286

286. Walls and Gates (Medium)

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.

Companies:
Facebook, DoorDash, Amazon, Uber, Google, Bloomberg

Related Topics:
Array, Breadth-First Search, Matrix

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Solution 1. BFS

// OJ: https://leetcode.com/problems/walls-and-gates/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    void wallsAndGates(vector<vector<int>>& A) {
       int M = A.size(), N = A[0].size(), dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}}; 
        queue<pair<int, int>> q;
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] == 0) q.emplace(i, j);
            }
        }
        while (q.size()) {
            int cnt = q.size();
            while (cnt--) {
                auto [x, y] = q.front();
                q.pop();
                for (auto &[dx, dy] : dirs) {
                    int a = x + dx, b = y + dy;
                    if (a < 0 || b < 0 || a >= M || b >= N || A[a][b] <= A[x][y] + 1) continue; 
                    A[a][b] = A[x][y] + 1;
                    q.emplace(a, b);
                }
            }
        }
    }
};