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Arrange consonants and vowels nodes in a linked list

  • Difficulty Level : Medium
  • Last Updated : 12 Jan, 2022

Given a singly linked list, we need to arrange the consonants and vowel nodes of it in such a way that all the vowel nodes come before the consonants while maintaining the order of their arrival.
Examples: 
 

Input : a -> b -> c -> e -> d -> 
        o -> x -> i
Output : a -> e -> o -> i -> b -> 
         c -> d -> x

 

Solution : 
The idea is to keep a marker of the latest vowel we found while traversing the list. If we find another vowel, we take it out of the chain and put it after the existing latest vowel. Example: For linked list:
 

a -> b -> c -> e -> d -> o -> x -> i

say our latestVowel reference references the ‘a’ node, and that we currently reached the ‘e’ node. We do: 
 

a -> e -> b -> c -> d -> o -> x -> i

So what was after the ‘a’ node is now after the ‘e’ node after deleting it, and linking ‘a’ directly to ‘e’.
To properly remove and add links, it’s best to use the node before the one you are checking. So if you have a curr, you will check curr->next node to see if it’s a vowel or not. If it is, we need to add it after the latestVowel node, and then it’s easy to remove it from the chain by assigning its next to curr’s next. Also, if a list only contains consonants, we simply return head. 
 

C




/* C program to arrange consonants and
vowels nodes in a linked list */
#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
/* A linked list node */
struct Node
{
    char data;
    struct Node *next;
};
 
/* Function to add new node to the List */
struct Node *newNode(char key)
{
    struct Node *temp = (struct Node*)malloc(sizeof(struct Node));
    temp->data = key;
    temp->next = NULL;
    return temp;
}
 
// utility function to print linked list
void printlist(struct Node *head)
{
    if (! head)
    {
        printf("Empty list \n");
        return;
    }
    while (head != NULL)
    {
        printf("%c",head->data);
        if (head->next)
        printf("->");
        head = head->next;
    }
    printf("\n");
}
 
// utility function for checking vowel
bool isVowel(char x)
{
    return (x == 'a' || x == 'e' || x == 'i' ||
            x == 'o' || x == 'u');
}
 
/* function to arrange consonants and
vowels nodes */
struct Node *arrange(struct Node *head)
{
    struct Node *newHead = head;
 
    // for keep track of vowel
struct     Node *latestVowel;
 
    struct Node *curr = head;
 
    // list is empty
    if (head == NULL)
        return NULL;
 
    // We need to discover the first vowel
    // in the list. It is going to be the
    // returned head, and also the initial
    // latestVowel.
    if (isVowel(head->data))
 
        // first element is a vowel. It will
        // also be the new head and the initial
        // latestVowel;
        latestVowel = head;
 
    else
    {
 
        // First element is not a vowel. Iterate
        // through the list until we find a vowel.
        // Note that curr points to the element
        // *before* the element with the vowel.
        while (curr->next != NULL &&
            !isVowel(curr->next->data))
            curr = curr->next;
 
 
        // This is an edge case where there are
        // only consonants in the list.
        if (curr->next == NULL)
            return head;
 
        // Set the initial latestVowel and the
        // new head to the vowel item that we found.
        // Relink the chain of consonants after
        // that vowel item:
        // old_head_consonant->consonant1->consonant2->
        // vowel->rest_of_list becomes
        // vowel->old_head_consonant->consonant1->
        // consonant2->rest_of_list
        latestVowel = newHead = curr->next;
        curr->next = curr->next->next;
        latestVowel->next = head;
    }
 
    // Now traverse the list. Curr is always the item
    // *before* the one we are checking, so that we
    // can use it to re-link.
    while (curr != NULL && curr->next != NULL)
    {
        if (isVowel(curr->next->data))
        {
            // The next discovered item is a vowel
            if (curr == latestVowel)
            {
                // If it comes directly after the
                // previous vowel, we don't need to
                // move items around, just mark the
                // new latestVowel and advance curr.
                latestVowel = curr = curr->next;
            }
            else
            {
 
