Friend Groups - Amazon Top Interview Questions

Problem Statement :

You are given an undirected graph friends as an adjacency list, where friends[i] is a list of people i is friends with. Friendships are two-way. Two people are in a friend group as long as there is some path of mutual friends connecting them.

Return the total number of friend groups.


n ≤ 250 where n is the length of friends

Example 1


friends = [
    [0, 2],




The three friend groups are

[0, 1, 2]
[3, 4]

Solution :


                        Solution in C++ :

int find(int x, vector<int>& parent) {
    if (parent[x] == -1) return x;

    return find(parent[x], parent);

void unionn(int x, int y, vector<int>& parent) {
    int a = find(x, parent);
    // cout<<a<<" ";
    int b = find(y, parent);
    // cout<<b<<endl;
    if (a == b) return;
    parent[b] = a;

int solve(vector<vector<int>>& f) {
    if (f.empty()) return 0;
    int maxx = INT_MIN;

    for (int i = 0; i < f.size(); i++) {
        for (int j = 0; j < f[i].size(); j++) {
            maxx = max(f[i][j], maxx);
    vector<int> p(maxx + 1, -1);
    vector<int> v;

    for (int i = 0; i < f.size(); i++) {
        if (f[i].size() == 0 || f.size() == 1) continue;

        for (int j = 1; j < f[i].size(); j++) {
            // cout<<f[i][0]<<" "<<f[i][j]<<endl;
            unionn(f[i][0], f[i][j], p);

    int count = 0;
    for (int i = 0; i < p.size(); i++) {
        if (p[i] == -1) count++;
        //  cout<<p[i]<<" ";

    return count;

                        Solution in Python : 
class DSU:
    def __init__(self, n):
        self.parent = [i for i in range(n)]
        self.rank = [0 for i in range(n)]

    def find(self, x):
        if self.parent[x] == x:
            return self.parent[x]
        self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, u, v):
        pu = self.find(u)
        pv = self.find(v)

        if self.rank[pu] < self.rank[v]:
            pu, pv = pv, pu

        self.parent[pv] = pu
        self.rank[pu] += self.rank[pv]

    def count(self):
        par_set = set()
        for parent in self.parent:
        return len(par_set)

class Solution:
    def solve(self, A):
        n = len(A)
        dsu = DSU(n)
        for user, friends in enumerate(A):
            for friend in friends:
                dsu.union(user, friend)
        return dsu.count()

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