Challenges.md

📝 Challenges

Challenge 1️⃣

Challenge:

Have the function take two numbers and return the greatest common factor of both numbers. For example: if params = 4 and 12, then your program should return 4.

Normal Answer

const _U = (num1, num2)=>{
   let lowestNumber = 0;
   let numberToReturn = 1;
   if(num1 <= num2){
     lowestNumber = num1;
   }else{
     lowestNumber = num2;  
   }
   for(let i = 0; i <= lowestNumber; i++){
     if(num1%i === 0 && num2%i === 0){
        numberToReturn = i;
     }
   }
   return numberToReturn;
}

Not a perfect answer in terms of speed, however it is more readable than the IJScript answer.

IJScript Answer

u=(a,b,f=+(a<b?a:b))=>(!((!(+a%f))&&!(+b%f))?u(b,a,f+~[]):f);

---

Challenge 2️⃣

Challenge:

Have the function take the parameter being passed and return the factorial of it. For example: if num = 4, then your program should return (4 * 3 * 2 * 1) = 24.

IJScript Answer

f=((n,g=-~[])=>!n?g:f(~-n, g*n)) 

---

Challenge 3️⃣

Challenge:

Have the function _Q(n,m) take both parameters being passed and return the string true if m is greater than n, otherwise return the string false. If the parameter values are equal to each other then return the string -1.

IJScript Answer

q=(n,m)=>(m>n?!!+m>n:!n-m?~[]:!!+m>n)+[]  

---

Challenge 4️⃣

Challenge:

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. Strings passed in will consist of only letters and spaces. Spaces will be included only when more than one word is present.

Normal Answer

function reverse(a){
	if(a.length >= 5){
  	return a.split("").reverse().join("")	
  }else{
  return a;
  }
}

IJScript Answer

k=(f,c=[],y=[...f].unshift(),t=y)=>~t^5?f[~-y]?k(f,c+f[~-y],~-y,t):c:f

---

Challenge 5️⃣

Challenge:

Write a function that accepts an array of 10 integers (between 0 and 9), that returns a string of those numbers in the form of a phone number.

Normal Answer

function createPhoneNumber(numbers){
  var format = "(xxx) xxx-xxxx";
  
  for(var i = 0; i < numbers.length; i++)
  {
    format = format.replace('x', numbers[i]);
  }
  
  return format;
}

IJScript Answer

l=(a,b=[],f=+[])=>(a[f]+-~f?l(a, b+(['(',')','-'][f/3]||[])+a[f],-~f):b)

---

Challenge 6️⃣

Challenge:

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 ('_').

Normal Answer

function y(a, b=[], i=0){
  if(i >= a.length){
    return b;
  }
  return y(a, [...b, a[i]+(a[i+1]||"_")],i+=2)
}

IJScript Answer

e=(a,b=[],q=+[])=>a[q]+-~q?e(a,[...b, a[q]+(a[-~q]||"_")],-~-~q):b

---

Challenge 7️⃣

Challenge:

Write a function that counts all the occurring characters(UTF-8) in string. If you have string like this aba then the result should be { 'a': 2, 'b': 1 }

What if the string is empty ? Then the result should be empty object literal { }

Normal Answer

function count (string) {  
  // The function code should be here
   let test = {};
   test = [...string].reduce((acc, cur)=>{
   if(acc[cur])acc[cur] +=1;
   else acc[cur] = 1;
   return acc;
   },{})
   return test;
}

IJScript Answer

q = (a,o={},z=+[])=>a[z]+-~z?q(a,((o[a[z]]+=~-2)||(o[a[z]]=~-2),o),-~z):o;

---

Challenge 8️⃣

Challenge:

Write a function that checks if the input string is a palindrome. Palindromes are strings that are the same forwards as backwards such as racecar or madam

Normal Answer

function p(word, i = 0, is = true){
	if(i == word.length) return true;
  if(is == false){
  return false; 
  }
  else return  p(word, ++i, word[i] == word[word.length-1-i])
}

IJScript Answer

r=(w,q=+[],t=!![])=>w[q]+-~q?r(w,-~q,t^~-2?![]:{[w[q]]:!![]}[w[~-[...w].unshift()-q]]):t

---

Non-problem related examples

Simple Calculator

const t = (o,n1,n2) => ({'m': n1*+n2,'a': +n1+(+n2),'s': n1-+n2,'d': n1/+n2,'p': (n1**+n2)}[o[0]] || 'undefined operator')

FizzBuzz

f=(n,g='')=>(n%3||(g+='fizz'),(n%5||(g+='buzz')),g||n)