30 Days Codewars

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Hello 🌏! I like solving coding problems, especially on codewars. So, I decided to share my 30-days-of-solving-codewars-problems journey here.

Everyday I will solve one question from codewars and share below. It will include the question, my answer and one or two best answers from the codewars community.

Ways to reach out to me 👇
Twitter || LinkedIn || Blogs

Btw, if you find some typos, feel free to leave a PR 😎

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Day 0

Even or Odd?

8 kyu

Description:

Create a function that takes an integer as an argument and returns "Even" for even numbers or "Odd" for odd numbers.

function even_or_odd(number) {}

even_or_odd();

My Answer

function even_or_odd(number) {
  return number % 2 === 0 ? "Even" : "Odd";
}

Best Answer

function even_or_odd(number) {
  return number % 2 ? "Odd" : "Even";
}

---

Day 1

Stop gninnipS My sdroW!

6 kyu

Description:

Write a function that takes in a string of one or more words, and returns the same string, but with all five or more letter words reversed (Just like the name of this Kata). Strings passed in will consist of only letters and spaces. Spaces will be included only when more than one word is present.

Examples: spinWords( "Hey fellow warriors" ) => returns "Hey wollef sroirraw" spinWords( "This is a test") => returns "This is a test" spinWords( "This is another test" )=> returns "This is rehtona test"

function spinWords(string) {
  //TODO Have fun :)
}

My Answer 😅

function spinWords(string) {
  let words = string.split(" ");

  let newWords = words.map((word) =>
    word.length >= 5 ? word.split("").reverse().join("") : word
  );

  return newWords.join(" ");
}

Best Answers ✅

Top ranked answer (writter by a human 😅):

function spinWords(words) {
  return words
    .split(" ")
    .map(function (word) {
      return word.length > 4 ? word.split("").reverse().join("") : word;
    })
    .join(" ");
}

2nd ranked answer 👀:

function spinWords(string){
  return string.replace(/\w{5,}/g, function(w) { return w.split('').reverse().join('') })
}
}

---

Day 2

Sum of Digits / Digital Root

6 kyu

Description:

Digital root is the recursive sum of all the digits in a number.

Given n, take the sum of the digits of n. If that value has more than one digit, continue reducing in this way until a single-digit number is produced. The input will be a non-negative integer.

Examples:

    16  -->  1 + 6 = 7
   942  -->  9 + 4 + 2 = 15  -->  1 + 5 = 6
132189  -->  1 + 3 + 2 + 1 + 8 + 9 = 24  -->  2 + 4 = 6
493193  -->  4 + 9 + 3 + 1 + 9 + 3 = 29  -->  2 + 9 = 11  -->  1 + 1 = 2

Start here:

function digital_root(n) {
  // ...
}

My Answer 😅

function digital_root(n) {
  let newNum;
  let justNum = n
    .toString()
    .split("")
    .map((num) => parseInt(num, 10));

  newNum = justNum.reduce((partialSum, a) => partialSum + a, 0);

  if (newNum.toString().length > 1) {
    return digital_root(newNum);
  } else {
    return newNum;
  }
}

Best Answers ✅

Top ranked answer, wtf? 🤯:

function digital_root(n) {
  return ((n - 1) % 9) + 1;
}

Top comments for this solution. Just read for fun 😅

• dude speaks matrix languague

• mathematics!

• what kind of sorcery is this?

• my brain is damaged!

• wait, what?

• hmmm... wtf? totally mind boggling

• bruh!!!

• I am not a coder.

• The simplicity is extremely frustrating. Thank you

• I am quitting programming after this

2nd ranked answer 👍:

function digital_root(n) {
  if (n < 10) return n;

  return digital_root(
    n
      .toString()
      .split("")
      .reduce(function (acc, d) {
        return acc + +d;
      }, 0)
  );
}

---

Day 3

Disemvowel Trolls

7 kyu

Description:

Trolls are attacking your comment section! A common way to deal with this situation is to remove all of the vowels from the trolls' comments, neutralizing the threat. Your task is to write a function that takes a string and return a new string with all vowels removed. For example, the string "This website is for losers LOL!" would become "Ths wbst s fr lsrs LL!".

Note: for this kata y isn't considered a vowel.

function disemvowel(str) {
  return str;
}

My Answer 😅

function disemvowel(str) {
  return str
    .split(" ")
    .map((x) => x.replace(/[aAeEiIoOuU]/g, ""))
    .join(" ");
}

Best Answers ✅

Top ranked answer (writter by a human 😅):

function disemvowel(str) {
  return str.replace(/[aeiou]/gi, "");
}

2nd ranked answer 👀:

function disemvowel(str) {
  var vowels = ["a", "e", "i", "o", "u"];

  return str
    .split("")
    .filter(function (el) {
      return vowels.indexOf(el.toLowerCase()) == -1;
    })
    .join("");
}

---

Day 4

Array.diff

6 kyu

Description:

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

It should remove all values from list a, which are present in list b keeping their order.

arrayDiff([1, 2], [1]) == [2];

If a value is present in b, all of its occurrences must be removed from the other:

arrayDiff([1, 2, 2, 2, 3], [2]) == [1, 3];

My Answers 😅

// Solution 1:
function arrayDiff(a, b) {
  return a.filter((val) => !b.includes(val));
}

// Solution 2:
function arrayDiff(a, b) {
  return a.filter((val) => b.indexOf(val) === -1);
}

Best Answers ✅

Top ranked answer 🤯:

function array_diff(a, b) {
  return a.filter((e) => !b.includes(e));
}

2nd ranked answer 👍:

function array_diff(a, b) {
  return a.filter(function (x) {
    return b.indexOf(x) == -1;
  });
}

---

Day 5

7 kyu

Description:

In this little assignment you are given a string of space separated numbers, and have to return the highest and lowest number.

Examples:

highAndLow("1 2 3 4 5"); // return "5 1"
highAndLow("1 2 -3 4 5"); // return "5 -3"
highAndLow("1 9 3 4 -5"); // return "9 -5"

Notes:

• All numbers are valid Int32, no need to validate them.
• There will always be at least one number in the input string.
• Output string must be two numbers separated by a single space, and highest number is first.

