# 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.

Constraints

n ≤ 250 where n is the length of friends

Example 1

Input

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

Output

3

Explanation

The three friend groups are

[0, 1, 2]
[3, 4]
[5]```

### 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()```
```

## Array-DS

An array is a type of data structure that stores elements of the same type in a contiguous block of memory. In an array, A, of size N, each memory location has some unique index, i (where 0<=i<N), that can be referenced as A[i] or Ai. Reverse an array of integers. Note: If you've already solved our C++ domain's Arrays Introduction challenge, you may want to skip this. Example: A=[1,2,3

## 2D Array-DS

Given a 6*6 2D Array, arr: 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 An hourglass in A is a subset of values with indices falling in this pattern in arr's graphical representation: a b c d e f g There are 16 hourglasses in arr. An hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in arr, then print t

## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

## Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

## Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu