Given an array of pairs arr[][] of length N, and an array queries[] of length M, and an integer R, where each query contains an integer from 1 to R, the task for every queries[i] is to find the set to which it belongs and find the total number of elements of the set.
Note: Initially every integer from 1 to R belongs to the distinct set.
Examples:
Input: R = 5, arr[][] = {{1, 2}, {2, 3}, {4, 5}}, queries[] = {2, 4, 1, 3}
Output: 3 2 3 3
Explanation:
After making the sets from the arr[] pairs, {1, 2, 3}, {4, 5}
For the first query: 2 belongs to the set {1, 2, 3} and the total number of elements is 3.
For the second query: 4 belongs to the set {4, 5} and the total number of elements is 2.
For the third query: 1 belongs to the set {1, 2, 3} and the total number of elements is 3.
For the fourth query: 3 belongs to the set {1, 2, 3} and the total number of elements is 3.Input: R = 6, arr[][] = {{1, 3}, {2, 4}}, queries[] = {2, 5, 6, 1}
Output: 2 1 1 2
Approach: The given problem can be solved using the Disjoint Set Union. Initially, all the elements are in different sets, process the arr[] and do union operation on the given pairs and in union update, the total[] value for the parent element. For each query do find operation and for the returned parent element find the size of the current set as the value of total[parent]. Follow the steps below to solve the problem:
- Initialize the vectors parent(R + 1), rank(R + 1, 0), total(R + 1, 1).
- Iterate over the range [1, R+1) using the variable i and set the value of parent[I] as I.
- Iterate over the range [1, N-1] using the variable i and perform the Union Operation as Union(parent, rank, total, arr[I].first, arr[I].second).
-
Iterate over the range [1, M – 1] using the variable i and perform the following steps:
- Call for function for finding the parent of the current element queries[i] as Find(parent, queries[I]).
- Print the value of total[i] as the size of the current set.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to perform the find operation // of disjoint set union int Find(vector< int >& parent, int a)
{ return parent[a]
= (parent[a] == a)
? a
: (Find(parent, parent[a]));
} // Function to find the Union operation // of disjoint set union void Union(vector< int >& parent,
vector< int >& rank,
vector< int >& total,
int a, int b)
{ // Find the parent of a and b
a = Find(parent, a);
b = Find(parent, b);
if (a == b)
return ;
// If the rank are the same
if (rank[a] == rank[b]) {
rank[a]++;
}
if (rank[a] < rank[b]) {
int temp = a;
a = b;
b = temp;
}
// Update the parent for node b
parent[b] = a;
// Update the total number of
// elements of a
total[a] += total[b];
} // Function to find the total element // of the set which belongs to the // element queries[i] void findTotNumOfSet(vector<pair< int , int > >& arr,
vector< int >& queries,
int R, int N, int M)
{ // Stores the parent elements
// of the sets
vector< int > parent(R + 1);
// Stores the rank of the sets
vector< int > rank(R + 1, 0);
// Stores the total number of
// elements of the sets
vector< int > total(R + 1, 1);
for ( int i = 1; i < R + 1; i++) {
// Update parent[i] to i
parent[i] = i;
}
for ( int i = 0; i < N; i++) {
// Add the arr[i].first and
// arr[i].second elements to
// the same set
Union(parent, rank, total,
arr[i].first,
arr[i].second);
}
for ( int i = 0; i < M; i++) {
// Find the parent element of
// the element queries[i]
int P = Find(parent, queries[i]);
// Print the total elements of
// the set which belongs to P
cout << total[P] << " " ;
}
} // Driver Code int main()
{ int R = 5;
vector<pair< int , int > > arr{ { 1, 2 },
{ 2, 3 },
{ 4, 5 } };
