

Any application that can be written in JavaScript, will eventually be written in JavaScript:

const dontGiveMeFive = (min, max) =>
  [...Array(max - min + 1).keys()]
  .map(n => n + min)
  .filter(n => !`${n}`.includes('5'))

Comment deleted

Jessica Castro • Mar 11 '22 • Edited

Cool solution! But this will timeout with large numbers!

Sebastian Martin Dicke • Nov 17 '18 • Edited

Another Haskell solution:

dont_give_me_five :: [Int] -> Int
dont_give_me_five = foldl (\x y -> if notElem '5' $ show y then x + y else x) 0

Or a little bit shorter in the same language:

dont_give_me_five :: [Int] -> Int
dont_give_me_five = foldl1 (\x y -> if notElem '5' $ show y then x + y else x)

Michael Wood • Nov 18 '18

Clojure

(defn no5? [num]
  (not-any? #{\5} (str num)))

(defn dont-give-me-five! [start end]
  (->>
    (range start (inc end))
    (filter no5?)
    count))

user=> (dont-give-me-five! 1 9)
8
user=> (dont-give-me-five! 4 17)
12

David Wickes • Nov 18 '18

Common Lisp

(defun dont-give-me-five (start end)
  (loop for x from start to end
     when (not (contains-5-p x)) collect x))

(defun contains-5-p (n)
  (position #\5 (write-to-string n)))

Paul Cook • Nov 19 '18

Another JS solution:

const dontGiveMeFive = (start, end) => range(start, end).filter(noFive).length;
const range = (start, end) => [...Array(end - start + 1).keys()].map(i => i + start);
const noFive = num => !`${num}`.includes('5');

Thomas Landin • Nov 17 '18

Here's my Nim solution :)

There are filter and keepIf functions defined in the sequtils standard library but that feels like cheating :P

$ is Nim's stringify operator, and the find procedure returns -1 if it doesn't find the substring :)

import strutils # for the 'find' procedure

proc my_filter(start, stop: int): seq[int] =
  for i in start..stop:
    if find($i, "5") == -1:
      result.add(i)

echo my_filter(1,9).len
echo my_filter(4,17).len

Jason Steinhauser • Nov 18 '18

Elixir:

no_fives = fn x, y -> x..y |> Enum.map(& Integer.to_string(&1)) |> Enum.filter(& !String.contains?(&1, "5")) |> Enum.count() end

And surprise! I wrote an F# solution in a similar fashion:

let no_fives x y =
  [x .. y]
  |> Seq.map string
  |> Seq.filter (fun x -> x.Contains("5") |> not)
  |> Seq.length

Sebastian Martin Dicke • Nov 17 '18 • Edited

Haskell

-- The first line is not necessary
dont_give_me_five :: [Int] -> Int
dont_give_me_five list = sum $ filter (\x -> notElem '5' $ show x) list

Ben Halpern • Nov 17 '18 • Edited

Ruby

def dont_give_me_five(start, end)
  (start..end).to_a.reject { |n| n.to_s.include? "5" }.size
end

Ali Spittel • Nov 17 '18

def dont_give_me_five(start, end):
    return len([n for n in range(start, end + 1) if '5' not in str(n)])

Jay • Nov 17 '18

Rust Solution:

fn dont_give_me_5(start: usize, end: usize) -> u8 {
    (start..=end)
        .map(|n| n.to_string())
        .filter(|n| !n.contains('5'))
        .count() as u8
}

Ali Spittel • Nov 17 '18

Wednesday - Format words into a sentence (6 KYU)

Complete the method so that it formats the words into a single comma separated value. The last word should be separated by the word 'and' instead of a comma.

codewars.com/kata/format-words-int...

Arne Mæhlum • Nov 17 '18

This is a fun way of doing it, though it might be a bit confusing at first glance 😂

const format_words = words =>
  words
    .filter(Boolean)
    .join(', ')
    .split('').reverse().join('')
    .replace(' ,', ' dna ')
    .split('').reverse().join('')

Jay • Nov 17 '18

Rust Solution:

fn format_words(words: &[&str]) -> String {
    let words: Vec<&str> = words
        .into_iter()
        .filter(|&w| !w.is_empty())
        .map(|&x| x)    // deref &&str -> &str
        .collect();
      words
        .iter()
        .enumerate()
        .fold(String::new(), |mut sent, (i, w)| {
            if i == 0 {
                sent += w
            } else if i == words.len() - 1 {
                sent += &format!(" and {}", w);
            } else {
                sent += &format!(", {}", w);
            }
            sent
        })
}

