Print the node with the maximum degree in the prufer sequence

• Last Updated : 22 Jun, 2022

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++

 `// C++ implementation of the approach``#include ``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

 `// 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`

Python3

 `    ` `# 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#

 `// 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`

Javascript

 ``

Output:

`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.

My Personal Notes arrow_drop_up