                // But if it comes after an intervening
                // chain of consonants, we need to chain
                // the newly discovered vowel right after
                // the old vowel. Curr is not changed as
                // after the re-linking it will have a
                // new next, that has not been checked yet,
                // and we always keep curr at one before
                // the next to check.
                struct Node *temp = latestVowel->next;
 
                // Chain in new vowel
                latestVowel->next = curr->next;
 
                // Advance latestVowel
                latestVowel = latestVowel->next;
 
                // Remove found vowel from previous place
                curr->next = curr->next->next;
 
                // Re-link chain of consonants after latestVowel
                latestVowel->next = temp;
            }
        }
        else
        {
 
            // No vowel in the next element, advance curr.
            curr = curr->next;
        }
    }
    return newHead;
}
 
// Driver code
int main()
{
    struct Node *head = newNode('a');
    head->next = newNode('b');
    head->next->next = newNode('c');
    head->next->next->next = newNode('e');
    head->next->next->next->next = newNode('d');
    head->next->next->next->next->next = newNode('o');
    head->next->next->next->next->next->next = newNode('x');
    head->next->next->next->next->next->next->next = newNode('i');
 
    printf("Linked list before :\n");
    printlist(head);
 
    head = arrange(head);
 
    printf("Linked list after :\n");
    printlist(head);
 
    return 0;
}

C++




/* C++ program to arrange consonants and
   vowels nodes in a linked list */
#include<bits/stdc++.h>
using namespace std;
 
/* A linked list node */
struct Node
{
    char data;
    struct Node *next;
};
 
/* Function to add new node to the List */
Node *newNode(char key)
{
    Node *temp = new Node;
    temp->data = key;
    temp->next = NULL;
    return temp;
}
 
// utility function to print linked list
void printlist(Node *head)
{
    if (! head)
    {
        cout << "Empty List\n";
        return;
    }
    while (head != NULL)
    {
        cout << head->data << " ";
        if (head->next)
           cout << "-> ";
        head = head->next;
    }
    cout << endl;
}
 
// utility function for checking vowel
bool isVowel(char x)
{
    return (x == 'a' || x == 'e' || x == 'i' ||
            x == 'o' || x == 'u');
}
 
/* function to arrange consonants and
   vowels nodes */
Node *arrange(Node *head)
{
    Node *newHead = head;
 
    // for keep track of vowel
    Node *latestVowel;
 
    Node *curr = head;
 
    // list is empty
    if (head == NULL)
        return NULL;
 
    // We need to discover the first vowel
    // in the list. It is going to be the
    // returned head, and also the initial
    // latestVowel.
    if (isVowel(head->data))
 
        // first element is a vowel. It will
        // also be the new head and the initial
        // latestVowel;
        latestVowel = head;
 
    else
    {
 
        // First element is not a vowel. Iterate
        // through the list until we find a vowel.
        // Note that curr points to the element
        // *before* the element with the vowel.
        while (curr->next != NULL &&
               !isVowel(curr->next->data))
            curr = curr->next;
 
 
        // This is an edge case where there are
        // only consonants in the list.
        if (curr->next == NULL)
            return head;
 
        // Set the initial latestVowel and the
        // new head to the vowel item that we found.
        // Relink the chain of consonants after
        // that vowel item:
        // old_head_consonant->consonant1->consonant2->
        // vowel->rest_of_list becomes
        // vowel->old_head_consonant->consonant1->
        // consonant2->rest_of_list
        latestVowel = newHead = curr->next;
        curr->next = curr->next->next;
        latestVowel->next = head;
    }
 
    // Now traverse the list. Curr is always the item
    // *before* the one we are checking, so that we
    // can use it to re-link.
    while (curr != NULL && curr->next != NULL)
    {
        if (isVowel(curr->next->data))
        {
            // The next discovered item is a vowel
            if (curr == latestVowel)
            {
                // If it comes directly after the
                // previous vowel, we don't need to
                // move items around, just mark the
                // new latestVowel and advance curr.
                latestVowel = curr = curr->next;
            }
            else
            {
 
