You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values 0, 1 or 2, where:

* Each 0 marks an empty land which you can pass by freely.

* Each 1 marks a building which you cannot pass through.

* Each 2 marks an obstacle which you cannot pass through.

Note:

* There will be at least one building. If it is not possible to build such house according to the above rules, return -1.

* Method 1: BFS

```Java

class Solution {

private static final int[][] dir = new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

public int shortestDistance(int[][] grid) {

Set<Integer> buildings = new HashSet<>();

int height = grid.length, width = grid[0].length;

for(int i = 0; i < grid.length; i++){

for(int j = 0; j < grid[0].length; j++){

if(grid[i][j] == 1) buildings.add(i * width + j);

}

}

int min = Integer.MAX_VALUE;

int count = buildings.size();

for(int i = 0; i < height; i++){

LABEL:

for(int j = 0; j < width; j++){

if(grid[i][j] != 0) continue;

Queue<int[]> q = new LinkedList<>();

q.offer(new int[]{i, j});

Set<Integer> visited = new HashSet<>();

int reach = 0;

int step = 0;

int temp = 0;

Set<Integer> check = new HashSet<>(buildings);

while(!q.isEmpty() && reach != count){

int size = q.size();

step++;

for(int k = 0; k < size; k++){

int[] cur = q.poll();

if(check.contains(cur[0] * width + cur[1])){

temp += step - 1;

if(temp >= min) continue LABEL;

check.remove(cur[0] * width + cur[1]);

reach++;

continue;

}

int tx = 0, ty = 0;

for(int d = 0; d < 4; d++){

tx = cur[0] + dir[d][0];

ty = cur[1] + dir[d][1];

if(tx >= 0 && tx < height && ty >= 0 && ty < width && !visited.contains(tx * width + ty) && grid[tx][ty] != 2){

visited.add(tx * width + ty);

q.offer(new int[]{tx, ty});

}

}

}

}

if(reach == count) min = Math.min(min, temp);

}

}

return min == Integer.MAX_VALUE ? -1: min;

}

}

```

Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.

Design an algorithm to serialize and deserialize an N-ary tree. An N-ary tree is a rooted tree in which each node has no more than N children. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that an N-ary tree can be serialized to a string and this string can be deserialized to the original tree structure.

For example, you may serialize the following 3-ary tree


as [1 [3[5 6] 2 4]]. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

Note:

* N is in the range of [1, 1000]

* Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

* Method 1: recursion

```Java

/*

class Codec {

private static final String split = ",";

private static final String start = "[";

private static final String end = "]";

private static final String NULL = "#";

// Encodes a tree to a single string.

public String serialize(Node root) {

StringBuilder sb = new StringBuilder();

serialize(root, sb);

return sb.toString();

}

private void serialize(Node node, StringBuilder sb){

if(node == null){

sb.append(NULL + split);

return;

}

sb.append(node.val);

if(node.children != null && node.children.size() > 0){

sb.append(start);

for(Node n : node.children){

serialize(n, sb);

}

sb.append(end);

}

sb.append(split);

}

private int index = 0;

private char[] arr;

// Decodes your encoded data to tree.

public Node deserialize(String data) {

if(data.equals(NULL + split)) return null;

this.arr = data.toCharArray();

return deserialize().get(0);

}

private List<Node> deserialize(){

Node node = null;

List<Node> list = new ArrayList<>();

while(index < arr.length){

char c = arr[index];

if(Character.isDigit(c)){

int num = 0;

while(index < arr.length && Character.isDigit(arr[index])){

num = num * 10 + arr[index++] - '0';

}

index--;

node = new Node(num, new ArrayList<>());

list.add(node);

}else if(("" + c).equals(start)){

index++;

node.children = deserialize();

}else if(("" + c).equals(end)){

index++;

return list;

}

index++;

}

return list;

}

}

// Your Codec object will be instantiated and called as such:

// Codec codec = new Codec();

// codec.deserialize(codec.serialize(root));

```

Given a string S, check if the letters can be rearranged so that two characters that are adjacent to each other are not the same.

If possible, output any possible result. If not possible, return the empty string.

Note:

* S will consist of lowercase letters and have length in range [1, 500].

* Method 1: PriorityQueue

```Java

class Solution {

private static final class Node{

char c;

int f;

public Node(char c, int f){

this.c = c;

this.f = f;

}

}

public String reorganizeString(String S) {

char[] arr = S.toCharArray();

int[] table = new int[26];

for(char c : arr){

if(++table[c - 'a'] > (S.length() + 1)/ 2) return "";

}

PriorityQueue<Node> pq = new PriorityQueue<>((a, b) -> {

return b.f - a.f;

});

for(int i = 0; i < 26; i++){

if(table[i] > 0){

pq.offer(new Node((char)(i + 'a'), table[i]));

}

}

StringBuilder sb = new StringBuilder();

while(pq.size() >= 2){

Node first = pq.poll();

Node second = pq.poll();

sb.append(first.c);

sb.append(second.c);

if(--first.f > 0) pq.offer(first);

if(--second.f > 0) pq.offer(second);

}

if(!pq.isEmpty()) sb.append(pq.poll().c);

return sb.toString();

}

}

```