Given a group of N people, each having a unique ID value from 0 to (N – 1) and an array arr[] of M elements of the form {U, V, time} representing that the person U will become acquainted with person V at the given time. Let’s say that person U is acquainted with person V if U is friends with V, or U is a friend of someone acquainted with V. The task is to find the earliest time at which every person became acquainted with each other.
Examples:
Input: N = 4, arr[] = {{0, 1, 2}, {1, 2, 3}, {2, 3, 4}}
Output: 4
Explanation: Initially, the number of people are 4, i.e, {0}, {1}, {2}, {3}.
- At time = 2, {0} and {1} became friends. Therefore, the group of acquainted people becomes {0, 1}, {2}, {3}.
- At time = 3, {1} and {2} became friends. Therefore, the group of acquainted people becomes {0, 1, 2}, {3}.
- At time = 4, {2} and {3} became friends. Therefore, the group of acquainted people becomes {0, 1, 2, 3}.
Hence, at time = 4, every person became acquainted with each other.
Input: N = 6, arr[] = {{0, 1, 4}, {3, 4, 5}, {2, 3, 14}, {1, 5, 24}, {2, 4, 12}, {0, 3, 42}, {1, 2, 41}, {4, 5, 11}}
Output: 24
Approach: The given problem can be solved using the Disjoint Set Union Data Structure. The idea is to perform the union operations between people in order of the increasing value of time. The required answer will be the time when all people belong to the same set. Follow the steps below to solve the given problem:
- Implement the Disjoint Set Union Data Structure with the union and findSet functions according to the algorithm discussed here.
- Initialize a variable time, which stores the value of the current timestamp of the DSU.
- Sort the given array arr[] in increasing order of time.
- Traverse the given array arr[] using variable i, and perform union operation between (Ui, Vi) and update the current timestamp to timeiif Ui and Vi belong to the different sets.
- If the total number of sets after completely traversing through the array arr[] is 1, return the value of the variable time, else return -1.
Below is the implementation of the above approach:
// C++ program for the above approach #include "bits/stdc++.h" using namespace std;
// Implementation of DSU class UnionFind {
vector< int > parent, rank;
// Stores the current timestamp
int time ;
// Stores the count of current sets
int count;
public :
// Constructor to create and
// initialize sets of N items
UnionFind( int N)
: parent(N), rank(N), count(N)
{
time = 0;
// Creates N single item sets
for ( int i = 0; i < N; i++) {
parent[i] = i;
rank[i] = 1;
}
}
// Function to find the set of
// given item node
int find( int node)
{
if (node == parent[node])
return node;
return parent[node]
= find(parent[node]);
}
// Function to perform the union of
// two sets represented by u and v,
// and update the value of time
void performUnion( int u, int v,
int updatedTime)
{
if (count == 1)
return ;
// Find current sets of u and v
int rootX = find(u);
int rootY = find(v);
if (rootX != rootY) {
// Put smaller ranked item under
// bigger ranked item if ranks
// are different
if (rank[rootX] < rank[rootY]) {
parent[rootX] = rootY;
}
else if (rank[rootX] > rank[rootY]) {
parent[rootY] = rootX;
}
else {
parent[rootX] = rootY;
rank[rootY] += 1;
}
// Update the value of the
// current timestamp
time = updatedTime;
// Update the value of
// set count
count--;
}
}
// Function to return current
// set count
int getCount() { return count; }
// Function to return current
// time stamp
int getTime() { return time ; }
}; // Comparator function to sort the array // in increasing order of 3rd element bool cmp(vector< int >& A, vector< int >& B)
{ return A[2] <= B[2];
} // Function to find the earliest time when // everyone became friends to each other int earliestTime(vector<vector< int > >& arr,
int N)
{ // Sort array arr[] in increasing order
sort(arr.begin(), arr.end(), cmp);
// Initialize a DSU with N elements
UnionFind unionFind(N);
// Iterate over array arr[] perform
// union operation for each element
for ( auto & it : arr) {
// Perform union operation on
// it[0] and it[1] and update
// timestamp to it[2]
unionFind.performUnion(
it[0], it[1], it[2]);
}
// Return Answer
if (unionFind.getCount() == 1) {
return unionFind.getTime();
}
else {
return -1;
}
} // Driver Code int main()
{ int N = 6;
