Checkerboard - FreeCodeCamp #130 Daily Challenge

2025-12-18 2 min

Table of Contents

Checkerboard - Analysis & Explanation

Problem Statement

Given an array with two numbers, the first indicates the number of rows and the second the number of columns. You must return a matrix (array of arrays) filled with alternating “X” and “O”, like a checkerboard. The top-left corner is always “X”.

Example for [3, 3]:

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

Initial Analysis

The challenge is to build a two-dimensional matrix that simulates a checkerboard, alternating “X” and “O” in each cell. The pattern must be maintained in both rows and columns, alternating characters in every position.

Test Cases

[3, 3]: [[“X”, “O”, “X”], [“O”, “X”, “O”], [“X”, “O”, “X”]]
[6, 1]: [[“X”], [“O”], [“X”], [“O”], [“X”], [“O”]]
[2, 10]: [[“X”, “O”, …], [“O”, “X”, …]]
[5, 4]: [[“X”, “O”, “X”, “O”], [“O”, “X”, “O”, “X”], …]

Solution Approach

The key is to iterate through the matrix and decide the value of each cell based on the sum of its indices: if the sum is even, place “X”; if odd, place “O”. This efficiently generates the alternating pattern.

Why does it work?

By adding the row and column indices, we get a value that determines the parity of the position. Even sums correspond to “X” and odd sums to “O”, creating the desired pattern. For example:

(0,0) → 0 + 0 = 0 (even) → “X”
(0,1) → 0 + 1 = 1 (odd) → “O”
(1,0) → 1 + 0 = 1 (odd) → “O”
(1,1) → 1 + 1 = 2 (even) → “X”

Implementation

function createBoard(dimensions) {
  // We use destructuring to get rows and columns
  const [rows, cols] = dimensions
  // Initialize an empty matrix with the given number of rows and columns
  const board = new Array(rows).fill(null).map(() => new Array(cols))
  // Iterate through each cell to assign "X" or "O"
  for (let i = 0; i < rows; i++) {
    for (let j = 0; j < cols; j++) {
      // Assign "X" if the sum of indices is even, "O" if odd
      board[i][j] = (i + j) % 2 === 0 ? 'X' : 'O'
    }
  }
  return board
}

Complexity

Time: $O(n \times m)$ , where $n$ is rows and $m$ is columns.
Space: $O(n \times m)$ , for the generated matrix.

Edge Cases

If any dimension is zero, return an empty matrix.
For [1, 1], return [[“X”]].
Works for any positive size.

Reflections

Use of nested loops and modular arithmetic.
Efficient initialization of 2D arrays in JavaScript.
The solution is optimal: each cell is calculated in constant time.

Resources

freecodecamp daily-challenge