Daily Coding Puzzles - Nov 4th - Nov 9th

Ali Spittel on November 09, 2018

Every day on Twitter, I post coding puzzles. These are quick coding challenges that increase in difficulty across the span of the week -- with Mo... [Read Full]
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Array.diff (6 KYU):

Your goal in this kata is to implement a difference function, which subtracts one list from another and returns the result.



On the other hand, Ruby probably allocated about 10,000 object references under the hood and used 1gb of memory to make that happen. πŸ˜„

Honestly, I just came out here to have a good time and I'm feeling so attacked right now πŸ˜…πŸ˜…πŸ˜…

Haha nice. Honestly, I'm just so used to copy/pasting emojis from slack and discord. Fun fact, when you do that, it copies the colon emoji syntax, not the unicode glyph πŸ˜…πŸ™ƒ

@bendhalpern some numbers (using /usr/bin/time -l on OSX):

Go: 0.0s, 1.5Mb
Go run: 0.10s, 24Mb
Node: 0.06s, 19Mb
Ruby: 0.08s, 11Mb
Crystal: 0.0s, 1.5Mb
Crystal run: 0.4s, 10.4Mb
Rust: 0.00s - 1.12MiB
Rust (nightly): 0.01s - 0.88MiB
Go: 0.00s - 1.64MiB
Go run: 0.25s - 24.11MiB
Python: 0.21s - 5.48MiB

Disclaimer: I'm inexperienced with Go; using the example from below πŸ˜…

Oh, I know almost nothing about Go. I'm specifically using these exercises to learn the syntax. No clue if and how what I'm doing can be optimized. Go is famously, weirdly restrictive and intentionally verbose.


Tried in Javascript

function arrayDiff(arr1, arr2) {
    return arr1.filter(val => !arr2.includes(val));


  • With Liftime. The returned Vec<&T> has elements borrowed and valid until the list1 is valid.
fn diff<'a, T>(list1: &'a [T], list2: &[T]) -> Vec<&'a T>
    T: PartialEq,
    list1.into_iter().filter(|e| !list2.contains(e)).collect()
  • Without lifetime We return a new Vec free from any list provided, by cloning the data. The generic data type only allows the data type that implements Clone trait to be passed.
fn diff2<T>(list1: &[T], list2: &[T]) -> Vec<T>
    T: PartialEq + Clone,
        .map(|x| x.clone())
        .filter(|e| !list2.contains(&e))


def array_diff(a, b):
    return [l for l in a if l not in b]

Or with sets!

def array_diff(a, b):
    return list(set(a) - set(b))


func Difference(a []int, b []int) (diff []int) {
    m := make(map[int]bool)

    for _, item := range b {
        m[item] = true

    for _, item := range a {
        if _, presence := m[item]; !presence {
            diff = append(diff, item)

Common Lisp

I'll use lists because... well, it's Lisp, right?

(set-difference '(1 2 2 2 3) '(2))
;; => (3 1)

Can't beat a good standard library ;)



let diff a b =
    a |> Array.filter (fun v -> not (b |> Array.contains v))

also sequence expression syntax

let diff a b =
        for value in a do
            if not (b |> Array.contains value) then
                yield value

Or (although it will also remove duplicates from a):

(a |> Set.ofSeq) - (b |> Set.ofSeq)

Yes, this was my first thought, but I was trying to retain dupes.


It wasn't as easy as doing a set difference as the a needed to keep the duplicate values.

Below is the answer in C#.

Where is the same as filter in JavaScript.

using System.Linq;
using System.Collections.Generic;

public class Kata
  public static int[] ArrayDiff(int[] a, int[] b)
    var hash = new HashSet<int>(b);
    return a.Where(_ => !hash.Contains(_)).ToArray();


 a ~ b 

Replied to the post instead of the comment! My bad!



array_diff :: Eq a => [a] -> [a] -> [a] 
array_diff a b = filter (\x -> notElem x b) a

Another Haskell solution:

array_diff :: Eq a => [a] -> [a] -> [a]
array_diff a b = [x | x <- a, x `notElem` b]


$a | ? { $_ -notin $b }

Expanding the ? alias:

$a = @(1,2,3,4)
$b = @(2,3,4,5)