vector<vector< int > > arr
= { { 0, 1, 4 }, { 3, 4, 5 },
{ 2, 3, 14 }, { 1, 5, 24 },
{ 2, 4, 12 }, { 0, 3, 42 },
{ 1, 2, 41 }, { 4, 5, 11 } };
cout << earliestTime(arr, N);
return 0;
} |
// Java program for the above approach import java.util.*;
// Implementation of DSU class UnionFind {
private int [] parent;
private int [] rank;
private int time;
private int count;
// Constructor to create and
// initialize sets of N items
public UnionFind( int N)
{
parent = new int [N];
rank = new int [N];
time = 0 ;
count = N;
// Creates N single item sets
for ( int i = 0 ; i < N; i++) {
parent[i] = i;
rank[i] = 1 ;
}
}
// Function to find the set of
// given item node
public int find( int node)
{
if (node == parent[node])
return node;
return parent[node] = find(parent[node]);
}
// Function to perform the union of
// two sets represented by u and v,
// and update the value of time
public void performUnion( int u, int v, int updatedTime)
{
if (count == 1 )
return ;
// Find current sets of u and v
int rootX = find(u);
int rootY = find(v);
// Put smaller ranked item under
// bigger ranked item if ranks
// are different
if (rootX != rootY) {
if (rank[rootX] < rank[rootY]) {
parent[rootX] = rootY;
}
else if (rank[rootX] > rank[rootY]) {
parent[rootY] = rootX;
}
else {
parent[rootX] = rootY;
rank[rootY] += 1 ;
}
// Update the value of the
// current timestamp
time = updatedTime;
// Update the value of
// set count
count--;
}
}
// Function to return current
// set count
public int getCount() { return count; }
// Function to return current
// time stamp
public int getTime() { return time; }
} class Main {
private static int earliestTime(List< int []> arr, int N)
{
arr.sort(Comparator.comparingInt(a -> a[ 2 ]));
UnionFind unionFind = new UnionFind(N);
for ( int [] it : arr) {
unionFind.performUnion(it[ 0 ], it[ 1 ], it[ 2 ]);
}
return unionFind.getCount() == 1
? unionFind.getTime()
: - 1 ;
}
public static void main(String[] args)
{
int N = 6 ;
List< int []> arr
= Arrays.asList( new int [][] { { 0 , 1 , 4 },
{ 3 , 4 , 5 },
{ 2 , 3 , 14 },
{ 1 , 5 , 24 },
{ 2 , 4 , 12 },
{ 0 , 3 , 42 },
{ 1 , 2 , 41 },
{ 4 , 5 , 11 } });
System.out.println(earliestTime(arr, N));
}
} |
# Python3 program for the above approach # Class to implement Union Find algorithm from typing import List
# Constructor to create and initialize sets of N items class UnionFind:
def __init__( self , N):
self .parent = [i for i in range (N)]
self .rank = [ 1 for i in range (N)]
self .time = 0
self .count = N
# Function to find the set of given item node def find( self , node):
if node = = self .parent[node]:
return node
self .parent[node] = self .find( self .parent[node])
return self .parent[node]
# Function to perform the union of two sets # represented by u and v, and update the value of time
def performUnion( self , u, v, updatedTime):
if self .count = = 1 :
return
rootX = self .find(u)
rootY = self .find(v)
if rootX ! = rootY:
if self .rank[rootX] < self .rank[rootY]:
self .parent[rootX] = rootY
elif self .rank[rootX] > self .rank[rootY]:
self .parent[rootY] = rootX
else :
self .parent[rootX] = rootY
self .rank[rootY] + = 1
self .time = updatedTime
self .count - = 1
def getCount( self ):
return self .count
def getTime( self ):
return self .time
# Function to find the earliest time when everyone became friends to each other def earliestTime(arr: List [ List [ int ]], N: int ) - > int :
arr = sorted (arr, key = lambda x: x[ 2 ])
unionFind = UnionFind(N)
for it in arr:
# Function to return current set count
unionFind.performUnion(it[ 0 ], it[ 1 ], it[ 2 ])
if unionFind.getCount() = = 1 :
# Function to return current time stamp
return unionFind.getTime()
else :
return - 1
# Driver Code if __name__ = = "__main__" :
N = 6
arr = [[ 0 , 1 , 4 ], [ 3 , 4 , 5 ],
[ 2 , 3 , 14 ], [ 1 , 5 , 24 ],
[ 2 , 4 , 12 ], [ 0 , 3 , 42 ],
[ 1 , 2 , 41 ], [ 4 , 5 , 11 ]]
print (earliestTime(arr, N))
|
// JavaScript code for the above approach
// Implementation of DSU
class UnionFind {
constructor(N) {
this .parent = new Array(N);
this .rank = new Array(N);
this .time = 0;
this .count = N;
// Creates N single item sets
for (let i = 0; i < N; i++) {
this .parent[i] = i;
this .rank[i] = 1;
}
}
// Function to find the set of
// given item node
find(node) {
if (node === this .parent[node]) {
return node;
}
