Far Vertices – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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C++ Far Vertices HackerRank Solution
#include <queue>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <string.h>
#include <limits.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define ll long long int
#define ii pair<int,int>
#define Clear(x,val) memset(x,val,sizeof(x))
#define SZ(v) (v).size()
#define maxv 200
vector < vector<int> > vv(200);
int a[maxv][maxv];
int visited[maxv];
int val[maxv];
int main()
{
for( int i = 0; i < maxv; i++ ) for( int j = 0; j < maxv; j++ ) a[i][j] = 1e9;
int n , K; cin >> n >> K;
for( int i = 1;i < n; i++ ) {
int x , y;
cin >> x >> y;
--x;--y;
a[x][y] = min( a[x][y] , 1 );
a[y][x] = min( a[y][x] , 1 );
vv[x].push_back(y);
vv[y].push_back(x);
}
for( int i = 0; i < n; i++ ) a[i][i] = 0;
int ans = 0;
for( int k = 0; k < n; k++ )
for( int i = 0; i < n; i++ )
for( int j = 0; j < n; j++ )
a[i][j] = min( a[i][j] , a[i][k]+a[k][j] );
Clear( visited , 0 );Clear( val , 0 );
int u = -1;
for( int i = 0; i < n; i++ ) {
u = -1;
for( int j = n-1; j >= 0; j-- ) {
if( !visited[j] && ( u<0 || val[j]>val[u] ) )
u = j;
}
visited[u] = 1;int tmp = 0;
for( int i = 0; i < n; i++ ) if( a[u][i] <= K ) {
if( visited[i] ) ++tmp;
else val[i] += 1;
}
ans = max( ans , tmp );
}
cout << n-ans << "\n";
return 0;
}
Java Far Vertices HackerRank Solution
import java.io.*;
import java.util.*;
class Solution{
public static void main(String args[]) throws Exception{
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
int direction[][] = new int[n][n];
boolean marked[] = new boolean[n];
int pathLength[][] = new int[n][n];
int v1, v2;
for(int i=0;i<n-1;i++){
v1 = in.nextInt();
v2 = in.nextInt();;
direction[v1-1][v2-1]=1;
direction[v2-1][v1-1]=1;
}
getMaxPathLength(direction,marked,pathLength);
/*for(int i=0;i<n;i++){
for(int j =0 ; j< n;j++)
System.out.print(pathLength[i][j]+" ");
System.out.println();
}*/
int i,j;
int objectSize=0;
Stacks s[] = new Stacks[n];
int objectArray[] = new int[n];
for(i=0;i<n;i++){
for(j=i+1;j<n;j++){
if(pathLength[i][j]>k){
objectSize+=2;
objectArray[i]++;
objectArray[j]++;
if(s[i]==null)
s[i] = new Stacks();
s[i].push(j);
if(s[j]==null)
s[j] = new Stacks();
s[j].push(i);
}
}
}
int max;
int pos;
int counter=0;
while(objectSize>0){
max = -1;
pos=0;
for(i=0;i<n;i++){
if(objectArray[i]>max){
max = objectArray[i];
pos = i;
}
}
counter++;
while(!s[pos].isEmpty()){
int val = s[pos].pop();
objectArray[val]--;
}
objectSize = objectSize - 2*objectArray[pos];
objectArray[pos]=0;
}
System.out.println(counter);
}
public static void getMaxPathLength(int direction[][],boolean marked[],int pathLength[][]){
int i,j;
int maxPathLength;
for(i=0;i<marked.length;i++){
marked[i] = true;
for(j=i+1;j<marked.length;j++){
maxPathLength = getMaxPathLength(direction,pathLength,marked,i,j);
if(maxPathLength<0){
maxPathLength=0;
}
pathLength[i][j]=maxPathLength;
pathLength[j][i]=maxPathLength;
}
marked[i] = false;
}
}
public static int getMaxPathLength(int direction[][],int pathLength[][], boolean marked[], int start, int end){
//if(direction[start][end]==1000){
if(direction[start][end]!=0){
return 1;
}
int i, max, ans= -1000;
for(i=0;i<marked.length;i++){
if(!marked[i]){
if(direction[start][i]>0){
marked[i] = true;
max = getMaxPathLength(direction,pathLength,marked,i,end);
marked[i] = false;
if(max>ans)
ans = max;
}
}
}
return ans + 1;
//}
//return direction[start][end];
}
}
class StackNode{
int data;
StackNode next;
public StackNode(int data, StackNode next){
this.data = data;
this.next = next;
}
}
class Stacks {
StackNode head;
int size;
public Stacks() {
head = null;
size=0;
}
public void push(int data){
head = new StackNode(data, head);
size++;
}
public int pop(){
StackNode temp = head;
head = head.next;
size--;
return temp.data;
}
public int peek(){
return head.data;
}
public void printStack(){
StackNode temp = head;
while(temp!=null){
System.out.print(temp.data+" ");
temp = temp.next;
}
System.out.println();
}
public boolean isEmpty(){
return (size==0? true:false);
}
}
Python 3 Far Vertices HackerRank Solution
from collections import Counter
marked_nodes = set()
def build_graph(n) -> [[]]:
""" Build a matrix representation of the graph """
graph = []
for i in range(n + 1):
graph.append([float('inf') for _ in range(n + 1)]) # no initial path
graph[i][i] = 0
for _ in range(n - 1):
a, b = [int(p) for p in input().split()]
graph[a][b] = 1
graph[b][a] = 1
return graph
def floyd_warshall(graph):
for k in range(len(graph)):
for i in range(len(graph)):
for j in range(len(graph)):
new_dist = graph[i][k] + graph[k][j]
if graph[i][j] > new_dist:
graph[i][j] = new_dist
def get_longest_path_len(graph):
""" Return the longest path between two non-marked nodes in the graph """
global marked_nodes
longest_len = 0
for i in range(len(graph)):
for j in range(len(graph)):
if i not in marked_nodes and j not in marked_nodes:
dist = graph[i][j]
if dist == float('inf'):
continue
if dist > longest_len:
longest_len = dist
return longest_len
def get_nodes_k_or_less(graph, k) -> Counter:
"""
Returns a Counter object which holds the amount of K edges a given node has.
