Given a 2D grid map, where -1 represents a wall, 0 represents a gate, and INF represents an empty room, fill each empty room with the distance to its nearest gate.
If it is impossible for an empty room to reach a gate, it should be filled with INF.
for the entire graph
if the room is empty
BFS(room) to find the nearest gate
Code
public void wallsAndGates(int[][] rooms) {
int m = rooms.length;
int n = rooms[0].length;
// search entire graph
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (rooms[i][j] != INF) {
continue;
}
int distance = bfsGate(rooms, new Pos(i,j));
if (distance > 0) {
rooms[i][j] = distance;
}
}
}
}
private int bfsGate(int[][] rooms, Pos start) {
// init queue
Queue<Pos> queue = new LinkedList<>();
boolean visited[][] = new boolean[rooms.length][rooms[0].length];
queue.offer(start);
visited[start.row][start.col] = true;
// bfs by layer
int layers = 0;
Pos current;
// stop when searched the connected component or found the gate
while (!queue.isEmpty()) {
// search by level
int queueSize = queue.size();
for (int i = 0; i < queueSize; i++) {
current = queue.poll();
for (int p = 0; p < 4; p++) {
Pos neighbor = new Pos(current.row + drow[p], current.col + dcol[p]);
if (!inBound(rooms, neighbor)) continue;
if (visited[neighbor.row][neighbor.col]) continue;
int roomType = rooms[neighbor.row][neighbor.col];
if (roomType == -1) continue; // wall
if (roomType == 0) return layers + 1; // gate
queue.offer(neighbor); // empty
visited[neighbor.row][neighbor.col] = true;
}
}
layers++;
}
return 0; // didn't find a gate
}
The last test case on LC gives this code a TLE. Let N = #nodes. The worst case time complexity is O(N). DFS doesn't help in this case. We are stuck with BFS. How can we do better?
From Dst to Src
We can reverse the starting points. If we start searching from the gates and update the empty rooms we can vastly improve the time complexity. Specifically, O(#gates * N) = O(N).
Step 1 Put all gates to queue
Step 2 BFS update empty rooms when the first one gets there
if an empty room is unreachable, it keeps INF
Notice that we don't need visited since we only put unprocessed empty rooms into the queue.
Code
static final int WALL = -1, INF = Integer.MAX_VALUE, GATE = 0;
public void wallsAndGates(int[][] rooms) {
int m = rooms.length;
int n = rooms[0].length;
// Step 1 init queue with dst
Queue<Pos> queue = new LinkedList<>();
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (rooms[i][j] == GATE) {
queue.offer(new Pos(i,j));
}
}
}
// Step 2 level BFS to reach src
Pos current;
int levels = 0;
while (!queue.isEmpty()) {
int queueSize = queue.size();
for (int i = 0; i < queueSize; i++) { // no need for this but keep it anyway
current = queue.poll();
// search neighbors
for (int p = 0; p < 4; p++) {
Pos neighbor = new Pos(current.row + drow[p], current.col + dcol[p]);
if (!inBound(rooms, neighbor)) continue;
if (rooms[neighbor.row][neighbor.col] == INF) {
rooms[neighbor.row][neighbor.col] = levels + 1;
queue.offer(neighbor);
}
}
}
levels++;
}
}
The question cleverly gives gate the value 0. We can tabulate the distance across the table so we don't really need the for loop that keeps tracks of the levels. We can use the current level rooms[neighbor.row][neighbor.col] = rooms[current.row][current.col] + 1 instead of a level variable.