return ( this .parent[node] = this .find( this .parent[node]));
}
// Function to perform the union of
// two sets represented by u and v,
// and update the value of time
performUnion(u, v, updatedTime) {
if ( this .count === 1) {
return ;
}
// Find current sets of u and v
let rootX = this .find(u);
let rootY = this .find(v);
if (rootX !== rootY) {
// Put smaller ranked item under
// bigger ranked item if ranks
// are different
if ( this .rank[rootX] < this .rank[rootY]) {
this .parent[rootX] = rootY;
} else if ( this .rank[rootX] > this .rank[rootY]) {
this .parent[rootY] = rootX;
} else {
this .parent[rootX] = rootY;
this .rank[rootY] += 1;
}
// Update the value of the
// current timestamp
this .time = updatedTime;
// Update the value of
// set count
this .count--;
}
}
// Function to return current
// set count
getCount() {
return this .count;
}
// Function to return current
// time stamp
getTime() {
return this .time;
}
}
// Comparator function to sort the array
// in increasing order of 3rd element
function cmp(A, B) {
return A[2] - B[2];
}
// Function to find the earliest time when
// everyone became friends to each other
function earliestTime(arr, N) {
// Sort array arr[] in increasing order
arr.sort(cmp);
// Initialize a DSU with N elements
let unionFind = new UnionFind(N);
// Iterate over array arr[] perform
// union operation for each element
for (let it of arr) {
// Perform union operation on
// it[0] and it[1] and update
// timestamp to it[2]
unionFind.performUnion(it[0], it[1], it[2]);
}
// Return Answer
if (unionFind.getCount() === 1) {
return unionFind.getTime();
} else {
return -1;
}
}
// Driver Code
let N = 6;
let arr = [
[0, 1, 4],
[3, 4, 5],
[2, 3, 14],
[1, 5, 24],
[2, 4, 12],
[0, 3, 42],
[1, 2, 41],
[4, 5, 11],
];
console.log(earliestTime(arr, N));
// This code is contributed by Potta Lokesh. |
// C# program for the above approach using System;
using System.Collections.Generic;
public class UnionFind
{ private int [] parent;
private int [] rank;
private int time;
private int count;
// Constructor to create and
// initialize sets of N items
public UnionFind( int N)
{
parent = new int [N];
rank = new int [N];
time = 0;
count = N;
// Creates N single item sets
for ( int i = 0; i < N; i++)
{
parent[i] = i;
rank[i] = 1;
}
}
// Function to find the set of
// given item node
public int Find( int node)
{
if (node == parent[node])
{
return node;
}
return parent[node] = Find(parent[node]);
}
// Function to perform the union of
// two sets represented by u and v,
// and update the value of time
public void PerformUnion( int u, int v, int updatedTime)
{
if (count == 1)
{
return ;
}
// Find current sets of u and v
int rootX = Find(u);
int rootY = Find(v);
// Put smaller ranked item under
// bigger ranked item if ranks
// are different
if (rootX != rootY)
{
if (rank[rootX] < rank[rootY])
{
parent[rootX] = rootY;
}
else if (rank[rootX] > rank[rootY])
{
parent[rootY] = rootX;
}
else
{
parent[rootX] = rootY;
rank[rootY] += 1;
}
// Update the value of the
// current timestamp
time = updatedTime;
// Update the value of
// set count
count--;
}
}
// Function to return current
// set count
public int GetCount() { return count; }
// Function to return current
// time stamp
public int GetTime() { return time; }
} class Program
{ // Function to find the earliest time when
// everyone became friends to each other
private static int EarliestTime(List< int []> arr, int N)
{
// Sort array arr[] in increasing order
arr.Sort((a, b) => a[2].CompareTo(b[2]));
UnionFind unionFind = new UnionFind(N);
// Iterate over array arr[] perform
// union operation for each element
foreach ( int [] it in arr)
{
// Perform union operation on
// it[0] and it[1] and update
// timestamp to it[2]
unionFind.PerformUnion(it[0], it[1], it[2]);
}
// Return Answer
return unionFind.GetCount() == 1
? unionFind.GetTime()
: -1;
}
// Driver Code
static void Main( string [] args)
{
int N = 6;
List< int []> arr = new List< int []>
{
new int [] { 0, 1, 4 },
new int [] { 3, 4, 5 },
new int [] { 2, 3, 14 },
new int [] { 1, 5, 24 },
new int [] { 2, 4, 12 },
new int [] { 0, 3, 42 },
new int [] { 1, 2, 41 },
new int [] { 4, 5, 11 }
};
Console.WriteLine(EarliestTime(arr, N));
}
} // This code is contributed by divyansh2212 |
24
Time Complexity: O(N*log(N))
Auxiliary Space: O(N)