$a | Where-Object { $_ -notin $b }

Here's one in Dart:

arrayDiff(List a, List b) => a..retainWhere((i) => !b.contains(i));
function array_diff(a, b) {
  let counter = {};
  for(let i of b) {
    counter[i] = true;
  let result = a.filter(item => counter[item] !== true);
  return result;


Scramblies (5 KYU):

Complete the function scramble(str1, str2) that returns true if a portion of str1 characters can be rearranged to match str2, otherwise returns false




let scramble str1 str2 =
    let letterLookup = str1 |> Seq.countBy id |> Map.ofSeq
    let requiredCounts = str2 |> Seq.countBy id
    requiredCounts |> Seq.forall (fun (letter, requiredCount) ->
        match letterLookup |> Map.tryFind letter with
        | None -> false // letter missing
        | Some count -> requiredCount <= count

Edit: I had previously posted a version that had ever-so-slightly more performance but was more code. I'm also including that version below since it solves the problem differently.

let scramble str1 str2 =
    let letterPool = str1 |> Seq.sort |> List.ofSeq
    let requiredLetters = str2 |> Seq.sort |> List.ofSeq
    let rec loop requiredLetters letterPool =
        match requiredLetters, letterPool with
        | [], _ -> true // found all
        | _ :: _, [] -> false // pool ran out
        | letter :: _, next :: pool when letter > next ->
            loop requiredLetters pool // skip next in pool
        | letter :: required, next :: pool when letter = next ->
            loop required pool
        | _, _ -> false // letter not available in the pool
    loop requiredLetters letterPool

Here are the tests (console).

        "rkqodlw", "world", true
        "cedewaraaossoqqyt", "codewars", true
        "katas", "steak", false
    |> List.iter (fun (str1, str2, expected) ->
        let actual = scramble str1 str2
        let e = if expected = actual then "√" else "X"
        printfn "%s %A ==> %b, %b" e (str1, str2) expected actual
    // √ ("rkqodlw", "world") ==> true, true
    // √ ("cedewaraaossoqqyt", "codewars") ==> true, true
    // √ ("katas", "steak") ==> false, false

Nice! yeah -- I only sometimes think about the efficiencies of these. They're contrived and we're doing them for fun. Part of me cares, part of me doesn't haha

I agree. I would have posted a much more expressive version, but you beat me to it! I did find an alternative way of solving the problem that was a little more expressive and nearly the same perf. I updated my post to include it.



func Scramble(str1 string, str2 string) (result bool) {
    m := make(map[string]int)

    for _, item := range strings.Split(str1, "") {
        m[item] += 1

    for _, item := range strings.Split(str2, "") {
        if val, ok := m[item]; !ok || val == 0 {
            return false
        } else {
            m[item] -= 1

    return true


fn scramble(str1: &str, str2: &str) -> bool {
    if str2.len() > str1.len() {
        return false;
    str2.chars().all(|c| {
        str1.chars().filter(|&ch| ch == c).count() >= str2.chars().filter(|&c2| c2 == c).count()

Solved it awhile ago in C#.

⚠ Warning: Really ugly...

using System;
using System.Collections.Generic;
using System.Linq;

  public class Scramblies
    public static bool Scramble(string s1, string s2)
      Func<string, Dictionary<char, int>> toMap = s =>
        s.GroupBy(c => c)
        .Select(g => new { g.Key, Count = g.Count() })
        .ToDictionary(pair => pair.Key, pair => pair.Count);

      var map1 = toMap(s1);
      var map2 = toMap(s2);

      foreach (KeyValuePair<char, int> p2 in map2)
        if (!map1.ContainsKey(p2.Key)) return false;

        if (map1[p2.Key] < p2.Value) return false;

      return true;

Common Lisp

(defun scramble (source target)
          (let ((source (coerce source 'list))
                (target (coerce target 'list)))
            (cond ((null target) t) 
                  ((member (first target) source)
                   (scramble (remove (first target) source :count 1)
                             (rest target))))))

(mapcar #'(lambda (args) (apply #'scramble args))
        '(("rkqodlw" "world")
          ("cedewaraaossoqqyt" "codewars")
          ("katas" "steak")
          ("scriptjavx" "javascript")
          ("scriptingjava" "javascript")
          ("scriptsjava" "javascripts")
          ("javscripts" "javascript")
          ("aabbcamaomsccdd" "commas")
          ("commas" "commas")
          ("sammoc" "commas")))
;; => (T T NIL NIL T T NIL T T T)


def scramble(s1,s2):
    for letter in set(s2):
        if s1.count(letter) < s2.count(letter):
            return False
    return True

