matrix-spiral.md

Matrix Spiral

Create a function that given integer n returns an n * n spiraling matrix.

//  matrix(3)
//
//  [[1, 2, 3],
//  [8, 9, 4],
//  [7, 6, 5]]

//  matrix(4)
//
//   [[1,   2,  3, 4],
//   [12, 13, 14, 5],
//   [11, 16, 15, 6],
//   [10,  9,  8, 7]]

Overview

Initiate the matrix by pushing n number of arrays into an array.

• Create a counter variable to keep track of the number we're adding to the matrix.
• Create variables for columns and rows boundaries at their start and end.
• Create an outer loop that runs while the starting column and starting row are less than or equal to their ending counterparts.
• Create an inner loop for each side of the matrix starting clockwise for the first side:
  • Loop and add the counter value to the side of the matrix. Increment the counter variable at each loop.
  • "Shrink" that side's boundary.

It can help to see an animation of this process in effect:

Code

function matrix(n) {
  const results = [];

  for (let i = 0; i < n; i++) {
    results.push([]);
  }

  let counter = 1;
  let startColumn = 0;
  let endColumn = n - 1;
  let startRow = 0;
  let endRow = n - 1;
  while (startColumn <= endColumn && startRow <= endRow) {
    // Top row
    for (let i = startColumn; i <= endColumn; i++) {
      results[startRow][i] = counter;
      counter++;
    }
    startRow++;

    // Right column
    for (let i = startRow; i <= endRow; i++) {
      results[i][endColumn] = counter;
      counter++;
    }
    endColumn--;

    // Bottom row
    for (let i = endColumn; i >= startColumn; i--) {
      results[endRow][i] = counter;
      counter++;
    }
    endRow--;

    // Start column
    for (let i = endRow; i >= startRow; i--) {
      results[i][startColumn] = counter;
      counter++;
    }
    startColumn++;
  }

  return results;
}