Hackerrank Roads and Libraries Solution
.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}
The Ruler of HackerLand believes that every citizen of the country should have access to a library. Unfortunately, HackerLand was hit by a tornado that destroyed all of its libraries and obstructed its roads! As you are the greatest programmer of HackerLand, the ruler wants your help to repair the roads and build some new libraries efficiently.
HackerLand has cities numbered from to . The cities are connected by bidirectional roads. A citizen has access to a library if:
- Their city contains a library.
- They can travel by road from their city to a city containing a library.
The following figure is a sample map of HackerLand where the dotted lines denote obstructed roads:
The cost of repairing any road is dollars, and the cost to build a library in any city is dollars. If in the above example and , we would build roads at a cost of and libraries for a cost of . We don't need to rebuild one of the roads in the cycle .
You are given queries, where each query consists of a map of HackerLand and value of and . For each query, find the minimum cost of making libraries accessible to all the citizens and print it on a new line.
Function Description
Complete the function roadsAndLibraries in the editor below. It must return the minimal cost of providing libraries to all, as an integer.
roadsAndLibraries has the following parameters:
- n: integer, the number of cities
- c_lib: integer, the cost to build a library
- c_road: integer, the cost to repair a road
- cities: 2D array of integers where each contains two integers that represent cities connected by an obstructed road .
Input Format.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}
The first line contains a single integer , that denotes the number of queries.
The subsequent lines describe each query in the following format:
- The first line contains four space-separated integers that describe the respective values of , , and , the number of cities, number of roads, cost of a library and cost of a road.
- Each of the next lines contains two space-separated integers, and , that describe a bidirectional road that connects cities and .
Constraints.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}
- Each road connects two distinct cities.
Output Format.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}
For each query, print an integer that denotes the minimum cost to make libraries accessible to all the citizens on a new line.
Sample Input.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue}
2
3 3 2 1
1 2
3 1
2 3
6 6 2 5
1 3
3 4
2 4
1 2
2 3
5 6
Sample Output.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue}
4
12
Explanation.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}
Perform the following queries:
HackerLand contains cities connected by bidirectional roads. The price of building a library is and the price for repairing a road is .
The cheapest way to make libraries accessible to all is to:
- Build a library in city at a cost of .
- Repair the road between cities and at a cost of .
- Repair the road between cities and at a cost of .This gives a total cost of . Note that the road between cities and does not need to be repaired each is connected to city .
In this scenario it is optimal to build a library in each city because the cost of building a library () is less than the cost of repairing a road ().
There are cities, so the total cost is .
Solution in java8
Approach 1.
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int q = in.nextInt();
for(int a0 = 0; a0 < q; a0++){
int n = in.nextInt();
Graph g=new Graph(n,false);
int m = in.nextInt();
int clib = in.nextInt();
int croad = in.nextInt();
for(int a1 = 0; a1 < m; a1++){
int city_1 = in.nextInt();
int city_2 = in.nextInt();
g.insertEdge(city_1-1,city_2-1);
}
System.out.println(getCost(g,clib,croad));
}
}
private static long getCost(Graph g,long clib,long croad){
if(clib<croad){
return clib*g.v;
}
boolean visited[]=new boolean[g.v];
long cost=0;
for(int i=0;i<g.v;i++){
if(!visited[i]){
int x=DFSUtil(g,visited,i);
cost+=(x-1)*croad;
cost+=clib;
}
}
return cost;
}
private static int DFSUtil(Graph g,boolean visited[],int current){
int count=1;
visited[current]=true;
for(int a:g.list[current]){
if(!visited[a]){
count+=DFSUtil(g,visited,a);
}
}
return count;
}
}
class Graph {
int v;
LinkedList<Integer> list[];
private final boolean directed;
/** Constructor of Graph, takes no. of vertex and boolean value to specify whether graph is directed or not */
Graph(int v,boolean directed){
this.v=v;
this.directed=directed;
list=new LinkedList[v];
for(int i=0;i<v;i++){
list[i]=new LinkedList<Integer>();
}
}
/* vertex numbering start with 0 as well as adjacency list no. start with 0 */
public void insertEdge(int a,int b){
list[a].add(b);
if(!directed){
list[b].add(a);
}
}
}
Approach 2.
