Given a Prufer sequence of a Tree, the task is to print the nodes with prime-degree in this tree.
Examples:
Input: arr[] = {4, 1, 3, 4} Output: 1 3 4 Explanation: The tree is: 2----4----3----1----5 | 6 Hence, the degree of 1, 3 and 4 are 2, 2 and 3 respectively which are prime. Input: a[] = {1, 2, 2} Output: 1 2
Approach:
- Since the length of prufer sequence is N – 2 if N is the number of nodes. Therefore, create an array degree[] of size 2 more than the length of the Prufer sequence.
- Initially, fill the degree array with 1.
- Iterate in the Prufer sequence and increase the frequency in the degree table for every element. This method works because the frequency of a node in the Prufer sequence is one less than the degree in the tree.
- Further, to check if the node degree is prime or not, we will use Sieve Of eratosthenes. Create a sieve which will help us to identify if the degree is prime or not in O(1) time.
- If a node has a prime degree, then print the node number.
Below is the implementation of the above approach:
C++
// C++ implementation to print the // nodes with prime degree from the // given prufer sequence #include <bits/stdc++.h> using namespace std;
// Function to create Sieve // to check primes void SieveOfEratosthenes(
bool prime[], int p_size)
{ // False here indicates
// that it is not prime
prime[0] = false ;
prime[1] = false ;
for ( int p = 2; p * p <= p_size; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p]) {
// Update all multiples of p,
// set them to non-prime
for ( int i = p * 2; i <= p_size;
i += p)
prime[i] = false ;
}
}
} // Function to print the nodes with // prime degree in the tree // whose Prufer sequence is given void PrimeDegreeNodes( int prufer[], int n)
{ int nodes = n + 2;
bool prime[nodes + 1];
memset (prime, true , sizeof (prime));
SieveOfEratosthenes(prime, nodes + 1);
// Hash-table to mark the
// degree of every node
int degree[n + 2 + 1];
// Initially let all the degrees be 1
for ( int i = 1; i <= nodes; i++)
degree[i] = 1;
// Increase the count of the degree
for ( int i = 0; i < n; i++)
degree[prufer[i]]++;
// Print the nodes with prime degree
for ( int i = 1; i <= nodes; i++) {
if (prime[degree[i]]) {
cout << i << " " ;
}
}
} // Driver Code int main()
{ int a[] = { 4, 1, 3, 4 };
int n = sizeof (a) / sizeof (a[0]);
PrimeDegreeNodes(a, n);
return 0;
} |
Java
// Java implementation to print the // nodes with prime degree from the // given prufer sequence import java.util.*;
class GFG{
// Function to create Sieve // to check primes static void SieveOfEratosthenes(
boolean prime[], int p_size)
{ // False here indicates
// that it is not prime
prime[ 0 ] = false ;
prime[ 1 ] = false ;
for ( int p = 2 ; p * p <= p_size; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p]) {
// Update all multiples of p,
// set them to non-prime
for ( int i = p * 2 ; i <= p_size;
i += p)
prime[i] = false ;
}
}
} // Function to print the nodes with // prime degree in the tree // whose Prufer sequence is given static void PrimeDegreeNodes( int prufer[], int n)
{ int nodes = n + 2 ;
boolean []prime = new boolean [nodes + 1 ];
Arrays.fill(prime, true );
SieveOfEratosthenes(prime, nodes + 1 );
// Hash-table to mark the
// degree of every node
int []degree = new int [n + 2 + 1 ];
// Initially let all the degrees be 1
for ( int i = 1 ; i <= nodes; i++)
degree[i] = 1 ;
// Increase the count of the degree
for ( int i = 0 ; i < n; i++)
degree[prufer[i]]++;
// Print the nodes with prime degree
for ( int i = 1 ; i <= nodes; i++) {
if (prime[degree[i]]) {
System.out.print(i+ " " );
}
}
} // Driver Code public static void main(String[] args)
{ int a[] = { 4 , 1 , 3 , 4 };
int n = a.length;
PrimeDegreeNodes(a, n);
} } // This code contributed by Princi Singh |
Python3
# Python3 implementation to print the # nodes with prime degree from the # given prufer sequence # Function to create Sieve # to check primes def SieveOfEratosthenes(prime, p_size):
# False here indicates
# that it is not prime
prime[ 0 ] = False
