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

classSolution {

privatestaticfinalint[][] dir=newint[][]{{0,1}, {0,-1}, {1,0}, {-1,0}};

publicintshortestDistance(int[][]grid) {

Set<Integer> buildings=newHashSet<>();

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++){

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

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

Queue<int[]> q=newLinkedList<>();

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

Set<Integer> visited=newHashSet<>();

int reach=0;

int step=0;

int temp=0;

Set<Integer> check=newHashSet<>(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)continueLABEL;

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(newint[]{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

classCodec {

privatestaticfinalString split=",";

privatestaticfinalString start="[";

privatestaticfinalString end="]";

privatestaticfinalStringNULL="#";

publicStringserialize(Noderoot) {

StringBuilder sb=newStringBuilder();

serialize(root, sb);

return sb.toString();

}

privatevoidserialize(Nodenode,StringBuildersb){

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);

}

privateint index=0;

privatechar[] arr;

publicNodedeserialize(Stringdata) {

if(data.equals(NULL+ split))returnnull;

this.arr= data.toCharArray();

return deserialize().get(0);

}

privateList<Node>deserialize(){

Node node=null;

List<Node> list=newArrayList<>();

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=newNode(num,newArrayList<>());

list.add(node);

}elseif((""+ c).equals(start)){

index++;

node.children= deserialize();

}elseif((""+ c).equals(end)){

index++;

return list;

}

index++;

}

return list;

}

}

```

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

classSolution {

privatestaticfinalclassNode{

char c;

int f;

publicNode(charc,intf){

this.c= c;

this.f= f;

}

}

publicStringreorganizeString(StringS) {

char[] arr=S.toCharArray();

int[] table=newint[26];

for(char c: arr){

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

}

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

return b.f- a.f;

});

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

if(table[i]>0){

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

}

}

StringBuilder sb=newStringBuilder();

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();

}

}

```