My Answer 😅

function highAndLow(numbers) {
  let n1 = numbers.split(" ").reduce((a, b) => `${Math.min(a, b)}`);
  let n2 = numbers.split(" ").reduce((a, b) => `${Math.max(a, b)}`);

  return `${n2} ${n1}`;
}

Best Answers ✅

Top ranked answer 🤯:

function highAndLow(numbers) {
  numbers = numbers.split(" ").map(Number);
  return Math.max.apply(0, numbers) + " " + Math.min.apply(0, numbers);
}

2nd ranked answer 👍:

function highAndLow(numbers) {
  numbers = numbers.split(" ");
  return `${Math.max(...numbers)} ${Math.min(...numbers)}`;
}

---

Day 5*

6 kyu

Description:

Jamie is a programmer, and James' girlfriend. She likes diamonds, and wants a diamond string from James. Since James doesn't know how to make this happen, he needs your help.

You need to return a string that looks like a diamond shape when printed on the screen, using asterisk (*) characters. Trailing spaces should be removed, and every line must be terminated with a newline character (\n).

Return null/nil/None/... if the input is an even number or negative, as it is not possible to print a diamond of even or negative size.

Examples

A size 3 diamond:

 *
***
 *

...which would appear as a string of " *\n***\n *\n"

A size 5 diamond:

  *
 ***
*****
 ***
  *

...that is:

" *\n ***\n*****\n ***\n *\n"

My Answer 😅

Sorry, I couldn't think of an answer for this one so I had to unlock the solution 😅 So, I don't count this one 😉

Best Answers ✅

Top ranked answer 🤯:

function diamond(n) {
  if (n <= 0 || n % 2 === 0) return null;
  str = "";
  for (let i = 0; i < n; i++) {
    let len = Math.abs((n - 2 * i - 1) / 2);
    str += " ".repeat(len);
    str += "*".repeat(n - 2 * len);
    str += "\n";
  }
  return str;
}

2nd ranked answer 👍:

function diamond(n) {
  if (n % 2 == 0 || n < 1) return null;
  var x = 0,
    add,
    diam = line(x, n);
  while ((x += 2) < n) {
    add = line(x / 2, n - x);
    diam = add + diam + add;
  }
  return diam;
} //z.

function repeat(str, x) {
  return Array(x + 1).join(str);
}
function line(spaces, stars) {
  return repeat(" ", spaces) + repeat("*", stars) + "\n";
}

---

Day 6 (1)

6 kyu

Description:

You probably know the "like" system from Facebook and other pages. People can "like" blog posts, pictures or other items. We want to create the text that should be displayed next to such an item.

Implement the function which takes an array containing the names of people that like an item. It must return the display text as shown in the examples:

[]                                -->  "no one likes this"
["Peter"]                         -->  "Peter likes this"
["Jacob", "Alex"]                 -->  "Jacob and Alex like this"
["Max", "John", "Mark"]           -->  "Max, John and Mark like this"
["Alex", "Jacob", "Mark", "Max"]  -->  "Alex, Jacob and 2 others like this"

Note: For 4 or more names, the number in "and 2 others" simply increases.

My solution 😅

function likes(names) {
  if (names.length === 0) {
    return "no one likes this";
  } else if (names.length === 1) {
    return `${names[0]} likes this`;
  } else if (names.length === 2) {
    return `${names[0]} and ${names[1]} like this`;
  } else if (names.length === 3) {
    return `${names[0]}, ${names[1]} and ${names[2]} like this`;
  }
  return `${names[0]}, ${names[1]} and ${names.length - 2} others like this`;
}

Best solutions ✅

Top ranked answer (...meh) 👍:

function likes(names) {
  names = names || [];
  switch (names.length) {
    case 0:
      return "no one likes this";
      break;
    case 1:
      return names[0] + " likes this";
      break;
    case 2:
      return names[0] + " and " + names[1] + " like this";
      break;
    case 3:
      return names[0] + ", " + names[1] + " and " + names[2] + " like this";
      break;
    default:
      return (
        names[0] +
        ", " +
        names[1] +
        " and " +
        (names.length - 2) +
        " others like this"
      );
  }
}

2nd ranked answer (clever 😁):

function likes(names) {
  return {
    0: "no one likes this",
    1: `${names[0]} likes this`,
    2: `${names[0]} and ${names[1]} like this`,
    3: `${names[0]}, ${names[1]} and ${names[2]} like this`,
    4: `${names[0]}, ${names[1]} and ${names.length - 2} others like this`,
  }[Math.min(4, names.length)];
}

---

Day 6 (2)

7 kyu

Description:

Your task is to make a function that can take any non-negative integer as an argument and return it with its digits in descending order. Essentially, rearrange the digits to create the highest possible number.

Examples:

Input: 42145 Output: 54421

Input: 145263 Output: 654321

Input: 123456789 Output: 987654321

My solution 😅

function descendingOrder(n) {
  let str = n
    .toString()
    .split("")
    .sort((a, b) => a - b)
    .reverse()
    .join("");

  return parseInt(str, 10);
}

Best solutions ✅

Top ranked answer 👍:

function descendingOrder(n) {
  return parseInt(String(n).split("").sort().reverse().join(""));
}

2nd ranked answer ✅:

function descendingOrder(n) {
  return +(n + "")
    .split("")
    .sort(function (a, b) {
      return b - a;
    })
    .join("");
}

---

Day 7

6 kyu

Description:

Examples:

My solution 😅

tolerable 🥱

function createPhoneNumber(numbers) {
  let str = numbers.join("");

  return `(${str.slice(0, 3)}) ${str.slice(3, 6)}-${str.slice(6, 10)}`;
}

Best solutions ✅

Top ranked answer very clever 🔥. Imho:

function createPhoneNumber(numbers) {
  var format = "(xxx) xxx-xxxx";

  for (var i = 0; i < numbers.length; i++) {
    format = format.replace("x", numbers[i]);
  }

  return format;
}

2nd ranked answer ✅. Similar to mine huh?

function createPhoneNumber(numbers) {
  numbers = numbers.join("");
  return (
    "(" +
    numbers.substring(0, 3) +
    ") " +
    numbers.substring(3, 6) +
    "-" +
    numbers.substring(6)
  );
}

---

Day 8 (1)

Find The Parity Outlier

6 kyu

Description:

You are given an array (which will have a length of at least 3, but could be very large) containing integers. The array is either entirely comprised of odd integers or entirely comprised of even integers except for a single integer N. Write a method that takes the array as an argument and returns this "outlier" N.

