One Pager Cheat Sheet

• Print the elements of an m * n matrix array in a spiral order, using O(1) space and O(n*m) time complexity, with all values between -1000000000 and 1000000000.
• Traversing an m * n matrix in spiral order requires transitioning from the outer rings to the inner rings, while keeping track of the outer left and right columns, and the upper and bottom rows, with colLower, colUpper, rowLower, and rowUpper variables.
• We can traverse an entire matrix in spiral order by following an algorithm of moving right, then down, then left, then up, and incrementing rowLower and decrementing colUpper after each iteration.
• The algorithm to traverse an m * n matrix in spiral order has a linear time complexity and fixed space complexity of O(1).

This is our final solution.

To visualize the solution and step through the below code, click Visualize the Solution on the right-side menu or the VISUALIZE button in Interactive Mode.

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}

var assert = require('assert');

​

function spiraltraverse(inmatrix) {

  if (!inmatrix.length) return [];

  const res = [];

  const dirs = [

    [0, 1],

    [1, 0],

    [0, -1],

    [-1, 0],

  ];

  const scope = [inmatrix[0].length, inmatrix.length - 1];

  let d = 0,

    r = 0,

    c = -1;

  while (scope[d % 2] > 0) {

    for (let i = 0; i < scope[d % 2]; i++) {

      r += dirs[d][0];

      c += dirs[d][1];

      res.push(inmatrix[r][c]);

    }

    scope[d % 2]--;

    d = (d + 1) % 4;

  }

  return res;

}

​

try {

  assert.equal(

    spiraltraverse([

      [1, 2, 3, 4, 5, 6],

      [7, 8, 9, 10, 11, 12],

That's all we've got! Let's move on to the next tutorial.

If you had any problems with this tutorial, check out the main forum thread here.