幅優先探索 (BFS) でグリッドの最短の道を数える

from collections import deque
STEPS = [
    (-1, -1), (-1, 0), (-1, 1),
    (0, -1), (0, 1),
    (1, -1), (1, 0), (1, 1),
]


def shortest_path(grid):
    n = len(grid)
    if grid[0][0] == 1:
        return -1
    goal = (n - 1, n - 1)
    grid[0][0] = 1
    queue = deque([(0, 0, 1)])
    push = queue.append
    pop = queue.popleft
    while queue:
        r, c, d = pop()
        if (r, c) == goal:
            return d
        for dr, dc in STEPS:
            a = r + dr
            b = c + dc
            if not 0 <= a < n:
                continue
            if not 0 <= b < n:
                continue
            if grid[a][b] == 1:
                continue
            grid[a][b] = 1
            push((a, b, d + 1))
    return -1

const STEPS = [
  [-1, -1], [-1, 0], [-1, 1],
  [0, -1], [0, 1],
  [1, -1], [1, 0], [1, 1],
];

function shortestPath(grid) {
  const n = grid.length;
  if (grid[0][0] === 1) {
    return -1;
  }
  grid[0][0] = 1;
  const queue = [[0, 0, 1]];
  let head = 0;
  while (head < queue.length) {
    const p = queue[head];
    head++;
    const [r, c, d] = p;
    if (r === n - 1) {
      if (c === n - 1) return d;
    }
    for (const s of STEPS) {
      const a = r + s[0];
      const b = c + s[1];
      if (grid[a]?.[b] !== 0) {
        continue;
      }
      grid[a][b] = 1;
      queue.push([a, b, d + 1]);
    }
  }
  return -1;
}

var steps = [8][2]int{
	{-1, -1}, {-1, 0}, {-1, 1},
	{0, -1}, {0, 1},
	{1, -1}, {1, 0}, {1, 1},
}

func shortestPath(
	grid [][]int,
) int {
	n := len(grid)
	if grid[0][0] == 1 {
		return -1
	}
	grid[0][0] = 1
	queue := [][3]int{{0, 0, 1}}
	head := 0
	for head < len(queue) {
		p := queue[head]
		head++
		r, c, d := p[0], p[1], p[2]
		if r == n-1 && c == n-1 {
			return d
		}
		for _, s := range steps {
			a, b := r+s[0], c+s[1]
			if a < 0 || a >= n {
				continue
			}
			if b < 0 || b >= n {
				continue
			}
			if grid[a][b] == 1 {
				continue
			}
			grid[a][b] = 1
			nx := [3]int{a, b, d + 1}
			queue = append(queue, nx)
		}
	}
	return -1
}