vector< int > queries{ 2, 4, 1, 3 };
int N = arr.size();
int M = queries.size();
findTotNumOfSet(arr, queries, R, N, M);
return 0;
} |
// Java program for the above approach class GFG
{ // Function to perform the find operation // of disjoint set union static int Find( int [] parent, int a)
{ return parent[a]
= (parent[a] == a)
? a
: (Find(parent, parent[a]));
} // Function to find the Union operation // of disjoint set union static void Union( int [] parent,
int [] rank,
int [] total,
int a, int b)
{ // Find the parent of a and b
a = Find(parent, a);
b = Find(parent, b);
if (a == b)
return ;
// If the rank are the same
if (rank[a] == rank[b]) {
rank[a]++;
}
if (rank[a] < rank[b]) {
int temp = a;
a = b;
b = temp;
}
// Update the parent for node b
parent[b] = a;
// Update the total number of
// elements of a
total[a] += total[b];
} // Function to find the total element // of the set which belongs to the // element queries[i] static void findTotNumOfSet( int [][] arr,
int [] queries,
int R, int N, int M)
{ // Stores the parent elements
// of the sets
int [] parent = new int [R + 1 ];
// Stores the rank of the sets
int [] rank = new int [R + 1 ];
// Stores the total number of
// elements of the sets
int [] total = new int [R + 1 ];
for ( int i = 0 ; i < total.length; i++) {
total[i] = 1 ;
}
for ( int i = 1 ; i < R + 1 ; i++) {
// Update parent[i] to i
parent[i] = i;
}
for ( int i = 0 ; i < N; i++) {
// Add the arr[i][0] and
// arr[i][1] elements to
// the same set
Union(parent, rank, total,
arr[i][ 0 ],
arr[i][ 1 ]);
}
for ( int i = 0 ; i < M; i++) {
// Find the parent element of
// the element queries[i]
int P = Find(parent, queries[i]);
// Print the total elements of
// the set which belongs to P
System.out.print(total[P]+ " " );
}
} // Driver Code public static void main(String[] args)
{ int R = 5 ;
int [][] arr = { { 1 , 2 },
{ 2 , 3 },
{ 4 , 5 } };
int [] queries = { 2 , 4 , 1 , 3 };
int N = arr.length;
int M = queries.length;
findTotNumOfSet(arr, queries, R, N, M);
} } // This code is contributed by 29AjayKumar. |
# Python3 program for the above approach # Function to perform the find operation # of disjoint set union def Find(parent, a):
if (parent[a] = = a):
return a
else :
return Find(parent, parent[a])
# Function to find the Union operation # of disjoint set union def Union(parent, rank, total, a, b):
# Find the parent of a and b
a = Find(parent, a)
b = Find(parent, b)
if (a = = b):
return
# If the rank are the same
if (rank[a] = = rank[b]):
rank[a] + = 1
if (rank[a] < rank[b]):
temp = a
a = b
b = temp
# Update the parent for node b
parent[b] = a
# Update the total number of
# elements of a
total[a] + = total[b]
# Function to find the total element # of the set which belongs to the # element queries[i] def findTotNumOfSet(arr, queries, R, N, M):
# Stores the parent elements
# of the sets
parent = [ None ] * (R + 1 )
# Stores the rank of the sets
rank = [ 0 ] * (R + 1 )
# Stores the total number of
# elements of the sets
total = [ 1 ] * (R + 1 )
for i in range ( 1 , R + 1 ):
# Add the arr[i].first and
# arr[i].second elements to
# the same set
parent[i] = i
for i in range (N):
Union(parent, rank, total, arr[i][ 0 ], arr[i][ 1 ])
for i in range (M):
# Find the parent element of
# the element queries[i]
P = Find(parent, queries[i])
# Print the total elements of
# the set which belongs to P
print (total[P], end = " " )
# Driver code R = 5
arr = [[ 1 , 2 ], [ 2 , 3 ], [ 4 , 5 ]]
queries = [ 2 , 4 , 1 , 3 ]
N = len (arr)
M = len (queries)
findTotNumOfSet(arr, queries, R, N, M) # This code is contributed by parthmanchanda81 |
// C# program for the above approach using System;
public class GFG