Not a fancy solution, but it seems to work πŸ˜ƒ

function scramble(s1, s2) {
    if (s1.length < s2.length) return false;

    const _s2 = s2.split("");

    s1.split("").forEach(val => {
        if (_s2.includes(val)) _s2.splice(_s2.indexOf(val), 1);

    return _s2.length == 0;
function scramble(str1, str2) {
 let letterHolder = {};
 for (let letter of str1) {
   if(letterHolder[letter]) letterHolder[letter]++;
   else letterHolder[letter] = 1;
 for (let letter of str2) {
   if(!letterHolder[letter]) return false
   else letterHolder[letter]--;
 return true;


Product of Array Items (7 KYU):

Calculate the product of all elements in an array.




let product arr =
    arr |> Array.reduce (*)

I probably wouldn't define a separate function for this in actual code.


Does F# require you to name the arr parameter explicitly or can you simply do the following:

let product = Array.reduce (*)

I've been curious about F# for some time but haven't really done a deep dive.

Yes, you can do that in F#. (You probably know this, but for the sake of onlookers) it is called point-free notation. It can be handy for small functions like this. But I noticed when I use it too much, my code can become hard to understand.

Especially for code examples, I rarely use it because it can confuse readers.


A simple aggregation of data (using reduce) worked like a charm.

C# answer.

namespace Kata {
  using System;
  using System.Linq;
  public class ArrayMath
    public static int Product(int[] values)
      return values.Aggregate((product, current) => product * current);

Common Lisp

as * in Common Lisp can take as many arguments as you like...

(apply #'* (list 1 2 3 4 5))
;; => 120


function arrayProduct(arr) {
    return arr.reduce((acc, val) => acc * val);


fn product(list: &[i32]) -> i32 {
    list.iter().fold(1, |p, n| p * n)

I decided to not use reduce so that I could return early if I hit a zero.

function product(values) {
  if(!values || values.length === 0) return null;
  let prod = values[0];
  for(let i=1; i<values.length; i++) {
    if(values[i] === 0) return 0;
    prod *= values[i];
  return result;


def product(numbers):
    if not numbers: return None
    running_product = 1
    for number in numbers:
        running_product *= number
    return running_product
def product (numbers):
  return reduce(lambda total, number: total * number, numbers)


const product = arr => arr.reduce((acc, x) => acc * x)


func Product(nums []int)(x int) {
    x = 1

    for _, num := range nums {
        x *= num



product' :: Num a => [a] -> a
product' = product


The Last Word (CodeJam):

You are the next contestant on this show, and the host has just showed you the string S. What's the winning last word that you should produce?




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

fn find_last_word(s: &str) -> String {
        .fold(vec![], |mut last, letter| {
            if let Some(&c) = last.last() {
                if (c as u8) > (letter as u8) {
                    last.insert(0, letter);
                } else {
            } else {

fn main() {
    let stdin = io::stdin();

    for (i, line) in stdin.lock().lines().skip(1).enumerate() {
        if let Ok(text) = line {
            match io::stdout()
                .write(format!("Case #{}: {}\n", i + 1, find_last_word(&text)).as_bytes())
                Ok(_) => (),
                Err(why) => panic!(why),

usage: last_word(.exe) < small.in > small.txt


Common Lisp

(defun last-word (word)
  (let ((cs (coerce word 'list)))
    (coerce (reduce #'(lambda (word c)
                        (if (char> (first word) c)
                            (append word (list c))
                            (cons c word)))
                    (rest cs)
                    :initial-value (list (first cs)))

Quick little test...

(mapcar #'last-word (list "CAB"


let lastWord s =
    let update (first, word) letter =
        if letter >= first then (letter, string letter + word)
        else                    (first, word + string letter)
    s |> Seq.fold update ('A', "") |> snd

Testing it (console)

        "CAB", "CAB"
        "JAM", "MJA"
        "CODE", "OCDE"
        "ABAAB", "BBAAA"
    |> List.iter (fun (input, expected) ->
        let actual = lastWord input
        let e = if expected = actual then "√" else "X"
        printfn "%s %s ==> %s, %s" e input expected actual
    // √ CAB ==> CAB, CAB
    // √ JAM ==> MJA, MJA
    // √ CODE ==> OCDE, OCDE
    // √ ABAAB ==> BBAAA, BBAAA


Awesome -- this one (for me) was a lot easier than they made it sound!

