Given an integer N, return a N x N spiral matrix.

For example, N = 4

The output should be:

```
[
[1, 2, 3, 4],
[12, 13, 14, 5],
[11, 16, 15, 6],
[10, 9, 8, 7]
]
```

We are going to start from the number 1 and go around clockwise in a spiral fashion. The last number we will print is N^2. In this example where N is 4, the last number is 16.

First, we need to create an empty array of N amount of subarrays.

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

If we return results, the output would be: `[[], [], [], []]`

.

We are going to create variables to keep track of counter, start column, end column, start row and end row. We are also going to create a `while`

loop and a `for`

loop from start column to end column to assign values for the top row:

```
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 – going from left to right
for (let i = startColumn; i <= endColumn; i++) {
results[startRow][i] = counter;
counter++;
}
startRow++;
}
```

We will create a for loop and iterate from start row to end row to assign values to the right column:

```
…
//Right column – going from up to down
for (let i = startRow; i <= endRow; i++) {
results[i][endColumn] = counter;
counter++;
}
endColumn--;
```

We are going to assign the values for the bottom row by iterating from the end column down to start column.

```
…
//Bottom row – going from right to left
for (let i = endColumn; i >= startColumn; i--) {
results[endRow][i] = counter;
counter++;
}
endRow--;
```

Next, we are going to iterate from end row down to start row to assign values on left column:

```
…
//Left column – going from down to up
for (let i = endRow; i >= startRow; i--) {
results[i][startColumn] = counter;
counter++;
}
startColumn++;
```

Here is the code put together:

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

The `while`

loop would execute again if the conditions are met and execute through the `for`

loops. If not, we would exit out of the `while`

loop and return the results.