                // But if it comes after an intervening
                // chain of consonants, we need to chain
                // the newly discovered vowel right after
                // the old vowel. Curr is not changed as
                // after the re-linking it will have a
                // new next, that has not been checked yet,
                // and we always keep curr at one before
                // the next to check.
                Node *temp = latestVowel->next;
 
                // Chain in new vowel
                latestVowel->next = curr->next;
 
                // Advance latestVowel
                latestVowel = latestVowel->next;
 
                // Remove found vowel from previous place
                curr->next = curr->next->next;
 
                // Re-link chain of consonants after latestVowel
                latestVowel->next = temp;
            }
        }
        else
        {
 
            // No vowel in the next element, advance curr.
            curr = curr->next;
        }
    }
    return newHead;
}
 
// Driver code
int main()
{
    Node *head = newNode('a');
    head->next = newNode('b');
    head->next->next = newNode('c');
    head->next->next->next = newNode('e');
    head->next->next->next->next = newNode('d');
    head->next->next->next->next->next = newNode('o');
    head->next->next->next->next->next->next = newNode('x');
    head->next->next->next->next->next->next->next = newNode('i');
 
    printf("Linked list before :\n");
    printlist(head);
 
    head = arrange(head);
 
    printf("Linked list after :\n");
    printlist(head);
 
    return 0;
}

Java




/* Java program to arrange consonants and
vowels nodes in a linked list */
class GfG
{
 
/* A linked list node */
static class Node
{
    char data;
    Node next;
}
 
/* Function to add new node to the List */
static Node newNode(char key)
{
    Node temp = new Node();
    temp.data = key;
    temp.next = null;
    return temp;
}
 
// utility function to print linked list
static void printlist(Node head)
{
    if (head == null)
    {
        System.out.println("Empty List");
        return;
    }
    while (head != null)
    {
        System.out.print(head.data +" ");
        if (head.next != null)
        System.out.print("-> ");
        head = head.next;
    }
    System.out.println();
}
 
// utility function for checking vowel
static boolean isVowel(char x)
{
    return (x == 'a' || x == 'e' || x == 'i' ||
            x == 'o' || x == 'u');
}
 
/* function to arrange consonants and
vowels nodes */
static Node arrange(Node head)
{
    Node newHead = head;
 
    // for keep track of vowel
    Node latestVowel;
 
    Node curr = head;
 
    // list is empty
    if (head == null)
        return null;
 
    // We need to discover the first vowel
    // in the list. It is going to be the
    // returned head, and also the initial
    // latestVowel.
    if (isVowel(head.data) == true)
 
        // first element is a vowel. It will
        // also be the new head and the initial
        // latestVowel;
        latestVowel = head;
 
    else
    {
 
        // First element is not a vowel. Iterate
        // through the list until we find a vowel.
        // Note that curr points to the element
        // *before* the element with the vowel.
        while (curr.next != null &&
            !isVowel(curr.next.data))
            curr = curr.next;
 
 
        // This is an edge case where there are
        // only consonants in the list.
        if (curr.next == null)
            return head;
 
        // Set the initial latestVowel and the
        // new head to the vowel item that we found.
        // Relink the chain of consonants after
        // that vowel item:
        // old_head_consonant->consonant1->consonant2->
        // vowel->rest_of_list becomes
        // vowel->old_head_consonant->consonant1->
        // consonant2->rest_of_list
        latestVowel = newHead = curr.next;
        curr.next = curr.next.next;
        latestVowel.next = head;
    }
 
    // Now traverse the list. Curr is always the item
    // *before* the one we are checking, so that we
    // can use it to re-link.
    while (curr != null && curr.next != null)
    {
        if (isVowel(curr.next.data) == true)
        {
            // The next discovered item is a vowel
            if (curr == latestVowel)
            {
                // If it comes directly after the
                // previous vowel, we don't need to
                // move items around, just mark the
                // new latestVowel and advance curr.
                latestVowel = curr = curr.next;
            }
            else
            {
 