Ali Spittel • Nov 17 '18

def format_words(words):
    sentence = ''
    if words:
        words = [word for word in words if word]
        for index, word in enumerate(words):
            if index == 0:
                sentence += word
            elif index == len(words) - 1:
                sentence += " and " + word
            else:
                sentence += ", " + word
    return sentence

Anthony Tomson • Nov 19 '18

My Js solution:

function formatWords(words) {
  if (!words || words.length < 1) return '';

  words = words.filter(val => !!val);
  const length = words.length;

  if (!length) return '';


  return words.reduce((acc, cumul, index) => {
    return index + 1 === length ?
      `${acc} and ${cumul}` :
      `${acc}, ${cumul}`
  });
}

Thomas Landin • Nov 17 '18 • Edited

My Nim solution :)

Nim supports specifying ranges both in absolute terms and in relative terms. ^2 would be the second to last position of a range, etc :) As you can see at the end of the procedure you can also supply just one end of the range and Nim will infer the other end, so filtered[^1] would be just the final element of the sequence.

import sequtils # for filter

proc format_words(words: seq[string]): string =
  result = ""
  if words.len > 0:
    var filtered = filter(words, proc(w: string): bool = w.len > 0)
    for word in filtered[0..^2]:
      result.add(word & ", ")
    return result[0..^3] & (" and " & filtered[^1])

echo format_words(@["ninja","samurai","ronin","leonardo","michelangelo","donatello","raphael"])
echo format_words(@["ninja","samurai","ronin"])
echo format_words(@["ninja","", "ronin"])
echo format_words(@[])

David Wickes • Nov 18 '18 • Edited

Common Lisp

Inhumane format version:

(defun format-words (words)
  (format nil "~{~#[~;~a~;~a and ~a~:;~@{~a~#[~;, and ~:;, ~]~}~]~}" words))

appropriated from Practical Common Lisp

More humane version (kinda like a string builder but using a stream):

(defun format-words (words)
  (with-output-to-string (s)
    (format s "~{~a~^, ~}" (butlast words))
    (unless (= 1 (length words))
      (format s " and "))
    (format s "~a" (car (last words)))))

Courtney • Apr 17 '19 • Edited

function formatWords(words) {
    if(!words) return '';
    words = words.filter((word) => word.length);
    let returnString = '';
    for(let i=0; i<words.length; i++) {
        if(i === words.length - 1 && i > 0) {
        returnString += ` and `;
        }
        else if (i > 0){
            returnString += `, `;
        }
        returnString += words[i];
    }
    return returnString;
}

E. Choroba • Nov 17 '18

Perl solution, tests included:

#! /usr/bin/perl
use warnings;
use strict;

sub format_words {
    my @words = grep length, @_;
    my $last = pop @words;
    return join(', ', @words) . (@words ? " and $last" : "")
}

use Test::More tests => 3;
is format_words('ninja', 'samurai', 'ronin'), 'ninja, samurai and ronin';
is format_words('ninja', '', 'ronin'), 'ninja and ronin';
is format_words(), "";

Ben Halpern • Nov 17 '18 • Edited

Ruby

def format_words(words)
  words.reject(&:blank?).join(", ").reverse.sub(',', 'dna ').reverse
end

Ali Spittel • Nov 17 '18

Monday - Transportation on vacation (8 KYU):

You will need a rental car in order for you to get around in your vacation. The manager of the car rental makes you some good offers.

codewars.com/kata/transportation-o...

Ali Spittel • Nov 17 '18

Java (!!) solution

public class Kata {
  public static int rentalCarCost(int d) {
    int total = d * 40;
    if(d >= 7)
      total -= 50;
    else if(d >= 3)
      total -= 20;
    return total;
  }
}

Sebastian Martin Dicke • Nov 17 '18 • Edited

Haskell solution:

transportation_on_vacation :: Int -> Int
transportation_on_vacation days
    | days > 6 = max_costs - 50
    | days > 3 = max_costs - 20
    | otherwise = max_costs
    where
    max_costs = 40 * days

E. Choroba • Nov 17 '18

Perl solution, tests included:

