Skip to content
Programmingoneonone
Programmingoneonone

LEARN EVERYTHING ABOUT PROGRAMMING

  • Home
  • CS Subjects
    • IoT ? Internet of Things
    • Digital Communication
    • Human Values
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
  • HackerRank Solutions
    • HackerRank Algorithms Solutions
    • HackerRank C problems solutions
    • HackerRank C++ problems solutions
    • HackerRank Java problems solutions
    • HackerRank Python problems solutions
Programmingoneonone

LEARN EVERYTHING ABOUT PROGRAMMING

HackerRank Rust & Murderer problem solution

YASH PAL, 31 July 2024

In this HackerRank Rust & Murderer problem, distance is calculated as several village roads (side lanes) between any two places in the city. Rust wants to calculate the shortest distance from his position (Node S) to all the other places in the city if he travels only using the village roads (side lanes).

HackerRank Rust & Murderer problem solution

Topics we are covering

Toggle
  • Problem solution in Python.
  • Problem solution in Java.
  • Problem solution in C++.
  • Problem solution in C.

Problem solution in Python.

tests = int(input())

for _ in range(tests):
    [n, e] = [int(i) for i in input().split(" ")]
    dists = [1] * n
    roads = {}
    for _ in range(e):
        [n1, n2] = [int(i) for i in input().split(" ")]
        if n1 not in roads:
            roads[n1] = set()
        if n2 not in roads:
            roads[n2] = set()
        roads[n1].add(n2)
        roads[n2].add(n1)
    start_loc = int(input())
    not_visited = roads[start_loc] if start_loc in roads else set()
    newly_visited = set()
    curr_dist = 2
    while len(not_visited) > 0:
        for i in not_visited:
            diff = not_visited | roads[i]
            if len(diff) < n:
                dists[i-1] = curr_dist
                newly_visited.add(i)
        not_visited = not_visited - newly_visited
        newly_visited = set()
        curr_dist += 1
    del dists[start_loc-1]
    print(" ".join(str(i) for i in dists))

{“mode”:”full”,”isActive”:false}

Problem solution in Java.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.StringJoiner;

public class Solution {

    public static void main(String [] args) throws IOException {

        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));

        StringBuilder resultBuilder = new StringBuilder();
        int numberOfTestCases = Integer.parseInt(bufferedReader.readLine());
        for(int t=0;t<numberOfTestCases;t++) {

            String [] parts = bufferedReader.readLine().split(" ");
            int numberOfNodes = Integer.parseInt(parts[0]);
            int numberOfMainRoads = Integer.parseInt(parts[1]);

            HashSet<Integer>[] mainRoadConnections = new HashSet[numberOfNodes + 1];
            for(int i=0;i<numberOfNodes + 1;i++) {
                mainRoadConnections[i] = new HashSet();
            }

            for(int i=0;i<numberOfMainRoads;i++) {
                parts = bufferedReader.readLine().split(" ");
                int from = Integer.parseInt(parts[0]);
                int to = Integer.parseInt(parts[1]);

                mainRoadConnections[from].add(to);
                mainRoadConnections[to].add(from);
            }

            int start = Integer.parseInt(bufferedReader.readLine());

            int [] distances = solve(numberOfNodes, mainRoadConnections, start);

            StringJoiner stringJoiner = new StringJoiner(" ");
            for(int i=1;i<numberOfNodes + 1;i++) {
                if (i!=start) {
                    stringJoiner.add(distances[i] + "");
                }
            }

            resultBuilder.append(stringJoiner.toString() +"n");
        }

        System.out.println(resultBuilder.toString());
    }

    private static int[] solve(int numberOfNodes, HashSet<Integer>[] mainRoadConnections, int start) {

        boolean visited [] = new boolean[numberOfNodes + 1];
        int result [] = new int [numberOfNodes + 1];
        Arrays.fill(result, Integer.MAX_VALUE);

        LinkedList<Node> queue = new LinkedList<>();
        queue.add(new Node(start, 0));

        HashSet<Integer> notUsed = new HashSet<>();
        for(int i=1;i<=numberOfNodes;i++) {
            notUsed.add(i);
        }
        notUsed.remove(start);

        while(!queue.isEmpty()) {

            Node top = queue.poll();

            result[top.index] = top.distance;

            HashSet<Integer> forRemove = new HashSet<>();
            for(Integer i : notUsed) {
                if (i != top.index && !visited[i] && !mainRoadConnections[top.index].contains(i)) {
                    visited[i] = true;
                    forRemove.add(i);
                    queue.add(new Node(i, top.distance + 1));
                }
            }

            notUsed.removeAll(forRemove);
        }

        return result;
    }

    private static class Node {
        int index;
        int distance;

        public Node(int index, int distance) {
            this.index = index;
            this.distance = distance;
        }
    }
}

{“mode”:”full”,”isActive”:false}

Problem solution in C++.