                // But if it comes after an intervening
                // chain of consonants, we need to chain
                // the newly discovered vowel right after
                // the old vowel. Curr is not changed as
                // after the re-linking it will have a
                // new next, that has not been checked yet,
                // and we always keep curr at one before
                // the next to check.
                Node temp = latestVowel.next;
 
                // Chain in new vowel
                latestVowel.next = curr.next;
 
                // Advance latestVowel
                latestVowel = latestVowel.next;
 
                // Remove found vowel from previous place
                curr.next = curr.next.next;
 
                // Re-link chain of consonants after latestVowel
                latestVowel.next = temp;
            }
        }
        else
        {
 
            // No vowel in the next element, advance curr.
            curr = curr.next;
        }
    }
    return newHead;
}
 
// Driver code
public static void main(String[] args)
{
    Node head = newNode('a');
    head.next = newNode('b');
    head.next.next = newNode('c');
    head.next.next.next = newNode('e');
    head.next.next.next.next = newNode('d');
    head.next.next.next.next.next = newNode('o');
    head.next.next.next.next.next.next = newNode('x');
    head.next.next.next.next.next.next.next = newNode('i');
 
    System.out.println("Linked list before : ");
    printlist(head);
 
    head = arrange(head);
 
    System.out.println("Linked list after :");
    printlist(head);
}
}
 
// This code is contributed by Prerna Saini.

C#




/* C# program to arrange consonants and
vowels nodes in a linked list */
using System;
 
class GfG
{
 
    /* A linked list node */
    public class Node
    {
        public char data;
        public Node next;
    }
 
    /* Function to add new node to the List */
    static Node newNode(char key)
    {
        Node temp = new Node();
        temp.data = key;
        temp.next = null;
        return temp;
    }
 
    // utility function to print linked list
    static void printlist(Node head)
    {
        if (head == null)
        {
            Console.WriteLine("Empty List");
            return;
        }
        while (head != null)
        {
            Console.Write(head.data +" ");
            if (head.next != null)
                Console.Write("-> ");
            head = head.next;
        }
        Console.WriteLine();
    }
 
    // utility function for checking vowel
    static bool isVowel(char x)
    {
        return (x == 'a' || x == 'e' || x == 'i' ||
                x == 'o' || x == 'u');
    }
 
    /* function to arrange consonants and
    vowels nodes */
    static Node arrange(Node head)
    {
        Node newHead = head;
 
        // for keep track of vowel
        Node latestVowel;
 
        Node curr = head;
 
        // list is empty
        if (head == null)
            return null;
 
        // We need to discover the first vowel
        // in the list. It is going to be the
        // returned head, and also the initial
        // latestVowel.
        if (isVowel(head.data) == true)
 
            // first element is a vowel. It will
            // also be the new head and the initial
            // latestVowel;
            latestVowel = head;
 
        else
        {
 
            // First element is not a vowel. Iterate
            // through the list until we find a vowel.
            // Note that curr points to the element
            // *before* the element with the vowel.
            while (curr.next != null &&
                !isVowel(curr.next.data))
                curr = curr.next;
 
 
            // This is an edge case where there are
            // only consonants in the list.
            if (curr.next == null)
                return head;
 
            // Set the initial latestVowel and the
            // new head to the vowel item that we found.
            // Relink the chain of consonants after
            // that vowel item:
            // old_head_consonant->consonant1->consonant2->
            // vowel->rest_of_list becomes
            // vowel->old_head_consonant->consonant1->
            // consonant2->rest_of_list
            latestVowel = newHead = curr.next;
            curr.next = curr.next.next;
            latestVowel.next = head;
        }
 