#! /usr/bin/perl
use warnings;
use strict;

my @discounts = ([7, 50], [3, 20], [0, 0]);
sub total {
    my ($days) = @_;
    my $total = $days * 40;
    for (@discounts) {
        my ($at_least, $discount) = @$_;
        return $total - $discount if $days >= $at_least;
    }
}

use Test::More tests => 8;
is total(1), 40;
is total(2), 80;
is total(3), 3 * 40 - 20;
is total(4), 4 * 40 - 20;
is total(5), 5 * 40 - 20;
is total(6), 6 * 40 - 20;
is total(7), 7 * 40 - 50;
is total(8), 8 * 40 - 50;

Thomas Landin • Nov 17 '18

My solution in Nim, my new favourite language :)

It provides an automatic default return variable called "result" that you can just start stuffing things into and it will automatically return it without you doing anything :)

let
  rate = 40.0
  weekly_discount = 50.0
  weekend_discount = 20.0
  test_data = [1, 2, 3, 4, 5, 6, 7, 8, 9]

proc total_cost(days: int): float =
  result = days.float * rate
  if days >= 7:
    result -= weekly_discount
  elif days >= 3:
    result -= weekend_discount

for n in test_data:
  echo "Cost for ", n, " days is: ", total_cost(n)

Jay • Nov 18 '18

I try to use the func keyword when defining methods, that way the compiler can guarantee that I have pure function with no side effects.
func is just sugar for proc {.noSideEffect.}.

func rental_car_cost(days: int): int =
  result = days * 40
  if days > 7:
    result -= 50
  elif days >= 3:
    result -= 20

Thomas Landin • Nov 18 '18

I haven't gotten into the habit of doing that yet. I know I should and I think it's a great idea in general, but years of habits are hard to break and the func keyword is still a fairly recent addition to the language :)

Courtney • Apr 16 '19 • Edited

function rentalCarCost(d : number) : number {
    let discount : number = 0;
    if(d >= 7) discount = 50;
    else if(d >= 3) discount = 20;
    return (d * 40) - discount;
}

//one-liner
function rentalCarCost2(d : number) : number {
    return d >= 7 ? ((d * 40) - 50) : (d >= 3 ? (d * 40) - 20 : d * 40);
}

Jay • Nov 17 '18

Rust Solution:

fn rental_car_cost(days: i32) -> i32 {
    (days * 40) - match days {
        n if n > 7 => 50,
        n if n >= 3 => 20,
        _ => 0,
    }
}

Ben Halpern • Nov 17 '18

Ruby

def rental_price(num_days)
  price = num_days * 40
  return price - 50 if num_days >= 7
  return price - 20 if num_days >= 3
  price
end

Ali Spittel • Nov 17 '18

Thursday - Queue from two stacks (Hacker Rank Medium)

Complete the put, pop, and peek methods in the editor below. They must perform the actions as described above.

hackerrank.com/challenges/ctci-que...

E. Choroba • Nov 17 '18

Perl solution. Passes all the tests on HackerRank.

#! /usr/bin/perl
use warnings;
use strict;
use feature qw{ say };

{   package Queue;

    use enum qw( LIFO FIFO );

    sub new  { bless [[], []], shift }
    sub Push { push @{ $_[0][LIFO] }, $_[1]; }
    sub Pop  { @{ $_[0][FIFO] } or $_[0]->_restack; pop @{ $_[0][FIFO] }; }
    sub Peek { @{ $_[0][FIFO] } or $_[0]->_restack; say $_[0][FIFO][-1]; }
    sub _restack {
        push @{ $_[0][FIFO] }, pop @{ $_[0][LIFO] } while @{ $_[0][LIFO] };
    }
}

my %dispatch = (
    1 => 'Push',
    2 => 'Pop',
    3 => 'Peek',
);

my $q = 'Queue'->new;

<>;  # skip the 1st line.
while (<>) {
    my ($operation, $argument) = split;
    $q->${\$dispatch{$operation}}($argument);
}

Pavel Susicky • Nov 17 '18

const Queue = {
    in: [],
    out: [],
    enqueue: function(x) {
        this.in.push(x);
    },
    dequeue: function() {
        if (this.out.length == 0) {
            while (this.in.length != 0) {
                this.out.push(this.in.pop());
            }
        }
        return this.out.length == 0 ? "Cannot dequeue" : this.out.pop();
    },
    front: function() {
        if (this.out.length == 0) {
            while (this.in.length != 0) {
                this.out.push(this.in.pop());
            }
        }
        return this.out.length == 0 ? "Cannot show front value" : this.out[this.out.length - 1];
    }
};