Same here. The hard part was understanding the problem. (It felt very much like "A train leaving SF at 50kph ..." word problems.) But after that the code wasn't so bad.

I think the possible "gotcha" here is that they do not want a reverse alphabetically sorted string, which, if you're not careful about the requirements, could be what you try to build and have it trip you up.



def get_last_word(word):
    last_word = word[0]
    for letter in word[1:]:
        if letter >= last_word[0]:
            last_word = letter + last_word
            last_word = last_word + letter
    return last_word

out_file = open('output.txt', 'w')
case_number = 0

for word in open('A-small-practice (3).in', 'r'):
    if case_number != 0:
        out_file.write("Case #{}: ".format(case_number) + get_last_word(word))
    case_number += 1


func Last(w string) (result []string) {
    l := strings.Split(w, "")
    result = []string{l[0]}

    for _, item := range l[1:] {
        if val := result[0]; item > val {
            result = append([]string{item}, strings.Join(result, ""))
        } else {
            result = append(result, item)


function processLastwordData(input :string) : void {
    const inputArray = input.split('\n');
    let resultStr = "";
    const cases = parseInt(inputArray[0]);
    for(let i=1; i <= cases; i++) {
        resultStr += `Case #${i}: ${lastWord(inputArray[i])}\n`

function lastWord(str : string) : string {
    let outStr = str[0];
    for(let i=1; i<str.length; i++) {
        if(str.charCodeAt(i) >= outStr.charCodeAt(0)) outStr = str[i] + outStr;
        else outStr = outStr + str[i];
    return outStr;

let _input = "";
process.stdin.on("data", function (input) {
    _input += input;
process.stdin.on("end", function () {


JS Solution: Takes two arrays, returns a single array with all of the items from array A, which do not exist in array B.

function diff(a, b){
  return a.filter(item => !b.includes(item))

Just realized @thesoreon posted almost the same solution already. Oh well, I guess great minds think alike 😜


That's nice! When i made this solution i was curious and searched for another approches and found the exact same solution on stack overflow πŸ˜†



Number Drills: Blue and red marbles (8 KYU):

You've decided to write a function, guess_blue() to help automatically calculate whether you should guess "blue" or "red". The function should take four arguments.




func Guess(bStart int, rStart int, bGone int, rGone int)(probability float32) {
    var b = bStart - bGone
    var r = rStart - rGone

    probability = float32(b) / float32(r + b)


fn guess_blue(blue_start: u32, red_start: u32, blue_pulled: u32, red_pulled: u32) -> f32 {
    let blue = (blue_start - blue_pulled) as f32;
    let red = (red_start - red_pulled) as f32;

    blue / (blue + red)

Common Lisp

(defun guess-blue (blue-in red-in blue-out red-out)
  (let ((blue (- blue-in blue-out))
        (red (- red-in red-out)))
    (/ blue (+ red blue)))

(guess-blue 5 5 2 3)
;; => 3/5

the fun of a language with rationals:

(guess-blue 1 2 0 0)
;; => 1/3


def guess_blue(blue_start, red_start, blue_pulled, red_pulled):
    blue = blue_start - blue_pulled
    red = red_start - red_pulled
    return blue / (red + blue)
function guessBlue(blueStart, redStart, bluePulled, redPulled) {
  let currentBlue = blueStart - bluePulled;
  let currentRed = redStart - redPulled;
  return currentBlue / (currentBlue + currentRed);


Ooh. I tried to do Advent of Code in Go last year, but AoC was probably too steep a challenge for a language I didn't know at all. This is a much more manageable entrypoint. Thanks for this! πŸ€—


For sure! I may try and incorporate those next month! We'll see! Go is so much fun, I should do more with it!


That's a great question -- I'm not totally sure, but it's the ranking system CodeWars uses, so I include it -- 8 KYU are the most beginner friendly, 1 KYU takes a really long time.


KYU is used for grading the difficulty levels, or degree of proficiency or experience.

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