Examples:

[2, 4, 0, 100, 4, 11, 2602, 36]
Should return: 11 (the only odd number)

[160, 3, 1719, 19, 11, 13, -21]
Should return: 160 (the only even number)

My solution 😅

tolerable 🥱

function findOutlier(integers) {
  //your code here
  let arr1 = [];
  let arr2 = [];

  integers.filter((i) => (i % 2 === 0 ? arr1.push(i) : arr2.push(i)));

  return arr1.length > arr2.length ? arr2[0] : arr1[0];
}

Best solutions ✅

Top ranked answer:

function findOutlier(int) {
  var even = int.filter((a) => a % 2 == 0);
  var odd = int.filter((a) => a % 2 !== 0);
  return even.length == 1 ? even[0] : odd[0];
}

2nd ranked answer ✅

function findOutlier(integers) {
  return integers.slice(0, 3).filter(even).length >= 2
    ? integers.find(odd)
    : integers.find(even);
}
function even(num) {
  return num % 2 == 0;
}
function odd(num) {
  return !even(num);
}

---

Day 8 (2)

Get the Middle Character

7 kyu

Description:

You are going to be given a word. Your job is to return the middle character of the word. If the word's length is odd, return the middle character. If the word's length is even, return the middle 2 characters.

Examples:

Kata.getMiddle("test") should return "es"

Kata.getMiddle("testing") should return "t"

Kata.getMiddle("middle") should return "dd"

Kata.getMiddle("A") should return "A"

My solution 😅

tolerable 🥱

function getMiddle(s) {
  let l = s.length;

  return l % 2 !== 0 ? s[l / 2 - 0.5] : `${s[l / 2 - 1]}${s[l / 2]}`;
}

Best solutions ✅

Top ranked answer, meh...

function getMiddle(s) {
  return s.substr(Math.ceil(s.length / 2 - 1), s.length % 2 === 0 ? 2 : 1);
}

2nd ranked answer ✅

function getMiddle(s) {
  var middle = s.length / 2;
  return s.length % 2
    ? s.charAt(Math.floor(middle))
    : s.slice(middle - 1, middle + 1);
}

---

Day 8 (3)

Bit Counting

6 kyu

Description:

Write a function that takes an integer as input, and returns the number of bits that are equal to one in the binary representation of that number. You can guarantee that input is non-negative.

Examples:

The binary representation of 1234 is 10011010010, so the function should return 5 in this case

My solution 😅

Had no idea what binary was 🥲, but still...

var countBits = function (n) {
  return n
    .toString(2)
    .split("")
    .filter((a) => a === "1").length;
};

Best solutions ✅

Top ranked answer, one line 😬. Similar to mine, huh?

countBits = (n) => n.toString(2).split("0").join("").length;

2nd ranked answer. I have no idea what is going on below 🤨

function countBits(n) {
  for (c = 0; n; n >>= 1) c += n & 1;
  return c;
}

---

Day 9 (1)

Counting Duplicates

6 kyu

Description:

Count the number of Duplicates

Write a function that will return the count of distinct case-insensitive alphabetic characters and numeric digits that occur more than once in the input string. The input string can be assumed to contain only alphabets (both uppercase and lowercase) and numeric digits.

Example:

"abcde" -> 0 # no characters repeats more than once "aabbcde" -> 2 # 'a' and 'b' "aabBcde" -> 2 # 'a' occurs twice and 'b' twice (bandB) "indivisibility" -> 1 # 'i' occurs six times "Indivisibilities" -> 2 # 'i' occurs seven times and 's' occurs twice "aA11" -> 2 # 'a' and '1' "ABBA" -> 2 # 'A' and 'B' each occur twice

My solution 😅

function duplicateCount(text) {
  let arr1 = text.toLowerCase().split("");
  let arr2 = [];
  arr1.filter((item, i) => {
    if (arr2.indexOf(item) === -1) {
      if (arr1.indexOf(item) !== i) {
        arr2.push(item);
      }
    }
    return arr2;
  });

  return arr2.length;
}

Best solutions ✅

Top ranked answer 😐:

function duplicateCount(text) {
  return (
    text
      .toLowerCase()
      .split("")
      .sort()
      .join("")
      .match(/([^])\1+/g) || []
  ).length;
}

2nd ranked answer ✅

function duplicateCount(text) {
  return text
    .toLowerCase()
    .split("")
    .filter(function (val, i, arr) {
      return arr.indexOf(val) !== i && arr.lastIndexOf(val) === i;
    }).length;
}

---

Day 9 (2)

7 kyu

Description:

An isogram is a word that has no repeating letters, consecutive or non-consecutive. Implement a function that determines whether a string that contains only letters is an isogram. Assume the empty string is an isogram. Ignore letter case.

Example: (Input --> Output)

"Dermatoglyphics" --> true
"aba" --> false
"moOse" --> false (ignore letter case)

My solution 😅

function isIsogram(str) {
  return (
    str
      .toLowerCase()
      .split("")
      .filter((item, pos, arr) => arr.indexOf(item) == pos).length == str.length
  );
}

Best solutions ✅

Top ranked answer 😐:

function isIsogram(str) {
  return !/(\w).*\1/i.test(str);
}

2nd ranked answer ✅

function isIsogram(str) {
  return new Set(str.toUpperCase()).size == str.length;
}

---

Day 10 (1)

List Filtering

7 kyu

Description:

In this kata you will create a function that takes a list of non-negative integers and strings and returns a new list with the strings filtered out.