        // Now traverse the list. Curr is always the item
        // *before* the one we are checking, so that we
        // can use it to re-link.
        while (curr != null && curr.next != null)
        {
            if (isVowel(curr.next.data) == true)
            {
                // The next discovered item is a vowel
                if (curr == latestVowel)
                {
                    // If it comes directly after the
                    // previous vowel, we don't need to
                    // move items around, just mark the
                    // new latestVowel and advance curr.
                    latestVowel = curr = curr.next;
                }
                else
                {
 
                    // But if it comes after an intervening
                    // chain of consonants, we need to chain
                    // the newly discovered vowel right after
                    // the old vowel. Curr is not changed as
                    // after the re-linking it will have a
                    // new next, that has not been checked yet,
                    // and we always keep curr at one before
                    // the next to check.
                    Node temp = latestVowel.next;
 
                    // Chain in new vowel
                    latestVowel.next = curr.next;
 
                    // Advance latestVowel
                    latestVowel = latestVowel.next;
 
                    // Remove found vowel from previous place
                    curr.next = curr.next.next;
 
                    // Re-link chain of consonants after latestVowel
                    latestVowel.next = temp;
                }
            }
            else
            {
 
                // No vowel in the next element, advance curr.
                curr = curr.next;
            }
        }
        return newHead;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        Node head = newNode('a');
        head.next = newNode('b');
        head.next.next = newNode('c');
        head.next.next.next = newNode('e');
        head.next.next.next.next = newNode('d');
        head.next.next.next.next.next = newNode('o');
        head.next.next.next.next.next.next = newNode('x');
        head.next.next.next.next.next.next.next = newNode('i');
 
        Console.WriteLine("Linked list before : ");
        printlist(head);
 
        head = arrange(head);
 
        Console.WriteLine("Linked list after :");
        printlist(head);
    }
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 program to remove vowels
# Nodes in a linked list
 
# A linked list node
class Node:
    def __init__(self, x):
        self.data = x
        self.next = None
 
# Utility function to print the
# linked list
def printlist(head):
    if (not head):
        print("Empty List")
        return
 
    while (head != None):
        print(head.data, end = " ")
        if (head.next):
            print(end = "-> ")
        head = head.next
    print()
 
# Utility function for checking vowel
def isVowel(x):
    return (x == 'a' or x == 'e' or x == 'i'
            or x == 'o' or x == 'u' or x == 'A'
            or x == 'E' or x == 'I' or x == 'O'
            or x == 'U')
 
#/* function to arrange consonants and
#   vowels nodes */
def arrange(head):
    newHead = head
 
    # for keep track of vowel
    latestVowel = None
 
    curr = head
 
    # list is empty
    if (head == None):
        return None
 
    # We need to discover the first vowel
    # in the list. It is going to be the
    # returned head, and also the initial
    # latestVowel.
    if (isVowel(head.data)):
 
        # first element is a vowel. It will
        # also be the new head and the initial
        # latestVowel
        latestVowel = head
 
    else:
 
        # First element is not a vowel. Iterate
        # through the list until we find a vowel.
        # Note that curr points to the element
        # *before* the element with the vowel.
        while (curr.next != None and
               not isVowel(curr.next.data)):
            curr = curr.next
 
 
        # This is an edge case where there are
        # only consonants in the list.
        if (curr.next == None):
            return head
 
        # Set the initial latestVowel and the
        # new head to the vowel item that we found.
        # Relink the chain of consonants after
        # that vowel item:
        # old_head_consonant.consonant1.consonant2.
        # vowel.rest_of_list becomes
        # vowel.old_head_consonant.consonant1.
        # consonant2.rest_of_list
        latestVowel = newHead = curr.next
        curr.next = curr.next.next
        latestVowel.next = head
 
    # Now traverse the list. Curr is always the item
    # *before* the one we are checking, so that we
    # can use it to re-link.
    while (curr != None and curr.next != None):
        if (isVowel(curr.next.data)):
             
