Spiral Matrix in Javascript

#javascript #webdev #programming #coding

This is one interesting interview question.

Generally, we know the matrix of any size but what is special about Spiral here.

2 * 2 Matrix can be:

[[1,2],
[3,4]]

So, it spiral version is:

[[1,2],
[4,3]]

3 * 3 Matrix can be:
[[1,2,3],
[4,5,6],
[7,8,9]]

So, it's spiral version is:
[[1,2,3],
[8,9,4],
[7,6,5]]

An image can explain better rather much long theories. I tried it using below image.

Ok. Now, I implemented this using javascript.

Here, is the code:



function spiralMatrix(n) {
    const arr = Array.from({ length: n }, () => []);
    let row = 0;
    let col = 0;
    let rowEnd = n - 1;
    let colEnd = n - 1;
    let counter = 1;
    while (col <= colEnd && row <= rowEnd) {

        // Top row & middle value (Where col === colEnd && row === rowEnd)
        for (let i = col; i <= colEnd; i++) {
            arr[row][i] = counter;
            counter++;
        }
        row++;

        // end column
        for (let i = row; i <= rowEnd; i++) {
            arr[i][colEnd] = counter;
            counter++;
        }
        colEnd--;

        // end row
        for (let i = colEnd; i >= col; i--) {
            arr[rowEnd][i] = counter;
            counter++;
        }
        rowEnd--;

        // First col from end
        for (let i = rowEnd; i >= row; i--) {
            arr[i][col] = counter;
            counter++;
        }
        col++;
    }
    return arr;
}

I would love to explain how I achieved it.

Here is the GIF that makes you get the Pseudo code behind this code.

Meanwhile, If some expert came across this article. Please do minify the above code.

Thanks.

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Top comments (6)

peerreynders • Apr 16 '22 • Edited

//           y
//           |
//    16 15 14 13 12
//    17  4  3  2 11
// -- 18  5  0  1 10 --- x
//    19  6  7  8  9
//    20 21 22 23 24
//           |
//
// Given coordinates (x,y) return "spiral index". Source:
// https://stackoverflow.com/questions/9970134/get-spiral-index-from-location/9971465#9971465
//
function spiralIndex(x, y) {
  const y2 = y * y;
  const x2 = x * x;
  if (y2 >= x2) {
    const value = 4 * y2 - y - x;
    return y >= x ? value : value - 2 * (y - x);
  }
  const value = 4 * x2 - y - x;
  return y >= x ? value : value + 2 * (y - x);
}

function makeToValue(n) {
  const n2 = n * n;
  if (n % 2 === 0) {
    const baseY = n / 2;
    const baseX = baseY - 1;
    return (row, column) => n2 - spiralIndex(column - baseX, baseY - row);
  }

  const base = (n - 1) / 2;
  return (row, column) => n2 - spiralIndex(base - column, row - base);
}

function spiralMatrix(n) {
  const toValue = makeToValue(n);
  const config = { length: n };
  return Array.from(config, (_vr, row) =>
    Array.from(config, (_vc, column) => toValue(row, column))
  );
}

console.log(spiralMatrix(3)); // [[1, 2, 3], [8, 9, 4], [7, 6, 5]]
console.log(spiralMatrix(4)); // [[1, 2, 3, 4], [12, 13, 14, 5], [11, 16, 15, 6], [10, 9, 8, 7]]

This seems to work as well

function spiralMatrix(n) {
  if (n < 1) return [];

  const matrix = Array.from({ length: n }, (_v, _i) => []);
  let row = 0, column = 0, value = 1;
  for (let low = 0, high = n - 1; low <= high; high -= 1) {
    for (; column < high; matrix[row][column] = value, column += 1, value += 1);
    for (; row < high; matrix[row][column] = value, row += 1, value += 1);
    for (; column > low; matrix[row][column] = value, column -= 1, value += 1);
    for (low += 1; row > low; matrix[row][column] = value, row -= 1, value += 1);
  }
  matrix[row][column] = value;
  return matrix;
}

console.log(spiralMatrix(3)); // [[1, 2, 3], [8, 9, 4], [7, 6, 5]]
console.log(spiralMatrix(4)); // [[1, 2, 3, 4], [12, 13, 14, 5], [11, 16, 15, 6], [10, 9, 8, 7]]}