# Forest Detection - (Binarysearch 527) - Coding Interview Question Mani
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You are given a list of lists edges representing an undirected graph. Each list contains (u, v) which means there is an undirected edge between nodes u and v. Return whether the graph is a collection of trees.

Constraints: n ≤ 100,000 where n is the length of edges

Source Link : https://binarysearch.com/problems/Forest-Detection

Solution:

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import java.util.*;

/**
* Approach: Group all the elements in the form of disjoint set.
* Iterate through all the edges and add keep adding nodes to the DJ .
* If at any instant if we know two nodes of the edge already exists in the set,
* then it means a cycle node is being detected.
* <p>
* One important step to note is in find(a) method,
* parent[a] = find(parent[a]).
* This step avoids time limit exceeded since all the nodes will eventually report
* to the parent.
* <p>
* Time Complexity: O(E)
* Space Complexity: O(N)
**/
class Solution {
public boolean solveUnionFind(int[][] edges) {
int n = 0;
for (int e[] : edges) {
n = Math.max(n, Math.max(e, e));
}

parent = new int[n + 1];

for (int i = 0; i < parent.length; i++) {
parent[i] = i;
}

for (int e[] : edges) {
if (union(e, e)) {
return false;
}
}

return true;
}

int[] parent;

boolean union(int a, int b) {
int x = find(a);
int y = find(b);

if (x == y) {
//Cycle detected.
return true;
}
parent[x] = y;
return false;
}

int find(int a) {
if (parent[a] == a) {
return a;
}
//Assigning parent[a] to the result makes the disjoint set flat.
//All children will report to root node eventually.
parent[a] = find(parent[a]);
return parent[a];
}

//Solution 2:
/**
* Approach: DFS Cycle detection:
* Perform a Dfs on each of the unvisited nodes.
* If no cycles found it means its a tree.
*
* Time Complexity: O(N)
* Space Compexity: O(N)
*
*/
boolean dfs(int i, int parent, Map<Integer, List<Integer>> map, Set<Integer> visited) {
for (Integer x : map.getOrDefault(i, new ArrayList<>())) {
if (x.equals(parent)) {
continue;
}
if (visited.contains(x)) {
return true;
}
if (dfs(x, i, map, visited)) {
return true;
}

}
return false;
}

public boolean solveDFS(int[][] edges) {
int max = 0;

Map<Integer, List<Integer>> map = new HashMap<>();
Set<Integer> visited = new HashSet<>();

for (int[] e : edges) {
max = Math.max(max, Math.max(e, e));
List<Integer> list = map.getOrDefault(e, new ArrayList<>());

List<Integer> list2 = map.getOrDefault(e, new ArrayList<>());

map.put(e, list);
map.put(e, list2);
}

for (int i = 0; i <= max; i++) {
if (!visited.contains(i)) {
if (dfs(i, -1, map, visited)) {
return false;
}
}
}

return true;
}
}
``````
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