JavaScript Katas: Split In Parts

Intro 🌐

Problem solving is an important skill, for your career and your life in general.

That's why I take interesting katas of all levels, customize them and explain how to solve them.

---

Understanding the Exercise❗

First, we need to understand the exercise!
If you don't understand it, you can't solve it!.

My personal method:

1. Input: What do I put in?
2. Output: What do I want to get out?

---

Today's exercise

Today, another 7 kyu kata,
meaning we slightly increase the difficulty.

Source: Codewars

Write a function splitInParts, that accepts two parameters: myString and partLength.

Given a string, e.g. "HelloDev",
and a number, e.g. 3,
return the input string split into partLength-long parts separated by a space, e.g. Hel loD ev.

---

Input: a string and a number.

Output: a string.

---

Thinking about the Solution 💭

I think I understand the exercise (= what I put into the function and what I want to get out of it).

Now, I need the specific steps to get from input to output.

I try to do this in small baby steps:

1. Get a partLength-long part of the input string and add a space
2. Do this for every part
3. Add the remaining part to the end
4. Return the string

---

Example:

• Input: "HelloDev", 3
• Get a 3-long part of the input string and add a space: "Hel "
• Get a 3-long part of the input string and add a space: "loD "
• Add the remaining part to the end: "ev"
• Return the string: "Hel loD ev"
• Output: "Hel loD ev" ✅

---

Implementation ⛑

function splitInParts(myString, partLength) {
  let remaining = myString;
  let result = "";

  // do it only if the remaining string is longer than the parts
  while (remaining.length >= partLength) {
    // add the next part and a space to the result
    result += remaining.slice(0, partLength) + " ";

    // remove the added part from the remaining string
    remaining = remaining.slice(partLength);
  }

  // add the last part that was smaller than the part length
  result += remaining;

  // remove a trailing space
  return result.trim();
}

Result

console.log(splitInParts("HelloDev", 3));
// "Hel loD ev" ✅

console.log(splitInParts("HelloDev", 1));
// "H e l l o D e v" ✅

---

Playground ⚽

You can play around with the code here

---

Next Part ➡️

Great work!

We learned how to use while, slice, trim.

I hope you can use your new learnings to solve problems more easily!

Next time, we'll solve another interesting kata. Stay tuned!

---

If I should solve a specific kata, shoot me a message here.

If you want to read my latest stuff, get in touch with me!

---

Further Reading 📖

• while
• slice
• trim

---

Questions ❔

• How often do you do katas?
• Which implementation do you like more? Why?
• Any alternative solution?

Comments

[Gerges Nady]

let splitInParts = (myString, partLength) => {
let res = ""
let count = 0

for (let i = 0; i < myString.length; i += partLength) {
    for (let j = i; j < (partLength + count); j++) {
        if (myString[j] !== undefined) {
            res += myString[j]
        }
    }
    count = (count + partLength)
    res += " "
}
console.log(res)

}

[Tim West]

Another golf-style solution that doesn't abuse regular expressions 😎

const splitInParts = (s, l) => {
    s.split('').map((c,i) => (i+1) % l ? c : `${c} `).join('').trim();
}

Here's the less abusive solution I would actually use:

const splitInParts = (s, partLength) => {
    const result = [];

    for (let start = 0; start < s.length; start += partLength)
        result.push(s.slice(start, start + partLength));

    return result.join(' ');
}

This works because String.prototype.slice() will stop at the end of the string if the second argument index is out of bounds.

[Joseph Mawa]

Another possible solution.

function splitInParts(myString, partLength) {
  let ans = "",
    i = 0;
  while (i < myString.length) {
    ans += myString.slice(i, i + partLength) + " ";
    i += partLength;
  }
  return ans.slice(0, -1);
}

[Kostia Palchyk]

Okay, this one is easy enough for me..

const splitInParts =
  (s, n) => s.replace(new RegExp(`.{${n}}`, 'g'), '$& ')