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 timei if 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++
// C++ program for the above approach#include "bits/stdc++.h"using namespace std;// Implementation of DSUclass 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 elementbool cmp(vector<int>& A, vector<int>& B){ return A[2] <= B[2];}// Function to find the earliest time when// everyone became friends to each otherint 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 Codeint 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
// Java program for the above approachimport java.util.*;// Implementation of DSUclass 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
# Python3 program for the above approach# Class to implement Union Find algorithmfrom typing import List# Constructor to create and initialize sets of N itemsclass 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 otherdef 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 Codeif __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
// 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#
// C# program for the above approachusing 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 |
Output:
24
Time Complexity: O(N*log(N))
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