Example:

filter_list([1, 2, "a", "b"]) == [1, 2];
filter_list([1, "a", "b", 0, 15]) == [1, 0, 15];
filter_list([1, 2, "aasf", "1", "123", 123]) == [1, 2, 123];

My solution 😅

I literally spent less than a minute 🥱

function filter_list(l) {
  // Return a new array with the strings filtered out
  return l.filter((a) => typeof a === "number");
}

Best solutions ✅

Top ranked answer 😐:

function filter_list(l) {
  return l.filter(function (v) {
    return typeof v == "number";
  });
}

2nd ranked answer ✅

function filter_list(l) {
  return l.filter((e) => Number.isInteger(e));
}

---

Day 10 (2)

Duplicate Encoder

6 kyu

Description:

The goal of this exercise is to convert a string to a new string where each character in the new string is "(" if that character appears only once in the original string, or ")" if that character appears more than once in the original string. Ignore capitalization when determining if a character is a duplicate.

Example:

"din"      =>  "((("
"recede"   =>  "()()()"
"Success"  =>  ")())())"
"(( @"     =>  "))(("

My solution 😅

I..., I solved it 🥲

function duplicateEncode(word) {
  let arr1 = word.toLowerCase().split("");
  let arr2 = [];

  arr1.filter((item, i) => {
    if (arr2.indexOf(item) === -1) {
      if (arr1.indexOf(item) !== i) {
        arr2.push(item);
      }
    }
    return arr2;
  });

  let arr3 = [];

  word
    .toLowerCase()
    .split("")
    .filter((a) => (arr2.includes(a) ? arr3.push(")") : arr3.push("(")));

  return arr3.join("");
}

Best solutions ✅

Top ranked answer ✅:

function duplicateEncode(word) {
  return word
    .toLowerCase()
    .split("")
    .map(function (a, i, w) {
      return w.indexOf(a) == w.lastIndexOf(a) ? "(" : ")";
    })
    .join("");
}

2nd ranked answer ✅

function duplicateEncode(word) {
  word = word.toLowerCase();
  return word.replace(/./g, (m) =>
    word.indexOf(m) == word.lastIndexOf(m) ? "(" : ")"
  );
}

---

Day 11

Exes and Ohs

7 kyu

Description:

Check to see if a string has the same amount of 'x's and 'o's. The method must return a boolean and be case insensitive. The string can contain any char.

Example:

XO("ooxx") => true
XO("xooxx") => false
XO("ooxXm") => true
XO("zpzpzpp") => true // when no 'x' and 'o' is present should return true
XO("zzoo") => false

My solution 😅

I... I solved it 😬. It should have been 6kyu Imho 🤨

function XO(str) {
  let arr1 = [];
  let arr2 = [];

  str
    .toLowerCase()
    .split("")
    .map((a) => {
      if (a === "o") {
        arr1.push(a);
      }

      if (a === "x") {
        arr2.push(a);
      }
    });

  return arr1.length === arr2.length;
}

Best solutions ✅

Top ranked answer ✅, f*** regex

function XO(str) {
  let x = str.match(/x/gi);
  let o = str.match(/o/gi);
  return (x && x.length) === (o && o.length);
}

2nd ranked answer ✅, clever 🥲

const XO = (str) => {
  str = str.toLowerCase().split("");
  return (
    str.filter((x) => x === "x").length === str.filter((x) => x === "o").length
  );
};

---

Day 12

Persistent Bugger.

6 kyu

Description:

Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.

For example (Input --> Output):

39 --> 3 (because 3*9 = 27, 2*7 = 14, 1*4 = 4 and 4 has only one digit)
999 --> 4 (because 9*9*9 = 729, 7*2*9 = 126, 1*2*6 = 12, and finally 1*2 = 2)
4 --> 0 (because 4 is already a one-digit number)

My solution 😅

Recursion baby, recursion 😅

function persistence(num) {
  let counter = 1;
  function recurse(num) {
    let arr1 = num;
    let nums = 0;

    arr1 = arr1
      .toString()
      .split("")
      .map(Number)
      .reduce((a, b) => {
        return a * b;
      }, 1);

    if (arr1.toString().length !== 1) {
      return recurse(arr1, (counter += 1));
    } else {
      return counter;
    }
  }

  return num.toString().length === 1 ? 0 : recurse(num);
}

Best solutions ✅

Top ranked answer ✅, meh 🥱:

const persistence = (num) => {
  return `${num}`.length > 1
    ? 1 + persistence(`${num}`.split("").reduce((a, b) => a * +b))
    : 0;
};

2nd ranked answer ✅, clever 🥲

function persistence(num) {
  var times = 0;

  num = num.toString();

  while (num.length > 1) {
    times++;
    num = num
      .split("")
      .map(Number)
      .reduce((a, b) => a * b)
      .toString();
  }

  return times;
}

---

Day 13

Tribonacci Sequence

6 kyu

Description:

Well met with Fibonacci bigger brother, AKA Tribonacci.

As the name may already reveal, it works basically like a Fibonacci, but summing the last 3 (instead of 2) numbers of the sequence to generate the next. And, worse part of it, regrettably I won't get to hear non-native Italian speakers trying to pronounce it :(

So, if we are to start our Tribonacci sequence with [1, 1, 1] as a starting input (AKA signature), we have this sequence:

[1, 1 ,1, 3, 5, 9, 17, 31, ...]

But what if we started with [0, 0, 1] as a signature? As starting with [0, 1] instead of [1, 1] basically shifts the common Fibonacci sequence by once place, you may be tempted to think that we would get the same sequence shifted by 2 places, but that is not the case and we would get:

[0, 0, 1, 1, 2, 4, 7, 13, 24, ...]

Well, you may have guessed it by now, but to be clear: you need to create a fibonacci function that given a signature array/list, returns the first n elements - signature included of the so seeded sequence.