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
static long roadsAndLibraries(int x, int y) {
long cost = 0;
for (int i = 0; i < adj.size(); i++) {
if (!visited[i]) {
count = 0;
dfs(i);
if (x > y) {
cost += x + y * (count - 1);
} else {
cost += x * count;
}
}
}
return cost;
}
static void dfs(int i) {
visited[i] = true;
count++;
List<Integer> list = adj.get(i);
for (int j = 0; j < list.size(); j++) {
if (!visited[list.get(j)]) {
dfs(list.get(j));
}
}
}
private static List<List<Integer>> adj;
private static boolean[] visited;
private static int count;
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int q = in.nextInt();
for(int a0 = 0; a0 < q; a0++){
int n = in.nextInt();
adj = new ArrayList<>();
for (int i = 0; i < n; i++) {
adj.add(new ArrayList<>());
}
visited = new boolean[n];
int m = in.nextInt();
int x = in.nextInt();
int y = in.nextInt();
for(int a1 = 0; a1 < m; a1++){
int city_1 = in.nextInt();
int city_2 = in.nextInt();
adj.get(city_1 - 1).add(city_2 - 1);
adj.get(city_2 - 1).add(city_1 - 1);
}
System.out.println(roadsAndLibraries(x, y));
}
in.close();
}
}
Approach 3.
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
private static List<List<Integer>> adj;
private static boolean[] visited;
private static int count;
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int q = in.nextInt();
for(int a0 = 0; a0 < q; a0++){
int n = in.nextInt();
adj = new ArrayList<>();
for (int i = 0; i < n; i++) {
adj.add(new ArrayList<>());
}
visited = new boolean[n];
int m = in.nextInt();
int x = in.nextInt();
int y = in.nextInt();
for(int a1 = 0; a1 < m; a1++){
int city_1 = in.nextInt();
int city_2 = in.nextInt();
adj.get(city_1 - 1).add(city_2 - 1);
adj.get(city_2 - 1).add(city_1 - 1);
}
System.out.println(roadsAndLibraries(x, y));
}
in.close();
}
private static long roadsAndLibraries(int x, int y) {
long cost = 0;
for (int i = 0; i < adj.size(); i++) {
if (!visited[i]) {
count = 0;
dfs(i);
if (x > y) {
cost += x + y * (count - 1);
} else {
cost += x * count;
}
}
}
return cost;
}
private static void dfs(int i) {
visited[i] = true;
count++;
List<Integer> list = adj.get(i);
for (int j = 0; j < list.size(); j++) {
if (!visited[list.get(j)]) {
dfs(list.get(j));
}
}
}
}
Solution in python3
Approach 1.
#!/bin/python3
def dfs(graph, start, visited):
stack = [start]
while stack:
vertex = stack.pop()
if vertex not in visited:
visited.add(vertex)
stack.extend(graph[vertex] - visited)
def roadsAndLibraries(n, cLib, cRoad, cities):
# construct graph as adjacency-list
graph = {i: set() for i in range(1,n+1)}
for city1, city2 in cities:
graph[city1].add(city2)
graph[city2].add(city1)
# if libraries are cheaper than roads, build
# library in every city and build no roads
if cLib < cRoad:
return cLib * n
# determine number of connected components (CC) in graph
visited = set()
nCC = 0
for city in range(1, n+1):
if city not in visited:
dfs(graph, city, visited)
nCC += 1
# 1 library per CC, nCitiesPerCC-1 roads per CC
return nCC*cLib + (n-nCC)*cRoad
if __name__ == '__main__':
q = int(input())
for _ in range(q):
n, m, cLib, cRoad = map(int, input().split())
cities = []
for _ in range(m):
cities.append(list(map(int, input().split())))
result = roadsAndLibraries(n, cLib, cRoad, cities)
print(result)Approach 2.
#!/bin/python3
import math
import os
import random
import re
import sys
from collections import defaultdict
def build_graph(n, cities):
g = defaultdict(set)
for x, y in cities:
g[x].add(y)
g[y].add(x)
return g
def get_cc_sizes(g, n):
visited = set()
sizes = []
for i in range(1, n+1):
if i in visited:
continue
visited.add(i)
cc_size = 1
to_visit = set(g[i])
while to_visit:
j = to_visit.pop()
if j in visited:
continue
visited.add(j)
cc_size += 1
to_visit.update(g[j])
sizes.append(cc_size)
return sizes
def cc_cost(size, c_road, c_lib):
assert size > 0
return (size - 1) * c_road + c_lib
# Complete the roadsAndLibraries function below.
def roadsAndLibraries(n, c_lib, c_road, cities):
if c_lib < c_road:
return n * c_lib
g = build_graph(n, cities)
cc_sizes = get_cc_sizes(g, n)
cost = sum(cc_cost(size, c_road, c_lib) for size in cc_sizes)
return cost
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
q = int(input())
for q_itr in range(q):
nmC_libC_road = input().split()
n = int(nmC_libC_road[0])
m = int(nmC_libC_road[1])
c_lib = int(nmC_libC_road[2])
c_road = int(nmC_libC_road[3])
cities = []
for _ in range(m):
cities.append(list(map(int, input().rstrip().split())))
result = roadsAndLibraries(n, c_lib, c_road, cities)
fptr.write(str(result) + '\n')
fptr.close()
Approach 3.
