JavaScript Katas: Draw Chessboard

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.

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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?

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Today's exercise

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

Source: Codewars

Write a function drawChessboard, that accepts two parameters: rows and columns.

Given a rows number, e.g. 3, and a columns number, e.g. 3,
return a chessboard as a two dimensional array, e.g.

[
  [ "O", "X", "O" ],
  [ "X", "O", "X" ],
  [ "O", "X", "O" ]
]

White should be represented by "O", black by "X", the first row starts with a "O".

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Input: two numbers.

Output: an array of array(s).

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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. Create a row with the correct length (= columns) that starts with "O"
2. Create a row with the correct length (= columns) that starts with "X"
3. Do this alternately until there are enough (= rows) rows
4. Return the array of array(s)

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Example:

• Input: 3, 3
• Create a row with the correct length (= columns) that starts with "O": [ "O", "X", "O" ]
• Create a row with the correct length (= columns) that starts with "X": [ "X", "O", "X" ]
• Do this alternately until there are enough (= rows) rows: 1 additional row (=> 3 - 2 = 1) : [ "O", "X", "O" ]
• Return the array of array(s): [ [ "O", "X", "O" ], [ "X", "O", "X" ], [ "O", "X", "O" ]]
• Output: [ [ "O", "X", "O" ], [ "X", "O", "X" ], [ "O", "X", "O" ]] ✅

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Implementation (Explicit) ⛑

function drawChessboard(rows, columns) {
  // Create an empty board with the correct amount of rows
  const board = [...Array(rows)].map((_) => []);

  // Create a row with the correct length that starts with "O"
  const rowO = [...Array(columns)].map((_, i) => (i % 2 ? "X" : "O"));

  // Create a row with the correct length that starts with "X"
  const rowX = [...Array(columns)].map((_, i) => (i % 2 ? "O" : "X"));

  // Add the proper row to each board row
  return board.map((_, i) => (i % 2 ? rowX : rowO));
}

Result

console.log(drawChessboard(3, 3));
/*
[
  [ "O", "X", "O" ],
  [ "X", "O", "X" ],
  [ "O", "X", "O" ]
]
*/
// ✅

console.log(drawChessboard(2, 4));
/*
[
  [ "O", "X", "O", "X" ],
  [ "X", "O", "X", "O" ]
]
*/
//  ✅

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Implementation (Implicit) ⛑

function drawChessboard(rows, columns) {
  return [...Array(rows)]
    .map((_) => [])
    .map((_, i) =>
      i % 2
        ? [...Array(columns)].map((_, i) => (i % 2 ? "O" : "X"))
        : [...Array(columns)].map((_, i) => (i % 2 ? "X" : "O"))
    );
}

Result

console.log(drawChessboard(3, 3));
/*
[
  [ "O", "X", "O" ],
  [ "X", "O", "X" ],
  [ "O", "X", "O" ]
]
*/
// ✅

console.log(drawChessboard(2, 4));
/*
[
  [ "O", "X", "O", "X" ],
  [ "X", "O", "X", "O" ]
]
*/
//  ✅

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Playground ⚽

You can play around with the code here

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Next Part ➡️

Great work!

We learned how to use ..., Array, map.

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

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

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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!

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Further Reading 📖

• ...
• Array
• map

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Questions ❔

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

Comments

[Kostia Palchyk]

👍

I think, your second solution can even be a bit shortened:

function drawChessboard(rows, columns) {
  return [...Array(rows)]
    .map((_, i) =>
      [...Array(columns)].map((_, j) => ((j+i) % 2 ? "X" : "O"))
    );
}

[Gerges Nady]

let arr = new Array()
let j = 0
let defArr = []
while (j <= m) {
defArr.push("o")
defArr.push("x")
j++
}
let x = defArr.slice(0, m)
let y = defArr.reverse().slice(0, m)

let count = 1
for (let i = 0; i < n; i += 2) {
    arr[i] = x
    if (count < n) {
        arr[count] = y
        count += 2
    }
}
console.log(arr)

[miku86]

Hey Gerges,

great work solving all these katas.
Looking forward to seeing some other solutions!

[Gerges Nady]

Ok, thanks

[Kostia Palchyk]

Couldn't let this kata go out of my mind, wanted to try something alternative:

const drawChessboard = (r, c) => [
    ...{ * [Symbol.iterator]() {
      for (let i = 0; i < r; i++) yield [
        ...{ * [Symbol.iterator]() {
          for (let j = i; j < c + i; j++) yield ['X', 'O'][j % 2]
        }}
      ]
    }}
  ]

[lbermudez]

@kosich I've fixed and simplified your approach (thanks because it help me to open mind with generators + spread operator):

const drawChessboard = (r, c) => [
    ...(function* () {
        for (let i = 0; i < r; i++)
            yield [
                ...(function* () {
                    for (let j = 0; j < c; j++)
                        yield ['O', 'X'][(i + j) % 2];
                })(),
            ];
    })(),
];