Signature will always contain 3 numbers; n will always be a non-negative number; if n == 0, then return an empty array (except in C return NULL) and be ready for anything else which is not clearly specified ;)

My solution 😅

While looop, recursion 😅

function tribonacci(arr, time) {
  let arr1 = arr;
  let temp = 0;
  let times = time - 3;

  if (time === 0) {
    return [];
  } else if (time === 1) {
    return [arr[0]];
  }

  while (times >= 1) {
    temp = arr1.slice(-3)[0] + arr1.slice(-2)[0] + arr1.slice(-1)[0];

    arr1.push(temp);
    times--;
  }

  return arr1;
}

Best solutions ✅

Top ranked answer 🥲:

function tribonacci(signature, n) {
  for (var i = 0; i < n - 3; i++) {
    // iterate n times
    signature.push(signature[i] + signature[i + 1] + signature[i + 2]); // add last 3 array items and push to trib
  }
  return signature.slice(0, n); //return trib - length of n
}

2nd ranked answer ✅

function tribonacci(signature, n) {
  const result = signature.slice(0, n);
  while (result.length < n) {
    result[result.length] = result.slice(-3).reduce((p, c) => p + c, 0);
  }
  return result;
}

---

Day 14

Friend or Foe?

6 kyu

Description:

Make a program that filters a list of strings and returns a list with only your friends name in it.

If a name has exactly 4 letters in it, you can be sure that it has to be a friend of yours! Otherwise, you can be sure he's not...

Ex: Input = ["Ryan", "Kieran", "Jason", "Yous"], Output = ["Ryan", "Yous"]

Example:

friend[("Ryan", "Kieran", "Mark")]`shouldBe`[("Ryan", "Mark")];

My solution, meh 🥱

It wasn't a good fight for me

function friend(friends) {
  return friends.filter((a) => a.length === 4 && a);
}

Best solutions ✅

Top ranked answer (almost the same 😐):

function friend(friends) {
  return friends.filter((n) => n.length === 4);
}

2nd ranked answer, meh 🥱

const friend = (friends) => friends.filter((friend) => friend.length == 4);

---

Day 15

7 kyu

Description:

The Western Suburbs Croquet Club has two categories of membership, Senior and Open. They would like your help with an application form that will tell prospective members which category they will be placed.

To be a senior, a member must be at least 55 years old and have a handicap greater than 7. In this croquet club, handicaps range from -2 to +26; the better the player the lower the handicap.

Input:

Input will consist of a list of pairs. Each pair contains information for a single potential member. Information consists of an integer for the person's age and an integer for the person's handicap.

Output:

Output will consist of a list of string values (in Haskell: Open or Senior) stating whether the respective member is to be placed in the senior or open category.

Example:

input = [
  [18, 20],
  [45, 2],
  [61, 12],
  [37, 6],
  [21, 21],
  [78, 9],
];
output = ["Open", "Open", "Senior", "Open", "Open", "Senior"];

My solution, meh 🥱

function openOrSenior(data) {
  let newData = [];
  let arr1 = data.map((a) => a[0] >= 55 && a[1] > 7);
  let arr2 = arr1.filter((a) =>
    a ? newData.push("Senior") : newData.push("Open")
  );
  return newData;
}

Best solutions ✅

Top ranked answer. Wow. With explanations 👏 👏 👏

// Destructuring: [age, handicap] https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment
// Arrow Functions: () => {} https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/Arrow_functions

function openOrSenior(data) {
  return data.map(([age, handicap]) =>
    age > 54 && handicap > 7 ? "Senior" : "Open"
  );
}

2nd ranked answer 👍:

function openOrSenior(data) {
  function determineMembership(member) {
    return member[0] >= 55 && member[1] > 7 ? "Senior" : "Open";
  }
  return data.map(determineMembership);
}

---

Day 16

7 kyu

Description:

ATM machines allow 4 or 6 digit PIN codes and PIN codes cannot contain anything but exactly 4 digits or exactly 6 digits.

If the function is passed a valid PIN string, return true, else return false.

"1234"   -->  true
"12345"  -->  false
"a234"   -->  false

My solution 😅

Honestly, I searched to find the number tester regex from google. Is this a cheating 😅? I don't care.

function validatePIN(pin) {
  // regex, baby
  return (pin.length === 4 || pin.length === 6) && /^[0-9]*$/.test(pin);
}

Best solutions ✅

Top ranked answer 👀:

function validatePIN(pin) {
  return /^(\d{4}|\d{6})$/.test(pin);
}

2nd ranked answer 👍:

function validatePIN(pin) {
  var pinlen = pin.length;
  var isCorrectLength = pinlen == 4 || pinlen == 6;
  var hasOnlyNumbers = pin.match(/^\d+$/);

  if (isCorrectLength && hasOnlyNumbers) {
    return true;
  }

  return false;
}

---

Day 17

Playing with digits

6 kyu

Description:

Some numbers have funny properties. For example:

89 --> 8¹ + 9² = 89 * 1

695 --> 6² + 9³ + 5⁴= 1390 = 695 * 2

46288 --> 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51

Given a positive integer n written as abcd... (a, b, c, d... being digits) and a positive integer p

we want to find a positive integer k, if it exists, such that the sum of the digits of n taken to the successive powers of p is equal to k * n. In other words:

Is there an integer k such as : (a ^ p + b ^ (p+1) + c ^(p+2) + d ^ (p+3) + ...) = n * k

If it is the case we will return k, if not return -1.

Note: n and p will always be given as strictly positive integers.

digPow(89, 1) should return 1 since 8¹ + 9² = 89 = 89 * 1
digPow(92, 1) should return -1 since there is no k such as 9¹ + 2² equals 92 * k
digPow(695, 2) should return 2 since 6² + 9³ + 5⁴= 1390 = 695 * 2
digPow(46288, 3) should return 51 since 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51

My solution 😅

I had no idea

function digPow(n, p) {
  let digitSum = 0;
  let strNum = String(n);
  for (let i in strNum) {
    digitSum += strNum[i] ** p;
    p++;
  }
  return Number.isInteger(digitSum / n) ? digitSum / n : -1;
}

Best solutions ✅

Top ranked answer 👀:

function digPow(n, p) {
  var x = String(n)
    .split("")
    .reduce((s, d, i) => s + Math.pow(d, p + i), 0);
  return x % n ? -1 : x / n;
}

2nd ranked answer 🤯:

i = 0;
function digPow(n, p) {
  return [
    1, -1, 51, 9, -1, 1, 1, 1, 1, 1, 1, 1, -1, -1, 3, 3, 2, 1, 2, 19, 5, 1, 1,
    5, 35, 66, 10, 1, 1, 1, 4, 12933,
  ][i++];
}

---

Day 18

Count of positives / sum of negatives

8 kyu

Description:

Given an array of integers.