            # The next discovered item is a vowel
            if (curr == latestVowel):
                # If it comes directly after the
                # previous vowel, we don't need to
                # move items around, just mark the
                # new latestVowel and advance curr.
                latestVowel = curr = curr.next
            else:
 
                # But if it comes after an intervening
                # chain of consonants, we need to chain
                # the newly discovered vowel right after
                # the old vowel. Curr is not changed as
                # after the re-linking it will have a
                # new next, that has not been checked yet,
                # and we always keep curr at one before
                # the next to check.
                temp = latestVowel.next
 
                # Chain in new vowel
                latestVowel.next = curr.next
 
                # Advance latestVowel
                latestVowel = latestVowel.next
 
                # Remove found vowel from previous place
                curr.next = curr.next.next
 
                # Re-link chain of consonants after latestVowel
                latestVowel.next = temp
 
        else:
 
            # No vowel in the next element, advance curr.
            curr = curr.next
 
    return newHead
 
# Driver code
if __name__ == '__main__':
     
    # Initialise the Linked List
    head = Node('a')
    head.next = Node('b')
    head.next.next = Node('c')
    head.next.next.next = Node('e')
    head.next.next.next.next = Node('d')
    head.next.next.next.next.next = Node('o')
    head.next.next.next.next.next.next = Node('x')
    head.next.next.next.next.next.next.next = Node('i')
 
    # Print the given Linked List
    print("Linked list before :")
    printlist(head)
 
    head = arrange(head)
 
    # Print the Linked List after
    # removing vowels
    print("Linked list after :")
    printlist(head)
 
# This code is contributed by mohit kumar 29

Javascript




<script>
 
/* JavaScript program to arrange consonants and
vowels nodes in a linked list */
     
     
 /* A linked list node */  
class Node
{
    /* Function to add new node to the List */
    constructor(key)
    {
        this.data=key;
        this.next = null;
    }
}
 
// utility function to print linked list
function printlist(head)
{
    if (head == null)
    {
        document.write("Empty List<br>");
        return;
    }
    while (head != null)
    {
        document.write(head.data +" ");
        if (head.next != null)
            document.write("-> ");
        head = head.next;
    }
    document.write("<br>");
}
 
// utility function for checking vowel
function isVowel(x)
{
    return (x == 'a' || x == 'e' || x == 'i' ||
            x == 'o' || x == 'u');
             
}
 
 
/* function to arrange consonants and
vowels nodes */
function arrange(head)
{
    let newHead = head;
  
    // for keep track of vowel
    let latestVowel;
  
    let curr = head;
  
    // list is empty
    if (head == null)
        return null;
  
    // We need to discover the first vowel
    // in the list. It is going to be the
    // returned head, and also the initial
    // latestVowel.
    if (isVowel(head.data) == true)
  
        // first element is a vowel. It will
        // also be the new head and the initial
        // latestVowel;
        latestVowel = head;
  
    else
    {
  
        // First element is not a vowel. Iterate
        // through the list until we find a vowel.
        // Note that curr points to the element
        // *before* the element with the vowel.
        while (curr.next != null &&
            !isVowel(curr.next.data))
            curr = curr.next;
  
  
        // This is an edge case where there are
        // only consonants in the list.
        if (curr.next == null)
            return head;
  
        // Set the initial latestVowel and the
        // new head to the vowel item that we found.
        // Relink the chain of consonants after
        // that vowel item:
        // old_head_consonant->consonant1->consonant2->
        // vowel->rest_of_list becomes
        // vowel->old_head_consonant->consonant1->
        // consonant2->rest_of_list
        latestVowel = newHead = curr.next;
        curr.next = curr.next.next;
        latestVowel.next = head;
    }
  
    // Now traverse the list. Curr is always the item
    // *before* the one we are checking, so that we
    // can use it to re-link.
    while (curr != null && curr.next != null)
    {
        if (isVowel(curr.next.data) == true)
        {
            // The next discovered item is a vowel
            if (curr == latestVowel)
            {
                // If it comes directly after the
                // previous vowel, we don't need to
                // move items around, just mark the
                // new latestVowel and advance curr.
                latestVowel = curr = curr.next;
            }
            else
            {
  
