coding-Interview/GraphValidTree.java at master · arjunmullick/coding-Interview · GitHub

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package com.leetcode;

import java.util.ArrayList;

import java.util.HashMap;

import java.util.LinkedList;

import java.util.List;

import java.util.Queue;

import java.util.Stack;

public class GraphValidTree {

One graph theory optimization on DFS and BFS below is if(edges.length != n-1) return false;

then check its connected by count visited == n

class SolutionBFS {

//Because its un-directional store parent for cyclic check

public boolean validTree(int n, int[][] edges) {

//in BFS if we visit all node and no cyclic in graph its a tree.

//check all nodes are part of tree

HashMap<Integer, List<Integer>> map = new HashMap<>();

HashMap<Integer,Integer> parent = new HashMap<>();

for(int[] edge : edges){

List<Integer> l1 = map.getOrDefault(edge[0],new ArrayList<>());

l1.add(edge[1]);

map.put(edge[0],l1);

List<Integer> l2 = map.getOrDefault(edge[1],new ArrayList<>());

l2.add(edge[0]);

map.put(edge[1],l2);

}

Queue<Integer> queue = new LinkedList<>();

boolean [] visited = new boolean[n];

queue.offer(0);

while(queue.size() > 0){

int node = queue.poll();

visited[node] = true;

List<Integer> childs = map.getOrDefault(node, new ArrayList<>());//to not give null pointer

for(int next : childs){

if(!visited[next]){

queue.offer(next);

parent.put(next,node);

}else if(parent.get(node) != next){//To ensure a node has only one parent . Check the node parent is market as child then then its ok 0-1 , 1-0 because of undirected

return false;//cyclic

}

// All visited

for(boolean e : visited){

if(!e) return false;

}

return true;

}

class SolutionDFS {

//DFS

public boolean validTree(int n, int[][] edges) {

//in BFS if we visit all node and no cyclic in graph its a tree.

//check all nodes are part of tree

HashMap<Integer, List<Integer>> map = new HashMap<>();

HashMap<Integer,Integer> parent = new HashMap<>();

for(int[] edge : edges){

List<Integer> l1 = map.getOrDefault(edge[0],new ArrayList<>());

l1.add(edge[1]);

map.put(edge[0],l1);

List<Integer> l2 = map.getOrDefault(edge[1],new ArrayList<>());

l2.add(edge[0]);

map.put(edge[1],l2);

}

Stack<Integer> stack = new Stack<>();

boolean [] visited = new boolean[n];

stack.push(0);

while(stack.size() > 0){

int node = stack.pop();

visited[node] = true;

List<Integer> childs = map.getOrDefault(node, new ArrayList<>());//to not give null pointer

for(int next : childs){

if(!visited[next]){

stack.push(next);

parent.put(next,node);

}else if(parent.get(node) != next){//check the node parent is market as child then then its ok 0-1 , 1-0 because of undirected

return false;//cyclic

}

// All visited

for(boolean e : visited){

if(!e) return false;

}

return true;

}

//O(N)

class SolutionDFSOptimized {

//DFS/ BFS optimized

public boolean validTree(int n, int[][] edges) {

if(edges.length != n-1) return false;

//in BFS if we visit all node and no cyclic in graph its a tree.

//check all nodes are part of tree

HashMap<Integer, List<Integer>> map = new HashMap<>();

for(int[] edge : edges){

List<Integer> l1 = map.getOrDefault(edge[0],new ArrayList<>());

l1.add(edge[1]);

map.put(edge[0],l1);

List<Integer> l2 = map.getOrDefault(edge[1],new ArrayList<>());

l2.add(edge[0]);

map.put(edge[1],l2);

}

Stack<Integer> stack = new Stack<>();

boolean [] visited = new boolean[n];

stack.push(0);

while(stack.size() > 0){

int node = stack.pop();

visited[node] = true;

List<Integer> childs = map.getOrDefault(node, new ArrayList<>());//to not give null pointer

for(int next : childs){

if(!visited[next]){

stack.push(next);

}

// All visited

for(boolean e : visited){

if(!e) return false;

}

return true;

}

//O(N) to verify .

// Complexity

/**

Time to build graph : O(nlog(n))

Space: O(n) parent and size array

n: the total number of nodes in the given graph

Find operation

Time: O(log(n)) log(n) is the height of the tree

Space: O(1)

Union operation

Time: O(1)

Space: O(1)

**/

class SolutionUnionFind {

public boolean validTree(int n, int[][] edges) {

if(edges.length != n -1) return false;

UnionFind unionFind = new UnionFind(n);

for (int[] edge : edges) {

int a = edge[0];

int b = edge[1];

if(!unionFind.union(a,b)){

return false;

}

// If we are successfully able to merge all edges then no cycle

return true;

}

class UnionFind{

int[] parent;

public UnionFind(int n){

parent = new int[n];

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

parent[i] = i;//n disjoint sets with each node parent to self

}

//find and merge the set

public int find(int node){

while(parent[node] != node){

node = parent[node];

}

return node;//trace the parent

}

public boolean union(int a , int b){

int parentA = find(a);

int parentB = find(b);

//if same set already

if(parentA == parentB){

return false;

}

//merge

parent[parentA] = parentB;

return true;

}

coding-Interview/GraphValidTree.java at master · arjunmullick/coding-Interview · GitHub

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