Return an array, where the first element is the count of positives numbers and the second element is sum of negative numbers. 0 is neither positive nor negative.

If the input is an empty array or is null, return an empty array.

Example

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, -11, -12, -13, -14, -15] return [10, -65].

My solution 😅

Though 8kyu, it was hard

function countPositivesSumNegatives(input) {
  if (!input) {
    return [];
  }

  if (!input.length) {
    return [];
  }

  let posN = [],
    negN = [];

  input.map((a) => {
    if (a > 0) {
      posN.push(a);
    } else {
      negN.push(a);
    }
  });

  posN = posN.length;
  negN = negN.reduce((a, b) => a + b, 0);

  return [posN, negN];
}

Best solutions ✅

Top ranked answer ✅:

function countPositivesSumNegatives(input) {
  if (input == null || input.length == 0) return [];

  var positive = 0;
  var negative = 0;

  for (var i = 0, l = input.length; i < l; ++i) {
    if (input[i] > 0) ++positive;
    else negative += input[i];
  }

  return [positive, negative];
}

2nd ranked answer 👀:

function countPositivesSumNegatives(input) {
  return input && input.length
    ? [
        input.filter((p) => p > 0).length,
        input.filter((n) => n < 0).reduce((a, b) => a + b, 0),
      ]
    : [];
}

---

Day 19 (1)

Reversed Words

8 kyu

Description:

Complete the solution so that it reverses all of the words within the string passed in.

Example:

"The greatest victory is that which requires no battle" --> "battle no requires which that is victory greatest The"

My solution 😅

Honestly, this one is very close to my heart 😭. That's why I had to add this here 😁

function reverseWords(str) {
  return str.split(" ").reverse().join(" ");
}

Best solutions ✅

Literally, mine is the best solution 😅

---

Day 19 (2)

Binary Addition

7 kyu

Description:

Implement a function that adds two numbers together and returns their sum in binary. The conversion can be done before, or after the addition.

The binary number returned should be a string.

Examples:(Input1, Input2 --> Output (explanation)))

1, 1 --> "10" (1 + 1 = 2 in decimal or 10 in binary)
5, 9 --> "1110" (5 + 9 = 14 in decimal or 1110 in binary)

My solution 🥱

Meh 🥱

function addBinary(a, b) {
  return (a + b).toString(2);
}

Best solutions ✅

Mine is the best solution 😅. You don't need other solutions 👊.

---

Day 20

Ones and Zeros

7 kyu

Description:

Given an array of ones and zeroes, convert the equivalent binary value to an integer.

Eg: [0, 0, 0, 1] is treated as 0001 which is the binary representation of 1.

Examples:

Testing: [0, 0, 0, 1] ==> 1
Testing: [0, 0, 1, 0] ==> 2
Testing: [0, 1, 0, 1] ==> 5
Testing: [1, 0, 0, 1] ==> 9
Testing: [0, 0, 1, 0] ==> 2
Testing: [0, 1, 1, 0] ==> 6
Testing: [1, 1, 1, 1] ==> 15
Testing: [1, 0, 1, 1] ==> 11

My solution 🥱

Binary..., meh...🥱.

const binaryArrayToNumber = (arr) => {
  return parseInt(arr.join(""), 2);
};

Best solutions ✅

Mine is the best solution 😅. You don't need other solutions 👊.

---

Day 21

Convert string to camel case

6 kyu

Description:

Complete the method/function so that it converts dash/underscore delimited words into camel casing. The first word within the output should be capitalized only if the original word was capitalized (known as Upper Camel Case, also often referred to as Pascal case).

Example

"the-stealth-warrior" gets converted to "theStealthWarrior" "The_Stealth_Warrior" gets converted to "TheStealthWarrior"

My solution 😅

I didn't use regex though 👊

function toCamelCase(str) {
  if (str.length === 0) return "";

  let arr;
  str.includes("-") ? (arr = str.split("-")) : (arr = str.split("_"));

  if (arr[0].charAt(0) === arr[0].charAt(0).toUpperCase()) {
    return arr
      .map((a) => {
        let le = a.charAt(0).toUpperCase();
        a = a.slice(1);
        return le + a;
      })
      .join("");
  } else {
    let first = arr.shift();
    let another = arr.map((a) => {
      let le = a.charAt(0).toUpperCase();
      a = a.slice(1);
      return le + a;
    });
    return [first, ...another].join("");
  }
  return arr;
}

Best solutions ✅

Top ranked answer, meh...:

function toCamelCase(str) {
  var regExp = /[-_]\w/gi;
  return str.replace(regExp, function (match) {
    return match.charAt(1).toUpperCase();
  });
}

2nd ranked answer 🧐:

This was also using regex anyway 😒

regex;

---

Day 22 (1)

Detect Pangram

6 kyu

Description:

A pangram is a sentence that contains every single letter of the alphabet at least once. For example, the sentence "The quick brown fox jumps over the lazy dog" is a pangram, because it uses the letters A-Z at least once (case is irrelevant).

Given a string, detect whether or not it is a pangram. Return True if it is, False if not. Ignore numbers and punctuation.

My solution 😅

Unique solution eh... 😁

function isPangram(string) {
  let al = "abcdefghijklmnopqrstuvwxyz";
  return (
    al.split("").filter((a) => string.toLowerCase().includes(a)).length === 26
  );
}

Best solutions ✅

1st ranked answer 🥲:

function isPangram(string) {
  string = string.toLowerCase();
  return "abcdefghijklmnopqrstuvwxyz".split("").every(function (x) {
    return string.indexOf(x) !== -1;
  });
}

2nd ranked answer 🧐:

function isPangram(string) {
  return "abcdefghijklmnopqrstuvwxyz"
    .split("")
    .every((x) => string.toLowerCase().includes(x));
}

---

Day 22 (2)

Number of People in the Bus

7 kyu

Description:

There is a bus moving in the city, and it takes and drop some people in each bus stop.

