How should I practice Spiral Matrix with Obstacles?

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Problem · Matrix

Spiral Matrix with Obstacles

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Problem statement

You are given a nonempty rectangular integer matrix grid. Every cell initially contains either 0, meaning it is fillable, or -1, meaning it is an obstacle.

Starting at the top-left cell and moving right, write consecutive positive integers into fillable cells. Before each move, inspect the next cell:

If it is inside the matrix, is not an obstacle, and has not yet been numbered, move into it.
Otherwise, rotate clockwise and inspect the next direction. Continue rotating, up to all four directions, until a valid move is available.

Obstacle cells remain -1. The input guarantees that the top-left cell is fillable and that this clockwise right-turn walk reaches every fillable cell. Stop after numbering every fillable cell, then return grid.

Function

fillSpiralWithObstacles(grid: int[][]) → int[][]

Examples

Example 1

grid = [[0,0,0,0],[0,0,0,0],[0,0,0,0]]return = [[1,2,3,4],[10,11,12,5],[9,8,7,6]]

With no obstacles, the matrix is numbered in the usual clockwise spiral order.

Example 2

grid = [[0,0,0,0],[0,0,-1,0],[0,0,0,0],[0,0,0,0]]return = [[1,2,3,4],[12,13,-1,5],[11,14,15,6],[10,9,8,7]]

The obstacle remains -1. Near the center, the traversal may reject several directions before finding the next open, unnumbered cell.

Example 3

grid = [[0,-1,0],[0,-1,0],[0,0,0]]return = [[1,-1,7],[2,-1,6],[3,4,5]]

The first move to the right is blocked, so the traversal rotates downward and later reaches the right column from below.

Constraints

1 <= grid.length and 1 <= grid[0].length.
grid is rectangular.
grid.length * grid[0].length <= 100000.
Every initial cell is either 0 or -1.
grid[0][0] == 0.
The clockwise right-turn walk reaches every fillable cell.