#include <cstdio>
#include <iostream>
#include <vector>
#include <set>
#include <queue>

using namespace std;
int dp[150000+1];
vector<int> graph[150002];
set<int> S;
queue<int> Q;

int main()
{
    int t,nodes,edges,start;
    int v1,v2;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&nodes,&edges);
        for(int i=1; i<= nodes; i++) graph[i].clear();
        for(int i=1; i<= edges; i++)
        {
            scanf("%d %d",&v1,&v2);
            graph[v1].push_back(v2);
            graph[v2].push_back(v1);
        }
        scanf("%d",&start);
        for(int i=1; i<=nodes; i++)
            S.insert(i);
        S.erase(start);
        dp[start] = 0;
        Q.push(start);
        while(!Q.empty())
        {
            int ele = Q.front();
            set<int> R;
            Q.pop();
            int sz = graph[ele].size();
            for(int i=0; i<sz; i++)
            {
                int x = graph[ele][i];
                if (S.find(x) != S.end())
                {
                    S.erase(x);
                    R.insert(x);
                }
            }
            set<int>::iterator itr;
            for(itr = S.begin(); itr != S.end(); itr++)
            {
                Q.push(*itr);
                dp[*itr] = dp[ele] + 1;
                S.erase(*itr);
            }
            S=R;
        }
        for(int i=1; i<=nodes; i++ )
        {
            if( i == start) continue;
            printf("%d ",dp[i]);
        }
        printf("n");
    }
    return 0;
}

{“mode”:”full”,”isActive”:false}

Problem solution in C.

#include <stdio.h>
#include <stdlib.h>
typedef struct _node{
  int x;
  struct _node *next;
} node;
void add_edge(int x,int y);
void clean();
int N,M;
int ans[500000],d[500000],unsolve[2][500000],unsolve_size[2],temp[500000];
node *list[500000]={0};

int main(){
  int T,S,x,y,solve,r_idx,w_idx,c,temp_size,count,i;
  node *p;
  scanf("%d",&T);
  while(T--){
    solve=c=w_idx=1;
    r_idx=0;
    scanf("%d%d",&N,&M);
    for(i=0;i<N;i++){
      d[i]=0;
      ans[i]=-1;
    }
    for(i=0;i<M;i++){
      scanf("%d%d",&x,&y);
      add_edge(x-1,y-1);
      d[x-1]++;
      d[y-1]++;
    }
    scanf("%d",&S);
    ans[S-1]=0;
    unsolve_size[r_idx]=0;
    for(i=0;i<N;i++)
      if(ans[i]==-1)
        unsolve[r_idx][unsolve_size[r_idx]++]=i;
    while(unsolve_size[r_idx]){
      temp_size=0;
      unsolve_size[w_idx]=0;
      for(i=0;i<unsolve_size[r_idx];i++)
        if(d[unsolve[r_idx][i]]<solve)
          temp[temp_size++]=unsolve[r_idx][i];
        else{
          for(count=0,p=list[unsolve[r_idx][i]];p;p=p->next)
            if(ans[p->x]!=-1)
              count++;
          if(count<solve)
            temp[temp_size++]=unsolve[r_idx][i];
          else
            unsolve[w_idx][unsolve_size[w_idx]++]=unsolve[r_idx][i];
        }
      for(i=0;i<temp_size;i++)
        ans[temp[i]]=c;
      solve+=temp_size;
      c++;
      r_idx=(r_idx+1)%2;
      w_idx=(w_idx+1)%2;
    }
    for(i=0;i<N;i++)
      if(i!=S-1)
        printf("%d ",ans[i]);
    printf("n");
    clean();
  }
  return 0;
}
void add_edge(int x,int y){
  node *p1,*p2;
  p1=(node*)malloc(sizeof(node));
  p2=(node*)malloc(sizeof(node));
  p1->x=x;
  p1->next=list[y];
  list[y]=p1;
  p2->x=y;
  p2->next=list[x];
  list[x]=p2;
  return;
}
void clean(){
  int i;
  node *p1,*p2;
  for(i=0;i<N;i++)
    if(list[i]){
      p1=list[i];
      while(p1){
        p2=p1;
        p1=p1->next;
        free(p2);
      }
      list[i]=NULL;
    }
  return;
}

{“mode”:”full”,”isActive”:false}

Algorithms coding problems solutions

Post navigation

Previous post
Next post
  • Automating Image Format Conversion with Python: A Complete Guide
  • HackerRank Separate the Numbers solution
  • How AI Is Revolutionizing Personalized Learning in Schools
  • GTA 5 is the Game of the Year for 2024 and 2025
  • Hackerrank Day 5 loops 30 days of code solution
How to download udemy paid courses for free

Pages

  • About US
  • Contact US
  • Privacy Policy

Programing Practice

  • C Programs
  • java Programs

HackerRank Solutions

  • C
  • C++
  • Java
  • Python
  • Algorithm

Other

  • Leetcode Solutions
  • Interview Preparation

Programming Tutorials

  • DSA
  • C

CS Subjects

  • Digital Communication
  • Human Values
  • Internet Of Things
  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2025 Programmingoneonone | WordPress Theme by SuperbThemes