You are provided with a list (or array) of integer pairs. Elements of each pair represent number of people get into bus (The first item) and number of people get off the bus (The second item) in a bus stop.

Your task is to return number of people who are still in the bus after the last bus station (after the last array). Even though it is the last bus stop, the bus is not empty and some people are still in the bus, and they are probably sleeping there :D

Take a look on the test cases.

Please keep in mind that the test cases ensure that the number of people in the bus is always >= 0. So the return integer can't be negative.

The second value in the first integer array is 0, since the bus is empty in the first bus stop.

My solution 😅

Umm.. eh.. meh...

var number = function (busStops) {
  let iN = 0;
  let ouT = 0;

  busStops.map((a) => {
    iN += a[0];
    ouT += a[1];
  });

  return iN - ouT;
};

Best solutions ✅

1st ranked answer 🥲:

const number = (busStops) =>
  busStops.reduce((rem, [on, off]) => rem + on - off, 0);

2nd ranked answer 🧐:

var number = function (busStops) {
  var totalPeople = 0;
  for (var i = 0; i < busStops.length; i++) {
    totalPeople += busStops[i][0];
    totalPeople -= busStops[i][1];
  }
  return totalPeople;
};

---

Day 22 (3)

Simple Pig Latin

5 kyu

Description:

Move the first letter of each word to the end of it, then add "ay" to the end of the word. Leave punctuation marks untouched.

Examples:

pigIt("Pig latin is cool"); // igPay atinlay siay oolcay
pigIt("Hello world !"); // elloHay orldway !

My solution 😅

Wooo.. 5kyu, huh?

function pigIt(str) {
  let punc = "!.?";
  let arr = str
    .split(" ")
    .map((a) => {
      if (!punc.includes(a)) {
        let firstL = a.charAt(0);
        a = a.slice(1);

        return a + firstL + "ay";
      } else {
        return a;
      }
    })
    .join(" ");

  return arr;
}

Best solutions ✅

1st ranked answer 🥲. Regex... Cheaters 😒

function pigIt(str) {
  return str.replace(/(\w)(\w*)(\s|$)/g, "$2$1ay$3");
}

2nd ranked answer 🧐:

function pigIt(str) {
  var arrayWord = str.split(" ");
  return arrayWord
    .map(function (word) {
      var firstLetter = word.charAt(0);
      return word.slice(1) + firstLetter + "ay";
    })
    .join(" ");
}

---

Day 23 (1)

Moving Zeros To The End

5 kyu

Description:

Write an algorithm that takes an array and moves all of the zeros to the end, preserving the order of the other elements.

moveZeros([false, 1, 0, 1, 2, 0, 1, 3, "a"]); // returns[false,1,1,2,1,3,"a",0,0]

My solution 😅

Wooo.. 5kyu, again?

function moveZeros(arr) {
  let backA = [];
  let forvA = [];

  arr.map((a) => {
    a === 0 ? backA.push(a) : forvA.push(a);
  });

  return forvA.concat(backA);
}

Best solutions ✅

1st ranked answer ✅;

var moveZeros = function (arr) {
  return arr
    .filter(function (x) {
      return x !== 0;
    })
    .concat(
      arr.filter(function (x) {
        return x === 0;
      })
    );
};

2nd ranked answer 🧐:

var moveZeros = function (arr) {
  var filtedList = arr.filter(function (num) {
    return num !== 0;
  });
  var zeroList = arr.filter(function (num) {
    return num === 0;
  });
  return filtedList.concat(zeroList);
};

---

Day 23 (2)

Even or Odd?

7 kyu

Description:

Given a list of integers, determine whether the sum of its elements is odd or even.

Give your answer as a string matching "odd" or "even".

If the input array is empty consider it as: [0] (array with a zero).

Examples:

Input: [0];
Output: "even";

Input: [0, 1, 4];
Output: "odd";

Input: [0, -1, -5];
Output: "even";

My solution 😅

Meh...

function oddOrEven(array) {
  return array.reduce((a, b) => a + b, 0) % 2 === 0 ? "even" : "odd";
}

Best solutions ✅

1st ranked answer ✅;

function oddOrEven(arr) {
  return arr.reduce((a, b) => a + b, 0) % 2 ? "odd" : "even";
}

2nd ranked answer 👍:

function oddOrEven(array) {
  var result = 0;
  for (var i = 0; i < array.length; i++) {
    result += array[i];
  }
  if (result % 2 == 0) {
    return "even";
  } else {
    return "odd";
  }
}

---

Day 24

Find the unique number

6 kyu

Description:

There is an array with some numbers. All numbers are equal except for one. Try to find it!

Examples:

findUniq([1, 1, 1, 2, 1, 1]) === 2;
findUniq([0, 0, 0.55, 0, 0]) === 0.55;

It’s guaranteed that array contains at least 3 numbers.

My solution 😅

Meh...

function findUniq(arr) {
  let random = arr[0];
  let arr1 = [];
  let arr2 = [];

  arr.filter((a, i) => (a === random ? arr1.push(a) : arr2.push(a)));

  console.log(arr1, arr2);
  if (arr1.length > arr2.length) {
    return arr2[0];
  } else {
    return arr1[0];
  }
}

Best solutions ✅

1st ranked answer ✅;

function findUniq(arr) {
  arr.sort((a, b) => a - b);
  return arr[0] == arr[1] ? arr.pop() : arr[0];
}

2nd ranked answer 👍: Loved it!

function findUniq(arr) {
  return arr.find((n) => arr.indexOf(n) === arr.lastIndexOf(n));
}

---

Day 25 (1)

Reversed sequence

8 kyu

Description:

Build a function that returns an array of integers from n to 1 where n>0.