{ // Function to perform the find operation // of disjoint set union static int Find( int [] parent, int a)
{ return parent[a]
= (parent[a] == a)
? a
: (Find(parent, parent[a]));
} // Function to find the Union operation // of disjoint set union static void Union( int [] parent,
int [] rank,
int [] total,
int a, int b)
{ // Find the parent of a and b
a = Find(parent, a);
b = Find(parent, b);
if (a == b)
return ;
// If the rank are the same
if (rank[a] == rank[b]) {
rank[a]++;
}
if (rank[a] < rank[b]) {
int temp = a;
a = b;
b = temp;
}
// Update the parent for node b
parent[b] = a;
// Update the total number of
// elements of a
total[a] += total[b];
} // Function to find the total element // of the set which belongs to the // element queries[i] static void findTotNumOfSet( int [,] arr,
int [] queries,
int R, int N, int M)
{ // Stores the parent elements
// of the sets
int [] parent = new int [R + 1];
// Stores the rank of the sets
int [] rank = new int [R + 1];
// Stores the total number of
// elements of the sets
int [] total = new int [R + 1];
for ( int i = 0; i < total.Length; i++) {
total[i] = 1;
}
for ( int i = 1; i < R + 1; i++) {
// Update parent[i] to i
parent[i] = i;
}
for ( int i = 0; i < N; i++) {
// Add the arr[i,0] and
// arr[i,1] elements to
// the same set
Union(parent, rank, total,
arr[i,0],
arr[i,1]);
}
for ( int i = 0; i < M; i++) {
// Find the parent element of
// the element queries[i]
int P = Find(parent, queries[i]);
// Print the total elements of
// the set which belongs to P
Console.Write(total[P]+ " " );
}
} // Driver Code public static void Main(String[] args)
{ int R = 5;
int [,] arr = { { 1, 2 },
{ 2, 3 },
{ 4, 5 } };
int [] queries = { 2, 4, 1, 3 };
int N = arr.GetLength(0);
int M = queries.GetLength(0);
findTotNumOfSet(arr, queries, R, N, M);
} } // This code is contributed by shikhasingrajput |
<script> // Javascript program for the above approach // Function to perform the find operation // of disjoint set union function Find(parent, a)
{ return (parent[a] = parent[a] == a ? a : Find(parent, parent[a]));
} // Function to find the Union operation // of disjoint set union function Union(parent, rank, total, a, b)
{ // Find the parent of a and b
a = Find(parent, a);
b = Find(parent, b);
if (a == b) return ;
// If the rank are the same
if (rank[a] == rank[b]) {
rank[a]++;
}
if (rank[a] < rank[b]) {
let temp = a;
a = b;
b = temp;
}
// Update the parent for node b
parent[b] = a;
// Update the total number of
// elements of a
total[a] += total[b];
} // Function to find the total element // of the set which belongs to the // element queries[i] function findTotNumOfSet(arr, queries, R, N, M)
{ // Stores the parent elements
// of the sets
let parent = new Array(R + 1);
// Stores the rank of the sets
let rank = new Array(R + 1).fill(0);
// Stores the total number of
// elements of the sets
let total = new Array(R + 1).fill(1);
for (let i = 1; i < R + 1; i++)
{
// Update parent[i] to i
parent[i] = i;
}
for (let i = 0; i < N; i++)
{
// Add the arr[i].first and
// arr[i].second elements to
// the same set
Union(parent, rank, total, arr[i][0], arr[i][1]);
}
for (let i = 0; i < M; i++)
{
// Find the parent element of
// the element queries[i]
let P = Find(parent, queries[i]);
// Print the total elements of
// the set which belongs to P
document.write(total[P] + " " );
}
} // Driver Code let R = 5; let arr = [ [1, 2],
[2, 3],
[4, 5],
]; let queries = [2, 4, 1, 3]; let N = arr.length; let M = queries.length; findTotNumOfSet(arr, queries, R, N, M); // This code is contributed by saurabh_jaiswal. </script> |
3 2 3 3
Time Complexity: O(M*log R)
Auxiliary Space: O(R)