Ali Spittel • Nov 17 '18

class MyQueue(object):
    def __init__(self):
        self.stack1 = []
        self.stack2 = []

    def peek(self):
        if not self.stack2:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
        return self.stack2[-1]

    def pop(self):
        if not self.stack2: 
            while self.stack1:
                self.stack2.append(self.stack1.pop())
        return self.stack2.pop()

    def put(self, value):
        self.stack1.append(value)

Courtney • Apr 18 '19

function processData(input) {
    class Node {...}
    class Stack {...}
    const inputArray = input.split('\n');
    let stackIn = new Stack();
    let stackOut = new Stack();
    let valArry, result = '';
    for(let i=1; i<=inputArray[0]; i++) {
        valArry = inputArray[i].split(' ');
        switch(valArry[0]) {
            case '1':
                enqueue(valArry[1]);
                break;
            case '2':
                dequeue();
                break;
            case '3':
                result += `${front()}\n`;
                break;
        }
    } 

    function enqueue(val) {
        stackIn.push(val);
    }

    function dequeue() {
        if(stackOut.size === 0) {
            while(stackIn.size > 0) {
                let item = stackIn.pop();
                stackOut.push(item);
            }
        }
        stackOut.pop();
    }

    function front() {
        if(stackOut.size === 0) {
            while(stackIn.size > 0) {
                let item = stackIn.pop();
                stackOut.push(item);
            }
        }
        return stackOut.peek();
    }
    return process.stdout.write(result);
}

Jay • Nov 17 '18

Rust Solution:

struct MyQueue {
    stack1: Vec<i32>,
    stack2: Vec<i32>,
}

impl MyQueue {
    pub fn new() -> Self {
        MyQueue {
            stack1: vec![],
            stack2: vec![],
        }
    }

    pub fn push(&mut self, x: i32) {
        self.stack1.push(x);
    }

    pub fn pop(&mut self) {
        if self.stack2.is_empty() {
            while !self.stack1.is_empty() {
                self.stack2.push(self.stack1.pop().unwrap());
            }
        }
        self.stack2.pop();
    }

    pub fn peek(&mut self) -> i32 {
        if self.stack2.is_empty() {
            while !self.stack1.is_empty() {
                self.stack2.push(self.stack1.pop().unwrap());
            }
        }
        *self.stack2.last().unwrap()
    }
}

Ali Spittel • Nov 17 '18 • Edited

Friday - Cruise Control (Code Jam):

Annie is a bus driver with a high-stress job. She tried to unwind by going on a Caribbean cruise, but that also turned out to be stressful, so she has recently taken up horseback riding...

codejam.withgoogle.com/codejam/con...

E. Choroba • Nov 17 '18

The Perl solution I submitted back in 2017:

#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };

my $test_count = <>;
for my $test_case (1 .. $test_count) {
    my ($destination, $horse_count) = split ' ', <>;
    my $other_max = 0;
    for my $horse (1 .. $horse_count) {
        my ($position, $max_speed) = split ' ', <>;
        my $speed = ($destination - $position) / $max_speed;
        $other_max = $speed if $other_max < $speed;
    }
    say "Case #$test_case: ", $destination / $other_max;
}

Jay • Nov 17 '18

Rust Solution:

use std::f64;
use std::io::{self, prelude::*};

fn cruise_control(distance: f64, horses: &[(f64, f64)]) -> f64 {
    let time = horses
        .iter()
        .map(|&(d, s)| (distance - d) / s)
        .fold(f64::NAN, f64::max);
    distance / time
}

fn parse_line(s: &str) -> (f64, f64) {
    let vec: Vec<f64> = s
        .to_string()
        .split_whitespace()
        .map(|c| c.to_string().parse().unwrap())
        .collect();
    (vec[0], vec[1])
}

fn main() -> Result<(), io::Error> {
    let mut buffer = String::new();
    io::stdin().read_to_string(&mut buffer)?;
    let mut buf_lines = buffer.lines();
    let cases: u8 = buf_lines.next().unwrap().parse().unwrap();
    for case in 0..cases {
        let (dist, number) = parse_line(buf_lines.next().unwrap());
        let mut horses_info = buf_lines.clone().take(number as usize);
        let mut horses = vec![];
        for horse in horses_info {
            buf_lines.next();
            horses.push(parse_line(horse));
        }
        match io::stdout()
            .write(format!("Case #{}: {}\n", case + 1, cruise_control(dist, &horses)).as_bytes())
        {
            Ok(_) => (),
            Err(why) => panic!(why),
        }
    }
    Ok(())
}