#!/bin/python3
import math
import os
import random
import re
import sys
from collections import deque, defaultdict
def bfs(graph, start_node, visited):
size = 0
queue = deque()
visited.add(start_node)
queue.append(start_node)
while len(queue) > 0:
current_node = queue.popleft()
size += 1
for node in graph[current_node]:
if node not in visited:
visited.add(node)
queue.append(node)
return size, visited
# Complete the roadsAndLibraries function below.
def roadsAndLibraries(n, c_lib, c_road, cities):
if c_lib <= c_road:
return c_lib * n
# create graph
graph = defaultdict(set)
for n1, n2 in cities:
graph[n1].add(n2)
graph[n2].add(n1)
allcities = set(range(1, n+1))
visited = set()
cost = 0
for node in graph.keys():
if node not in visited:
size, visited = bfs(graph, node, visited)
cost += (size - 1) * c_road + c_lib
standalone = allcities - visited
cost += len(standalone) * c_lib
return cost
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
q = int(input())
for q_itr in range(q):
nmC_libC_road = input().split()
n = int(nmC_libC_road[0])
m = int(nmC_libC_road[1])
c_lib = int(nmC_libC_road[2])
c_road = int(nmC_libC_road[3])
cities = []
for _ in range(m):
cities.append(list(map(int, input().rstrip().split())))
result = roadsAndLibraries(n, c_lib, c_road, cities)
fptr.write(str(result) + '\n')
fptr.close()
Solution in cpp
Approach 1.
#include <bits/stdc++.h>
using namespace std;
vector<int> p;
int f(int a){return p[a]==a?a:p[a]=f(p[a]);}
void u(int a, int b){p[f(a)] = f(b);}
signed main()
{
int T;cin >> T;while(T--){
int N, M, a, b;
long long c, d;
cin >> N >> M >> c >> d;
p.clear();p.resize(N);
iota(p.begin(), p.end(), 0);
while(M--){
cin >> a >> b;
--a, --b;
u(a, b);
}
int comp=0;
for(int i=0;i<N;++i){
if(p[i]==i)++comp;
}
cout << (comp*c+(N-comp)*min(c, d)) << "\n";
}
return 0;
}Approach 2.
#include <bits/stdc++.h>
using namespace std;
void dfs(vector<int> v[], bool arr[], int s){
arr[s] = true;
for(int i = 0; i <v[s].size(); i++){
if(arr[v[s][i]] == false){
dfs(v,arr,v[s][i]);
}
}
}
int main() {
int q;
cin >> q;
for(int a0 = 0; a0 < q; a0++){
int n;
int m;
long x;
long y;
cin >> n >> m >> x >> y;
vector<int> v[n];
bool arr[n];
for(int a1 = 0; a1 < m; a1++){
int city_1;
int city_2;
cin >> city_1 >> city_2;
v[city_1-1].push_back(city_2-1);
v[city_2-1].push_back(city_1-1);
}
for(int i = 0; i<n; i++){
arr[i] = false;
}
int j = 0;
for(int i = 0; i<n; i++){
if(arr[i] == false){
dfs(v,arr,i);
j++;
}
}
long long int ans = j * x;
ans += ((x < y) ? x : y)*(n-j);
cout<<ans<<endl;
}
return 0;
}
Approach 3.
#include <bits/stdc++.h>
using namespace std;
long long dfs(vector<int> &visited,list<int> *adj,int s)
{
visited[s]=1;
list<int> ::iterator it;
long long sum=1;
for(it=adj[s].begin();it!=adj[s].end();it++)
{
if(visited[*it]==0)
sum+=dfs(visited,adj,*it);
}
return sum;
}
int main() {
int q;
cin >> q;
for(int a0 = 0; a0 < q; a0++){
int n;
int m;
long x;
long y;
long long sum=0;
cin >> n >> m >> x >> y;
list<int> *adj=new list<int>[n];
for(int a1 = 0; a1 < m; a1++){
int city_1;
int city_2;
cin >> city_1 >> city_2;
adj[city_1-1].push_back(city_2-1);
adj[city_2-1].push_back(city_1-1);
}
vector<int> visited(n,0);
for(int i=0;i<n;i++)
{
if(visited[i]==0)
{
int n=dfs(visited,adj,i);
sum+= (long long)x+(long long)((n-1)*(min(x,y)));
}
}
cout<<sum<<endl;
}
return 0;
}