Given a Prufer sequence of a Tree, the task is to print the node with the maximum degree in the tree whose Prufer sequence is given. In case there are many nodes with maximum degree, print the node with the smallest number.
Examples:
Input: a[] = {4, 1, 3, 4} Output: 4 The tree is: 2----4----3----1----5 | 6 Input: a[] = {1, 2, 2} Output: 2
A simple approach is to create the tree using the Prufer sequence and then find the degree of all the nodes and then find the maximum among them.
Efficient approach: Create a degree[] array of size 2 more than the length of the Prufer sequence, since the length of prufer sequence is N – 2 if N is the number of nodes. 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. Now iterate in the degree array and find the node with the maximum frequency which will be our answer.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the node with // the maximum degree in the tree // whose Prufer sequence is given int findMaxDegreeNode( int prufer[], int n)
{ int nodes = n + 2;
// 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]]++;
int maxDegree = 0;
int node = 0;
// Find the node with maximum degree
for ( int i = 1; i <= nodes; i++) {
if (degree[i] > maxDegree) {
maxDegree = degree[i];
node = i;
}
}
return node;
} // Driver code int main()
{ int a[] = { 1, 2, 2 };
int n = sizeof (a) / sizeof (a[0]);
cout << findMaxDegreeNode(a, n);
return 0;
} |
// Java implementation of the approach import java.io.*;
class GFG
{ // Function to return the node with
// the maximum degree in the tree
// whose Prufer sequence is given
static int findMaxDegreeNode( int prufer[], int n)
{
int nodes = n + 2 ;
// 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]]++;
int maxDegree = 0 ;
int node = 0 ;
// Find the node with maximum degree
for ( int i = 1 ; i <= nodes; i++)
{
if (degree[i] > maxDegree)
{
maxDegree = degree[i];
node = i;
}
}
return node;
}
// Driver code
public static void main (String[] args)
{
int []a = { 1 , 2 , 2 };
int n = a.length;
System.out.println(findMaxDegreeNode(a, n));
}
} // This code is contributed by ajit_00023 |
# Python implementation of the approach # Function to return the node with # the maximum degree in the tree # whose Prufer sequence is given def findMaxDegreeNode(prufer, n):
nodes = n + 2 ;
# 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 (n):
degree[prufer[i]] + = 1 ;
maxDegree = 0 ;
node = 0 ;
# Find the node with maximum degree
for i in range ( 1 ,nodes + 1 ):
if (degree[i] > maxDegree):
maxDegree = degree[i];
node = i;
return node;
# Driver code a = [ 1 , 2 , 2 ];
n = len (a);
print (findMaxDegreeNode(a, n));
# This code has been contributed by 29AjayKumar |
// C# implementation of the approach using System;
class GFG
{ // Function to return the node with
// the maximum degree in the tree
// whose Prufer sequence is given
static int findMaxDegreeNode( int []prufer, int n)
{
int nodes = n + 2;
// 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]]++;
int maxDegree = 0;
int node = 0;
// Find the node with maximum degree
for ( int i = 1; i <= nodes; i++)
{
if (degree[i] > maxDegree)
{
maxDegree = degree[i];
node = i;
}
}
return node;
}
// Driver code
static public void Main ()
{
int []a = { 1, 2, 2 };
int n = a.Length;
Console.WriteLine(findMaxDegreeNode(a, n));
}
} // This code is contributed by AnkitRai01 |
<script> // Javascript implementation of the approach
// Function to return the node with
// the maximum degree in the tree
// whose Prufer sequence is given
function findMaxDegreeNode(prufer, n)
{
let nodes = n + 2;
// Hash-table to mark the
// degree of every node
let degree = new Array(n + 2 + 1);
degree.fill(0);
// 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]]++;
let maxDegree = 0;
let node = 0;
// Find the node with maximum degree
for (let i = 1; i <= nodes; i++)
{
if (degree[i] > maxDegree)
{
maxDegree = degree[i];
node = i;
}
}
return node;
}
let a = [ 1, 2, 2 ];
let n = a.length;
document.write(findMaxDegreeNode(a, n));
</script> |
2
Time Complexity: O(N), as we are using a loop to traverse N times. Where N is the number of elements in the array.
Auxiliary Space: O(N), as we are using extra space for the degree array. Where N is the number of elements.