Table Of Contents

• Sort array by string length
  • 📃 Description
  • 🤔 Thinking
  • 👨‍💻 Code
• Descending Order
  • 📃 Description
  • 🤔 Thinking
  • 👨‍💻 Code
• Playing with digits
  • 📃 Description
  • 🤔 Thinking
  • 👨‍💻 Code
  • 🐞 Bugs
  • 🏁 Finally

Sort array by string length: ✍ by Steve hopkinson

📃 Description

Write a function that takes an array of strings as an argument and returns a sorted array according to the string length Ascendingly

🤔 Thinking

I didn't have to think that much.
I know how sort works it takes a compare function or two arguments the first is the first element and the second is the second element and it keeps increment the index like the map or the filter methods

👨‍💻 Code

const sortByLength = arr => arr.sort((x, y) => x.length - y.length);

Descending Order : ✍ by TastyOs

📃 Description

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.

🤔 Thinking

It's pretty much like the last one
I will make the number an array then sort it then make it a number again It's easy

👨‍💻 Code

const descendingOrder = num => {
  const digitArr = Array.from(String(num), Number);
  const sortedArr = digitArr.sort((x, y) => y - x);
  const sorted = Number(sortedArr.join(""));

  return sorted;
}

Playing with digits : ✍ by g964

📃 Description

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 as 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.

🤔 Thinking

I will make the first number n an array, like in the last kata.
Then I will map over it and power it with p + index index is the second argument in the map method.
Then Add it all.
Then check if it's equal to n * k

👨‍💻 Code

const digPow = (n, p) => {
  const k = n * p;
  const numArr = Array.from(String(n), Number);
  const sum = numArr.reduce((acc, elm, i) => (elm ** (p + i)) + acc);  
  const result = sum === k ? p : -1;

  return result
}

🐞 Bugs

I don't understand what he want from this description

🏁 Finally

After many times trying and trying and reading the discussion section I gave up and googled it but I said to myself if I keep doing that I will never be better and I won't be better, So I didn't look at the solutions and reminded myself that I don't have to cheat so when I write this post it will be bugless

If I didn't have any bugs, I didn't learn anything new

So I unlocked the solutions and looked at them and then I submit this

const digPow = (n, p) => {
  const numArr = Array.from(String(n), Number);
  const sum = numArr.reduce((acc, elm, i) => acc + (elm ** (p + i)), 0);  
  const result = sum === (n * p) ? p : -1;

  return sum % n ? -1 : sum / n;
}

Sorry for the similar katas, But I will take it as today was my num to arr / sort day