Spiral Matrix v-1

Write a function that accepts an integer N and returns a NxN spiral matrix.’

Example:
matrix(2)
[[1, 2],
[4, 3]]
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]]

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;
}

We create a function called matrix with parameter called n and variable called results=[]

We start looping with for loop for all the numbers and if found it is used .push to push into the results array

We create 5 variable to keep track with our rows and columns
let counter = 1;
let startColumn = 0;
let endColumn = n - 1;
let startRow = 0;
let endRow = n - 1;

Then we use the while loop like this while (startColumn <= endColumn && startRow <= endRow)

First we check the TOP ROW and we create for loop where i = startColumn and is less than endColumn and also we increment by 1;

Inside the loop we check results of startRow and i and equal to counter then we increment counter by 1; after that we increment startRow by 1 with startRow++ results[startRow][i] = counter;
counter++; } startRow++;

Then we check the Righ Column and we create for loop where i = startRow i<= endRow and i++
results[i][endColumn] = counter; counter++; } endColumn–;

Then we check the Bottom Row and we create for loop where i = endColumn i>= startRow and i–
results[endRow][i] = counter; counter++; } endRow–;

Then we check the Start Column and we create for loop where i = endRow i>= startRow and i–
results[i][startColumn] = counter; counter++; } startColumn++;

In the end we return the return results