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.


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

Example 1


edges = [

    [1, 2],

    [1, 3],

    [1, 4],

    [4, 5]




Solution :


                        Solution in C++ :

int solve(vector<vector<int>>& edges) {
    int n = edges.size() + 1;
    if (n == 1) return 0;
    unordered_map<int, vector<int>> adj;
    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) {
        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) {
            List<Integer> peelNodes = degreeMap.keySet()
                                          .filter(key -> degreeMap.get(key) == 1)
            peelNodes.forEach(node -> {
                if (graph.containsKey(node)) {
                    for (int adj : graph.get(node)) {
                        if (degreeMap.containsKey(adj)) {
                            degreeMap.put(adj, degreeMap.get(adj) - 1);
        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)]
        return 0

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