                // But if it comes after an intervening
                // chain of consonants, we need to chain
                // the newly discovered vowel right after
                // the old vowel. Curr is not changed as
                // after the re-linking it will have a
                // new next, that has not been checked yet,
                // and we always keep curr at one before
                // the next to check.
                let temp = latestVowel.next;
  
                // Chain in new vowel
                latestVowel.next = curr.next;
  
                // Advance latestVowel
                latestVowel = latestVowel.next;
  
                // Remove found vowel from previous place
                curr.next = curr.next.next;
  
                // Re-link chain of consonants
                // after latestVowel
                latestVowel.next = temp;
            }
        }
        else
        {
  
            // No vowel in the next element, advance curr.
            curr = curr.next;
        }
    }
    return newHead;
}
 
// Driver code
let head = new Node('a');
head.next = new Node('b');
head.next.next = new Node('c');
head.next.next.next = new Node('e');
head.next.next.next.next = new Node('d');
head.next.next.next.next.next = new Node('o');
head.next.next.next.next.next.next = new Node('x');
head.next.next.next.next.next.next.next = new Node('i');
 
document.write("Linked list before : <br>");
printlist(head);
 
head = arrange(head);
 
document.write("Linked list after :<br>");
printlist(head);
     
 
// This code is contributed by unknown2108
 
</script>

Output:  

Linked list before :
a -> b -> c -> e -> d -> o -> x -> i 
Linked list after :
a -> e -> o -> i -> b -> c -> d -> x

TIME COMPLEXITY:- O(n)

SPACE COMPLEXITY:- O(1)

References: 
Stackoverflow
This article is contributed by Gaurav Miglani. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Another approach:-  

Another approach to solve the above problem is to create two separate Linked List one containing only vowels and the other one containing only consonant.

Remember the below approach is simple to understand BUT IT REQUIRES A SPACE COMPLEXITY OF O(N) 

While traversing the given input Linked List if the node data is vowel then add it to the vowel Linked List and if the node data is consonant then add it to the consonant Linked List.

After traversal you have to just link the last vowel node to the first consonant node of the respective Liked List. 

C++




/* C++ program to arrange consonants and
vowels nodes in a linked list */
#include <bits/stdc++.h>
using namespace std;
 
/* A linked list node */
struct Node {
    char data;
    struct Node* next;
    Node(int x)
    {
        data = x;
        next = NULL;
    }
};
 
/* Function to add new node to the List */
void append(struct Node** headRef, char data)
{
    struct Node* new_node = new Node(data);
    struct Node* last = *headRef;
    if (*headRef == NULL) {
        *headRef = new_node;
        return;
    }
    while (last->next != NULL)
        last = last->next;
    last->next = new_node;
    return;
}
 
// utility function to print linked list
void printlist(Node* head)
{
    if (!head) {
        cout << "Empty List\n";
        return;
    }
    while (head != NULL) {
        cout << head->data << " ";
        if (head->next)
            cout << "-> ";
        head = head->next;
    }
    cout << endl;
}
 
/* function to arrange consonants and
vowels nodes */
 
struct Node* arrange(Node* head)
{
    Node *vowel = NULL, *consonant = NULL, *start = NULL,
         *end = NULL;
    while (head != NULL) {
        char x = head->data;
        // Checking the the current node data is vowel or
        // not
        if (x == 'a' || x == 'e' || x == 'i' || x == 'o'
            || x == 'u') {
            if (!vowel) {
                vowel = new Node(x);
                start = vowel;
            }
            else {
                vowel->next = new Node(x);
                vowel = vowel->next;
            }
        }
        else {
            if (!consonant) {
                consonant = new Node(x);
                end = consonant;
            }
            else {
                consonant->next = new Node(x);
                consonant = consonant->next;
            }
        }
        head = head->next;
    }
    // In case when there is no vowel in the incoming LL
    // then we have to return the head of the consonant LL
    if (start == NULL)
        return end;
    // Connecting the vowel and consonant LL
    vowel->next = end;
    return start;
}
 
