Solve the Medium-level Spiral Matrix coding interview problem online in Python & JavaScript. Run your code against real test cases in the browser and study worked examples. It covers arrays and matrix. Read the step-by-step solution and complexity analysis. Java starter code is also included for reference (no in-browser JVM yet, so it can't be run here).

Spiral Matrix – Solve Online in Python & JavaScript

given number n return matrix n by n with numbers in spiral order

## Problem Description

Given an integer `n`, you need to implement a function `generate_spiral_matrix(n: int) -> List[List[int]]` that generates a matrix of size `n` x `n` containing elements in a spiral order starting from the top-left corner and moving clockwise.

- Create an `n x n` matrix and place values from `1` to `n²`.
- The direction starts moving right, then turns down, left, and up.
- After each side is filled, the active boundary moves inward.

Understand the Spiral Fill

- Keep `top`, `bottom`, `left`, and `right` boundaries for the unfilled rectangle.
- Fill the top row left-to-right, then move `top` down.
- Fill the right column top-to-bottom, then move `right` left, and continue around.

Track Four Boundaries

- One loop iteration fills the outer ring of the current rectangle.
- After a ring is finished, the boundaries describe the next inner ring.
- Continue while `top <= bottom` and `left <= right`.

Fill One Ring at a Time

- Odd-sized matrices eventually have a single center cell, which the boundary loop fills naturally.
- Some test inputs may provide `n` as a string after JSON parsing.
- Convert `n` to an integer before creating the matrix.

Handle Center Cells and Input Type

- The final solution fills each cell exactly once while shrinking the boundaries.
- Time complexity is O(n²).
- Space complexity is O(n²) for the returned matrix.