prime[ 1 ] = False
p = 2
while (p * p < = p_size):
# If prime[p] is not changed,
# then it is a prime
if (prime[p]):
# Update all multiples of p,
# set them to non-prime
for i in range (p * 2 , p_size + 1 , p):
prime[i] = False
p + = 1
# Function to print the nodes with # prime degree in the tree # whose Prufer sequence is given def PrimeDegreeNodes(prufer, n):
nodes = n + 2
prime = [ True ] * (nodes + 1 )
SieveOfEratosthenes(prime, nodes + 1 )
# Hash-table to mark the
# degree of every node
degree = [ 0 ] * (n + 2 + 1 );
# Initially let all the degrees be 1
for i in range ( 1 , nodes + 1 ):
degree[i] = 1 ;
# Increase the count of the degree
for i in range ( 0 , n):
degree[prufer[i]] + = 1
# Print the nodes with prime degree
for i in range ( 1 , nodes + 1 ):
if prime[degree[i]]:
print (i, end = ' ' )
# Driver Code if __name__ = = '__main__' :
a = [ 4 , 1 , 3 , 4 ]
n = len (a)
PrimeDegreeNodes(a, n)
# This code is contributed by rutvik_56 |
C#
// C# implementation to print the // nodes with prime degree from the // given prufer sequence using System;
class GFG{
// Function to create Sieve // to check primes static void SieveOfEratosthenes( bool []prime,
int p_size)
{ // False here indicates
// that it is not prime
prime[0] = false ;
prime[1] = false ;
for ( int p = 2; p * p <= p_size; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p])
{
// Update all multiples of p,
// set them to non-prime
for ( int i = p * 2; i <= p_size;
i += p)
prime[i] = false ;
}
}
} // Function to print the nodes with // prime degree in the tree // whose Prufer sequence is given static void PrimeDegreeNodes( int []prufer, int n)
{ int nodes = n + 2;
bool []prime = new bool [nodes + 1];
for ( int i = 0; i < prime.Length; i++)
prime[i] = true ;
SieveOfEratosthenes(prime, nodes + 1);
// Hash-table to mark the
// degree of every node
int []degree = new int [n + 2 + 1];
// Initially let all the degrees be 1
for ( int i = 1; i <= nodes; i++)
degree[i] = 1;
// Increase the count of the degree
for ( int i = 0; i < n; i++)
degree[prufer[i]]++;
// Print the nodes with prime degree
for ( int i = 1; i <= nodes; i++)
{
if (prime[degree[i]])
{
Console.Write(i + " " );
}
}
} // Driver Code public static void Main(String[] args)
{ int []a = { 4, 1, 3, 4 };
int n = a.Length;
PrimeDegreeNodes(a, n);
} } // This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript implementation to print the // nodes with prime degree from the // given prufer sequence // Function to create Sieve // to check primes function SieveOfEratosthenes(prime, p_size)
{ // False here indicates
// that it is not prime
prime[0] = false ;
prime[1] = false ;
for (let p = 2; p * p <= p_size; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p]) {
// Update all multiples of p,
// set them to non-prime
for (let i = p * 2; i <= p_size;
i += p)
prime[i] = false ;
}
}
} // Function to print the nodes with // prime degree in the tree // whose Prufer sequence is given function PrimeDegreeNodes(prufer, n)
{ let nodes = n + 2;
let prime = new Array(nodes + 1);
prime.fill( true );
SieveOfEratosthenes(prime, nodes + 1);
// Hash-table to mark the
// degree of every node
let degree = new Array(n + 2 + 1);
// Initially let all the degrees be 1
for (let i = 1; i <= nodes; i++)
degree[i] = 1;
// Increase the count of the degree
for (let i = 0; i < n; i++)
degree[prufer[i]]++;
// Print the nodes with prime degree
for (let i = 1; i <= nodes; i++) {
if (prime[degree[i]]) {
document.write(i + " " );
}
}
} // Driver Code let a = [ 4, 1, 3, 4 ]; let n = a.length PrimeDegreeNodes(a, n); // This code is contributed by gfgking </script> |
Output:
1 3 4
Time Complexity: O(n*log(log(n))) + O(n) which is asymptotically equal to O(n*log(log(n))).
Space Complexity: O(n) as arrays has been created to store values.