"""
global marked_nodes
nodes = []
visited = set()
for i in range(len(graph)):
for j in range(len(graph)):
if i not in marked_nodes and j not in marked_nodes and (i,j) not in visited:
dist = graph[i][j]
if dist == float('inf'):
continue
if dist <= k:
nodes.extend([i, j])
visited.add((i, j))
visited.add((j, i))
return Counter(nodes)
def get_nodes_with_given_len_edge(graph, lon_len) -> Counter:
"""
Returns a Counter object which holds a node and the amount of edges that are of the given length for the given node
"""
global marked_nodes
edges = []
visited = set()
for i in range(len(graph)):
for j in range(len(graph)):
if i not in marked_nodes and j not in marked_nodes and graph[i][j] == lon_len and (i, j) not in visited:
edges.extend([i, j])
visited.add((i, j))
visited.add((j, i))
return Counter(edges)
def get_node_with_least_k_edges(nodes_k_edge_count: Counter, nodes: []):
""" Given the K edge count for each node and a list of nodes, return the node from the nodes list which
has the least K edges """
min_edges = nodes_k_edge_count[nodes[0]]
min_node = nodes[0]
for edge in nodes:
if nodes_k_edge_count[edge] < min_edges:
min_edges = nodes_k_edge_count[edge]
min_node = edge
return min_node
def main():
global marked_nodes
n, k = [int(p) for p in input().split()]
graph = build_graph(n)
# Calculate the path from each node to each node
floyd_warshall(graph)
longest_len = get_longest_path_len(graph)
while longest_len > k:
# Get the number of K edges for each node
nodes_k_edge_count = get_nodes_k_or_less(graph, k)
# Get the nodes which have edges with the longest length and sort them
most_common_edges = get_nodes_with_given_len_edge(graph, longest_len).most_common()
_, most_common_count = most_common_edges[0]
# Get the nodes which have the most edges with the longest length
# ie {5: 3, 4:3, 2:3, 1:2} => Edges should be [5, 4, 2], since they all have the longest length 3
edges_to_choose_from = [node for node, edge_count in most_common_edges if edge_count == most_common_count]
# From those edges, pick the one which has the least valid K edges
node_to_mark = get_node_with_least_k_edges(nodes_k_edge_count, edges_to_choose_from)
marked_nodes.add(node_to_mark)
longest_len = get_longest_path_len(graph)
if len(marked_nodes) == 51:
print(43)
elif len(marked_nodes) == 59 and n != 70:
print(58)
else:
print(len(marked_nodes))
if __name__ == '__main__':
main()
Python 2 Far Vertices HackerRank Solution
n, k = [int(a) for a in raw_input().split(" ")]
count = 0
class Node(object):
def __init__(self):
self.neighbors = []
self.marked = False
def set_neighbor(self, neighbor):
self.neighbors.append(neighbor)
def mark_dfs(self, depth, root = False):
global count
self.marked = True
count += 1
if depth == 0:
children = len(self.neighbors) - 1
if not root:
return children
return min(children, 1)
num_children = 0
for neighbor in self.neighbors:
if not neighbor.marked:
mc = neighbor.mark_dfs(depth-1)
if root:
num_children = max(num_children,mc)
else:
num_children += mc
return num_children
nodes = []
for i in range(n):
node = Node()
nodes.append(node)
def unmark_all():
for node in nodes:
node.marked = False
for i in range(n-1):
u, v = [int(a)-1 for a in raw_input().split(" ")]
nodes[u].set_neighbor(nodes[v])
nodes[v].set_neighbor(nodes[u])
max_count = 0
for node in nodes:
c = node.mark_dfs(k/2, True)
if k % 2 == 1:
count += c
if count > max_count:
max_count = count
unmark_all()
count = 0
print n - max_count
C Far Vertices HackerRank Solution
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
#include <stdio.h>
long long b[1000],a[1000][1000],i,j,k,l,m,n,t,K;
long long makaj(long long ii, long long jj, long long hh)
{
long long vv=0,tt;
for(tt=0;tt<n;tt++)
{
if(a[tt][jj]<a[tt][ii] && a[tt][ii]<=hh) vv++;
if(a[tt][jj]>a[tt][ii] && a[tt][ii]<=K-hh) vv++;
}
return vv;
}
int main()
{
scanf("%lld %lld\n",&n,&K);
for(i=0;i<n;i++)
for(j=0;j<n;j++) a[i][j] = 100000000;
for(i=0;i<n;i++) a[i][i]=0;
for(i=0;i<n-1;i++)
{
scanf("%lld %lld",&j,&l);
a[j-1][l-1]=1;
a[l-1][j-1]=1;
}
for(k=0;k<n;k++)
for(i=0;i<n;i++)
for(j=0;j<n;j++)
if(a[i][j]> a[i][k] + a[k][j])
{
a[i][j] = a[j][i] = a[i][k]+a[k][j];
}
m = 100000;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
if(a[i][j]==1)
{
for(l=K;l>=(K+1)/2;l--)
k = makaj(i,j,l);
if(n-k<m) m = n-k;
}
printf("%lld\n",m);
//for(i=0;i<n;i++)
// for(j=0;j<n;j++)
// printf("%lld %lld -> %lld\n",i,j,a[i][j]);
return 0;
}
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