HackerRank Merging Communities problem solution YASH PAL, 31 July 2024 In this tutorial, we are going to solve or make a solution to the Merging Communities problem. so here we have n queries that representing the n communities. we need to print the size of the community to which person belongs. Problem solution in Python programming. #!/bin/python import sys from collections import deque N, Q = map(int,input().strip().split(' ')) s = [i for i in range(N+1)] cnt = [0]+[1 for i in range(N)] for i in range(Q): inpt = input().strip().split(' ') qry = inpt[0] a = sorted(map(lambda x: int(x),inpt[1:])) i0 = a[0] if qry == 'M' and s[i0] != s[a[1]]: i1 = a[1] tmp = deque() while i0 != s[i0]: tmp.append(i0) i0 = s[i0] while i1 != s[i1]: tmp.append(i1) i1 = s[i1] if s[i0] != s[i1]: cnt[s[i0]] += cnt[s[i1]] tmp.append(i1) for ix in tmp: s[ix] = s[i0] elif qry == 'Q': tmp = deque() while i0 != s[i0]: tmp.append(i0) i0 = s[i0] for ix in tmp: s[ix] = s[i0] print(cnt[i0]) Problem solution in Java Programming. import java.io.*; import java.util.*; public class Solution { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); String[] line = scanner.nextLine().split(" "); int n = Integer.valueOf(line[0]); int q = Integer.valueOf(line[1]); int[] arr = new int[n+1]; for(int i=0; i<=n; i++) { arr[i] = -1; } for(int i=0; i<q; i++) { line = scanner.nextLine().split(" "); if(line.length == 2) { int a = Integer.valueOf(line[1]); int root = getRoot(arr, a); System.out.println(size(arr, root)); } else { int a = Integer.valueOf(line[1]); int b = Integer.valueOf(line[2]); int aroot = getRoot(arr, a); int broot = getRoot(arr, b); if(aroot == broot) continue; if(size(arr, aroot) > size(arr, broot)) { arr[aroot] += arr[broot]; arr[broot] = aroot; } else { arr[broot] += arr[aroot]; arr[aroot] = broot; } } } } static int getRoot(int[] arr, int a) { int root = a; while(arr[root] > 0) { root = arr[root]; } if(a != root) arr[a] = root; return root; } static int size(int[] arr, int a) { return -1 * arr[a]; } } Problem solution in C++ programming. #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("%ldn",val[find(e)]); } } return 0; } Problem solution in C programming. #include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> typedef struct Node{ int cnt; int par; }node; node *com_array; int find_parent(int n){ while(com_array[n].par != -1){ n=com_array[n].par; } return n; } void mergeCom(int p1, int p2){ int max,low,par1,par2; if(p1==p2) return; par1=find_parent(p1); par2=find_parent(p2); if((par1>=0) && ( par1 == par2)) return; if(com_array[par1].cnt>=com_array[par2].cnt){ max=par1; low=par2; } else{ max=par2; low=par1; } com_array[max].cnt+=com_array[low].cnt; com_array[low].par=max; } int main() { int q,*res,n; char o; int i,p1,p2,qcnt=0; scanf("%d %d",&n,&q); com_array=(node*)malloc(n*sizeof(node)); for(i=0;i<n;i++){ com_array[i].par=-1; com_array[i].cnt=1; } res=(int*)calloc(q,sizeof(int)); for(i=0;i<q;i++){ scanf(" %c",&o); if(o=='Q'){ int par; scanf("%d",&p1); par=find_parent(p1-1); res[qcnt++]=com_array[par].cnt; } else{ scanf("%d %d",&p1,&p2); mergeCom(p1-1,p2-1); } } for(i=0;i<qcnt;i++) printf("%dn",res[i]); return 0; } Problem solution in JavaScript programming. function processData(input) { //Enter your code here var arr = input.split("n"); var [n, queries] = arr[0].split(" "); var uf = new UF(); for (var i = 1; i <= n; i++) { uf.add(i); } var result = ""; for (var i = 1; i < arr.length; i++) { var list = arr[i].split(" "); if (list[0] === "Q") { // result.push(uf.count[uf.find(list[1])]); if (result !== "") { result += "n"; } result += uf.count[uf.find(list[1])]; } else { uf.union(list[1], list[2]); } } console.log(result); } function UF() { this.parent = []; this.count = []; } UF.prototype.add = function(i) { this.parent[i] = i; this.count[i] = 1; } UF.prototype.find = function(i) { var j = i; while (this.parent[i] !== i) { i = this.parent[i]; } while (j !== i) { var pj = this.parent[j]; this.parent[j] = i; j = pj; } return i; } UF.prototype.union = function (i, j) { var pi = this.find(i); var pj = this.find(j); if (pi !== pj) { this.parent[pj] = pi; this.count[pi] += this.count[pj]; } } process.stdin.resume(); process.stdin.setEncoding("ascii"); _input = ""; process.stdin.on("data", function (input) { _input += input; }); process.stdin.on("end", function () { processData(_input); }); coding problems data structure