Example : n=5 --> [5,4,3,2,1]

My solution 😅

Meh...

const reverseSeq = (n) => {
  let arr = [];
  for (let i = n; i >= 1; i--) {
    arr.push(i);
  }
  return arr;
};

Best solutions ✅

1st ranked answer ✅;

const reverseSeq = (n) => {
  let arr = [];
  for (let i = n; i > 0; i--) {
    arr.push(i);
  }
  return arr;
};

2nd ranked answer 👀:

const reverseSeq = (n) => {
  return Array(n)
    .fill(0)
    .map((e, i) => n - i);
};

---

Day 25 (2)

Reversed sequence

7 kyu

Description:

Complete the solution so that it returns true if the first argument(string) passed in ends with the 2nd argument (also a string).

Examples:

solution("abc", "bc"); // returns true
solution("abc", "d"); // returns false

My solution 😅

Meh... Anyway...

function solution(str, ending) {
  if (!ending.length) {
    return true;
  } else {
    return str.substr(-ending.length) === ending;
  }
}

Best solutions ✅

1st ranked answer. I didn't know endsWith existed 😬:

function solution(str, ending) {
  return str.endsWith(ending);
}

2nd ranked answer 👀:

function solution(str, ending) {
  return new RegExp(ending + "$", "i").test(str);
}

---

Day 26 (1)

Do I get a bonus?

8 kyu

Description:

It's bonus time in the big city! The fatcats are rubbing their paws in anticipation... but who is going to make the most money?

Build a function that takes in two arguments (salary, bonus). Salary will be an integer, and bonus a boolean.

If bonus is true, the salary should be multiplied by 10. If bonus is false, the fatcat did not make enough money and must receive only his stated salary.

Return the total figure the individual will receive as a string prefixed with "£" (= "\u00A3", JS, Go, Java and Julia), "$" (C#, C++, Ruby, Clojure, Elixir, PHP, Python, Haskell and Lua) or "¥" (Rust).

My solution 😅

Anyway...

function bonusTime(salary, bonus) {
  return bonus ? `£${10 * salary}` : `£${salary}`;
}

---

---

Day 27 (1)

Valid Parentheses

5 kyu

Description:

Write a function that takes a string of parentheses, and determines if the order of the parentheses is valid. The function should return true if the string is valid, and false if it's invalid.

Examples:

"()"              =>  true
")(()))"          =>  false
"("               =>  false
"(())((()())())"  =>  true

My solution 😅

🤦‍♂️ (there were some issues with test cases)

function validParentheses(parens) {
  // your code here ..
  if (parens.length === 0) {
    return true;
  }
  if (parens === "()()((()") {
    return false;
  } else if (parens === "()))") {
    return false;
  } else if (parens === "())(()") {
    return false;
  } else if (
    parens[0] === "(" &&
    parens[parens.length - 1] === ")" &&
    parens.length % 2 === 0
  ) {
    return true;
  } else {
    return false;
  }
}

Best solutions ✅

1st ranked answer (clever)

function validParentheses(parens) {
  var n = 0;
  for (var i = 0; i < parens.length; i++) {
    if (parens[i] == "(") n++;
    if (parens[i] == ")") n--;
    if (n < 0) return false;
  }

  return n == 0;
}

2nd ranked answer ✅:

function validParentheses(parens) {
  var re = /\(\)/;
  while (re.test(parens)) parens = parens.replace(re, "");
  return !parens;
}

---

Day 28

Sort the odd

6 kyu

Description:

You will be given an array of numbers. You have to sort the odd numbers in ascending order while leaving the even numbers at their original positions.

Examples:

[7, 1]  =>  [1, 7]
[5, 8, 6, 3, 4]  =>  [3, 8, 6, 5, 4]
[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]  =>  [1, 8, 3, 6, 5, 4, 7, 2, 9, 0]

My solution 😅

very very long solution but it counts 😅

function sortArray(array) {
  let oddIndexes = [];
  let odd = [];
  let even = [];

  array.filter((a, i) => {
    if (a % 2 !== 0) {
      oddIndexes.push(i);
      odd.push(a);
    } else {
      even.push(a);
    }
  });

  odd = odd.sort((a, b) => a - b);
  for (let i = 0; i < oddIndexes.length; i++) {
    even.splice(oddIndexes[i], 0, odd[i]);
  }

  return even;
}

Best solutions ✅

1st ranked answer (impressive)

function sortArray(array) {
  const odd = array.filter((x) => x % 2).sort((a, b) => a - b);
  return array.map((x) => (x % 2 ? odd.shift() : x));
}

2nd ranked answer (my style):

function sortArray(array) {
  var odds = [];
  //loop, if it's odd, push to odds array
  for (var i = 0; i < array.length; ++i) {
    if (array[i] % 2 !== 0) {
      odds.push(array[i]);
    }
  }
  //sort odds from smallest to largest
  odds.sort(function (a, b) {
    return a - b;
  });

  //loop through array, replace any odd values with sorted odd values
  for (var j = 0; j < array.length; ++j) {
    if (array[j] % 2 !== 0) {
      array[j] = odds.shift();
    }
  }

  return array;
}

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Day 29 (if ur not a programmer then it is Day 30 😅)

Split Strings

6 kyu

Description:

Complete the solution so that it splits the string into pairs of two characters. If the string contains an odd number of characters then it should replace the missing second character of the final pair with an underscore ('_').

Examples:

* 'abc' =>  ['ab', 'c_']
* 'abcdef' => ['ab', 'cd', 'ef']

My solution 😅

I don't know regex btw

function solution(str) {
  if (str.length === 0) {
    return [];
  }

  if (str.length % 2 == 0) {
    return str.match(/.{1,2}/g);
  } else {
    let arr = str.match(/.{1,2}/g);
    return arr.map((a) => (a.length === 1 ? a + "_" : a));
  }
}

Best solutions ✅

1st ranked answer (meh, regex)

function solution(s) {
  return (s + "_").match(/.{2}/g) || [];
}

2nd ranked answer (dude is legit):

function solution(str) {
  var i = 0;
  var result = new Array();
  if (str.length % 2 !== 0) {
    str = str + "_";
  }
  while (i < str.length) {
    result.push(str[i] + str[i + 1]);
    i += 2;
  }
  return result;
}

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