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0286 - Walls and Gates (Medium)

https://leetcode.com/problems/walls-and-gates/

Problem Statement​

You are given an m x n grid rooms initialized with these three possible values.

  • -1 A wall or an obstacle.
  • 0 A gate.
  • INF Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.

Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.

Example 1:

Input: rooms = [[2147483647,-1,0,2147483647],[2147483647,2147483647,2147483647,-1],[2147483647,-1,2147483647,-1],[0,-1,2147483647,2147483647]]
Output: [[3,-1,0,1],[2,2,1,-1],[1,-1,2,-1],[0,-1,3,4]]

Example 2:

Input: rooms = [[-1]]
Output: [[-1]]

Constraints:

  • m == rooms.length
  • n == rooms[i].length
  • 1 <= m, n <= 250
  • rooms[i][j] is -1, 0, or 231 - 1.

Approach 1: Multi-Source BFS​

Written by @wingkwong
class Solution:
    def wallsAndGates(self, rooms: List[List[int]]) -> None:
        """
        Do not return anything, modify rooms in-place instead.
        """
        # Multi-source BFS
        # We can see each gate as a source, then we start from each gate
        # If the next room `rooms[next_x][next_y` is empty, 
        # we update the value rooms[x][y] + 1 and
        # we push (next_x, next_y) to the queue
        
        n, m = len(rooms), len(rooms[0])
        q = deque()
        for i in range(n):
            for j in range(m):
                if rooms[i][j] == 0:
                    q.append((i, j))
        while q:
            x, y = q.popleft()
            for dx, dy in (0, 1), (1, 0), (0, -1), (-1, 0):
                next_x, next_y = dx + x, dy + y
                if 0 <= next_x < n and 0 <= next_y < m and rooms[next_x][next_y] == 2147483647:
                    rooms[next_x][next_y] = rooms[x][y] + 1
                    q.append((next_x, next_y))