317-Shortest-Distance-From-All-Buildings

Sat 17 May 2025

https://leetcode.com/problems/shortest-distance-from-all-buildings

import pyutil as pyu
pyu.get_local_pyinfo()
print(pyu.ps2("python-dotenv"))
from typing import List
class Solution:
  def shortestDistance(self, grid: List[List[int]]) -> int:
    m = len(grid)
    n = len(grid[0])
    dirs = [0, 1, 0, -1, 0]
    nBuildings = sum(a == 1 for row in grid for a in row)
    ans = math.inf
    # dist[i][j] := total distance of grid[i][j] (0) to reach all buildings (1)
    dist = [[0] * n for _ in range(m)]
    # reachCount[i][j] := # Of buildings (1) grid[i][j] (0) can reach
    reachCount = [[0] * n for _ in range(m)]

    def bfs(row: int, col: int) -> bool:
      q = deque([(row, col)])
      seen = {(row, col)}
      depth = 0
      seenBuildings = 1

      while q:
        depth += 1
        for _ in range(len(q)):
          i, j = q.popleft()
          for k in range(4):
            x = i + dirs[k]
            y = j + dirs[k + 1]
            if x < 0 or x == m or y < 0 or y == n:
              continue
            if (x, y) in seen:
              continue
            seen.add((x, y))
            if not grid[x][y]:
              dist[x][y] += depth
              reachCount[x][y] += 1
              q.append((x, y))
            elif grid[x][y] == 1:
              seenBuildings += 1

      # True if all buildings (1) are connected
      return seenBuildings == nBuildings

    for i in range(m):
      for j in range(n):
        if grid[i][j] == 1:  # Bfs from this building
          if not bfs(i, j):
            return -1

    for i in range(m):
      for j in range(n):
        if reachCount[i][j] == nBuildings:
          ans = min(ans, dist[i][j])

    return -1 if ans == math.inf else ans
new Solution().shortestDistance()

Score: 5

Category: leetcode