# Length of the Longest Path in an N-Ary Tree - Facebook Top Interview Questions

### Problem Statement :

```You are given a two-dimensional list of integers edges representing an n-ary tree.

Each element in edges contains [u, v] meaning that u is the parent of v.

Return the length of the longest path in the tree.

Constraints

1 ≤ n ≤ 100,000 where n is the length of edges

Example 1

Input

edges = [

[1, 2],

[1, 3],

[1, 4],

[4, 5]

]

Output

4```

### Solution :

```                        ```Solution in C++ :

int solve(vector<vector<int>>& edges) {
int n = edges.size() + 1;
if (n == 1) return 0;
unordered_map<int, int> ind;
for (auto& e : edges) {
ind[e[0]] += 1;
ind[e[1]] += 1;
}
vector<int> leaves;
int dist = 0;
for (auto [element, indegree] : ind)
if (indegree == 1) leaves.push_back(element);
while (n > 2) {
vector<int> newleaves;
n -= leaves.size();
dist += 1;
for (auto k : leaves) {
ind[k] -= 1;
for (auto j : adj[k]) {
if (--ind[j] == 1) {
newleaves.push_back(j);
}
}
}
leaves = newleaves;
if (leaves.size() == 0) break;
}
if (n == 1) return dist * 2 + 1;
return 2 * dist + 2;
}```
```

```                        ```Solution in Java :

import java.util.*;
import java.util.stream.Collectors;

class Solution {
private Set<Integer> visited = new HashSet();
private Map<Integer, List<Integer>> graph = new HashMap();
private Map<Integer, Integer> degreeMap = new HashMap();
private int maxnodes = 0;
private Queue<int[]> queue = new LinkedList();
public int solve(int[][] edges) {
if (edges == null || edges.length == 0)
return 0;
for (int[] edge : edges) {
int u = edge[0];
int v = edge[1];
degreeMap.put(u, degreeMap.getOrDefault(u, 0) + 1);
degreeMap.put(v, degreeMap.getOrDefault(v, 0) + 1);
graph.computeIfAbsent(u, key -> new ArrayList()).add(v);
graph.computeIfAbsent(v, key -> new ArrayList()).add(u);
maxnodes = Math.max(maxnodes, u);
maxnodes = Math.max(maxnodes, v);
}

int level = 0;
while (degreeMap.size() > 2) {
level++;
List<Integer> peelNodes = degreeMap.keySet()
.stream()
.filter(key -> degreeMap.get(key) == 1)
.collect(Collectors.toList());
peelNodes.forEach(node -> {
if (graph.containsKey(node)) {
for (int adj : graph.get(node)) {
}
}
}
degreeMap.remove(node);
});
}
return 2 * level + degreeMap.size();
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, edges):
g = {}
for u, v in edges:
if u not in g:
g[u] = []
g[u] += (v,)
if v not in g:
g[v] = []
g[v] += (u,)
d = {}

def bfs(o):
d[o] = 1
f = o
q = [o]
for x in q:
for y in g[x]:
if y not in d:
d[y] = d[x] + 1
if d[y] > d[f]:
f = y
q += (y,)
return f

for o in g:
f = bfs(o)
d = {}
return d[bfs(f)]
break
return 0```
```

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