Ali Spittel • Nov 17 '18

def clean_input_file(file_name):
    inp = open(file_name)
    inp = inp.read()
    inp = inp.split('\n')
    cases = int(inp[0])
    inp.pop(0)
    inp.pop()
    return inp, cases


def cruise_control(distance, horse):
    return float(distance - horse[0]) / horse[1]


def get_speeds(distance, horses):
    speeds = []
    for horse in horses:
        speeds.append(cruise_control(distance, horse))
    return distance / max(speeds)

input_file, cases = clean_input_file('test.in')
write_file = open('solution.txt', 'w+')

row = 0

for case in range(cases):
    case_info = input_file[row].split(' ')
    speed = int(case_info[0])
    n_horses = int(case_info[1])

    horses = []
    for _ in range(n_horses):
        row += 1
        horse = input_file[row].split(' ')
        horses.append((int(horse[0]), int(horse[1])))
    write_file.write("Case #{}: {}\n".format(case + 1, "{0:.2f}".format(get_speeds(speed, horses))))
    row += 1

Courtney • Apr 18 '19

I had to resort to following along with a youtube video of the challenge, but I understand it now, so...learning?

But seriously, tons of new things learned this week, like handling stdin and stdout data, reading test data from a file and writing answers to a file. Glad to have these challenges to follow!

function processData(input) {
    let resultStr = '';
    const inputArray = input.split('\n');
    const cases = parseInt(inputArray.shift());
    for(let i=0; i<cases; i++) {
        resultStr += `Case # ${i+1}: `;
        let me = inputArray.shift().split(' ');
        let destination = parseFloat(me[0]);
        let numHorses = parseInt(me[1]);
        let horses = [];
        for(let j=0; j<numHorses; j++) {
            let horse = inputArray.shift().split(' ');
            horses.push({ velocity: parseFloat(horse[1]), location: parseFloat(horse[0])});
        }
        let sortedHorses = horses.sort((a, b) => b.location-a.location);
        resultStr += `${cruiseControl(destination, sortedHorses)}\n`;
    }
    return process.stdout.write(resultStr);
}

function cruiseControl(dest, horseArry) {
    let B = horseArry[0].location, vB = horseArry[0].velocity;
    for (let i=1; i<horseArry.length; i++) {
        let A = horseArry[i].location, vA = horseArry[i].velocity;
        if(vB === vA) B = A, vB = vA;
        else {
            let x = (vB * A - vA * B) / (vB - vA);
            if(x > dest || x < A) {
                B = A, vB = vA;
            } 
        }
    }
    let tB = (dest - B) / vB;
    return dest / tB;
}

Eljay-Adobe • Nov 17 '18

Python 3

def format_words(words):
    if len(words) < 2:
        return "".join(words)
    return ", ".join(words[:-1]) + " and " + words[-1]


words = ["ninja", "samurai", "ronin", "leonardo", "michelangelo", "donatello", "raphael"]
print(f"{format_words(words)}")
words = ["alpha", "beta"]
print(f"{format_words(words)}")
words = ["alone"]
print(f"{format_words(words)}")
words = []
print(f"{format_words(words)}")

Sam J Watkins • Nov 18 '19

Don't give me 5 answer.
If a number ends in 5 it is divisible by five but not by 10, in java...

void printNumbersNotEndingIn5 (int beginning, int last){
try {
for (int i=beginning;i <= last; i++){
if ((i%5!=0) || ((i%5==0) && (i%10==0))) System.out.println (i);
}
}
catch (Exception e){
}
}

printNumbersNotEndingIn5 (out, 1, 30);

Ali Spittel • Nov 17 '18

DEV Community

Daily Coding Puzzles - Nov 11th - Nov 16th

Top comments (52)

Tuesday - Don't give me five! (7 KYU)

Common Lisp

Wednesday - Format words into a sentence (6 KYU)

Common Lisp

Monday - Transportation on vacation (8 KYU):

Thursday - Queue from two stacks (Hacker Rank Medium)

Friday - Cruise Control (Code Jam):

Meta

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