# Merging Communities

### Problem Statement :

```People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to.

At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 will belong to the same community.

There are two type of queries:

1. M I J =>  communities containing person  and  merged (if they belong to different communities).

2. Q I =>  print the size of the community to which person  belongs.

Input Format

The first line of input will contain integers N and Q, i.e. the number of people and the number of queries.
The next Q lines will contain the queries.

Constraints :

1  <=  N  <= 10^5
1  <=  Q   <= 2 * 10^5

Output Format

The output of the queries.```

### Solution :

```                            ```Solution in C :

In C ++ :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int f[100005], s[100005];
int find(int x) {
if (x == f[x]) return x;
return f[x] = find(f[x]);
}
void merge(int x, int y) {
int fx = find(x), fy = find(y);
if (fx == fy) return;
if (s[fx] > s[fy]) {
f[fy] = fx;
s[fx] += s[fy];
} else {
f[fx] = fy;
s[fy] += s[fx];
}
}
int Q, N;
char c[5];
int main() {
scanf("%d%d", &N, &Q);
for (int i = 1; i <= N; ++i) f[i] = i, s[i] = 1;
for (int i = 0; i < Q; ++i) {
scanf("%s", c);
if (c[0] == 'M') {
int x, y; scanf("%d%d", &x, &y);
merge(x, y);
} else {
int x; scanf("%d", &x);
printf("%d\n", s[find(x)]);
}
}
return 0;
}

In Java :

import java.util.Arrays;
import java.util.Scanner;

public class MergingCommunities {
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int q = sc.nextInt();

parent = new int[n];
for (int i = 0; i < n; i++)
parent[i] = i;

size = new int[n];
Arrays.fill(size, 1);

while (q-- != 0) {
// System.out.println(Arrays.toString(parent));
// System.out.println(Arrays.toString(size));
// System.out.println();
String type = sc.next();
if (type.equals("Q"))
System.out.println(size[findSet(sc.nextInt() - 1)]);
else
union(sc.nextInt() - 1, sc.nextInt() - 1);
}

}

static int[] parent;
static int[] size;

static int findSet(int x) {
return parent[x] == x ? x : (parent[x] = findSet(parent[x]));
}

static void union(int x, int y) {
if (findSet(x) != findSet(y))
size[findSet(x)] = size[findSet(y)] = size[findSet(x)]
+ size[findSet(y)];
parent[findSet(x)] = findSet(y);
}
}

In  C :

#include<stdio.h>
#define mem(a,v) memset(a,v,sizeof(a))
long int parent[100005],rank[100005],val[100005];
long int find(long int x)
{
if(parent[x]==x)
return x;
return find(parent[x]);
}
void _union(long int x,long int y)
{
long int vv,dd=find(x);
long int hh=find(y);
if(dd!=hh)
{
if(rank[hh]>rank[dd])
{
parent[dd]=hh;
val[hh]+=val[dd];
}
else if(rank[hh]<rank[dd])
{
parent[hh]=dd;
val[dd]+=val[hh];
}
else
{
parent[hh]=dd;
val[dd]+=val[hh];
rank[dd]++;
}
}
}
int main()
{
long int c,d,t,i,e;
char ff[3];

for(i=0;i<=100005;i++)
{
parent[i]=i;
val[i]=1;
rank[i]=1;
}
scanf("%ld%ld",&c,&d);
for(i=0;i<=c;i++)
parent[i]=i;
while(d--)
{
scanf("%s",ff);
if(ff[0]=='M')
{
scanf("%ld%ld",&e,&t);
_union(e,t);
}
else
{
scanf("%ld",&e);
printf("%ld\n",val[find(e)]);
}
}
return 0;
}

In Python3 :

[N,Q]=[int(x) for x in input().split()]
ds = [[i,1] for i in range(0,N+1)]
for q in range(Q):
inp = input().split()
if inp[0] == 'M':
ind1 = int(inp[1])
ind2 = int(inp[2])
ind3 = ind2
while ds[ind1][0] != ind1:
ind1 = ds[ind1][0]
while ds[ind2][0] != ind2:
ind2 = ds[ind2][0]
if ind1 == ind2:
continue
while ind3 != ind2:
ind4 = ds[ind3][0]
ds[ind3][0] = ind1
ind3 = ind4
ds[ind2][0] = ind1
ds[ind1][1] += ds[ind2][1]
else:
ind1 = int(inp[1])
while ds[ind1][0] != ind1:
ind1 = ds[ind1][0]
print(ds[ind1][1])```
```

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