// Driver code
int main()
{
    struct Node* head = NULL;
    append(&head, 'a');
    append(&head, 'b');
    append(&head, 'c');
    append(&head, 'e');
    append(&head, 'd');
    append(&head, 'o');
    append(&head, 'x');
    append(&head, 'i');
 
    printf("Linked list before :\n");
    printlist(head);
 
    head = arrange(head);
 
    printf("Linked list after :\n");
    printlist(head);
 
    return 0;
}
 
// This code is contributed by Aditya Kumar

Java




/* Java program to arrange consonants and
vowels nodes in a linked list */
import java.util.*;
 
class GFG{
 
/* A linked list node */
static class Node {
    char data;
    Node next;
    Node(char x)
    {
        this.data = x;
        this.next = null;
    }
};
 
/* Function to add new node to the List */
static Node append(Node headRef, char data)
{
    Node new_node = new Node(data);
    Node last = headRef;
    if (headRef == null) {
        headRef = new_node;
        return headRef;
    }
    while (last.next != null)
        last = last.next;
    last.next = new_node;
    return headRef;
}
 
// utility function to print linked list
static void printlist(Node head)
{
    if (head == null) {
        System.out.print("Empty List\n");
        return;
    }
    while (head != null) {
        System.out.print(head.data+ " ");
        if (head.next!=null)
            System.out.print("-> ");
        head = head.next;
    }
    System.out.println();
}
 
/* function to arrange consonants and
vowels nodes */
 
@SuppressWarnings("null")
static Node arrange(Node head)
{
    Node vowel = null, consonant = null, start = null,
         end = null;
    while (head != null) {
        char x = head.data;
       
        // Checking the the current node data is vowel or
        // not
        if (x == 'a' || x == 'e' || x == 'i' || x == 'o'
            || x == 'u') {
            if (vowel==null) {
                vowel = new Node(x);
                start = vowel;
            }
            else {
                vowel.next = new Node(x);
                vowel = vowel.next;
            }
        }
        else {
            if (consonant == null) {
                consonant = new Node(x);
                end = consonant;
            }
            else {
                consonant.next = new Node(x);
                consonant = consonant.next;
            }
        }
        head = head.next;
    }
   
    // In case when there is no vowel in the incoming LL
    // then we have to return the head of the consonant LL
    if (start == null)
        return end;
   
    // Connecting the vowel and consonant LL
    vowel.next = end;
    return start;
}
 
// Driver code
public static void main(String[] args)
{
    Node head = null;
    head = append(head, 'a');
    head = append(head, 'b');
    head = append(head, 'c');
    head = append(head, 'e');
    head = append(head, 'd');
    head = append(head, 'o');
    head = append(head, 'x');
    head = append(head, 'i');
 
    System.out.printf("Linked list before :\n");
    printlist(head);
 
    head = arrange(head);
 
    System.out.printf("Linked list after :\n");
    printlist(head);
 
}
}
 
// This code is contributed by Rajput-Ji

C#




/* C# program to arrange consonants and
vowels nodes in a linked list */
using System;
 
public class GFG {
 
  /* A linked list node */
  public    class Node {
    public    char data;
    public    Node next;
 
    public    Node(char x) {
      this.data = x;
      this.next = null;
    }
  };
 
  /* Function to add new node to the List */
  static Node append(Node headRef, char data) {
    Node new_node = new Node(data);
    Node last = headRef;
    if (headRef == null) {
      headRef = new_node;
      return headRef;
    }
    while (last.next != null)
      last = last.next;
    last.next = new_node;
    return headRef;
  }
 
  // utility function to print linked list
  static void printlist(Node head) {
    if (head == null) {
      Console.Write("Empty List\n");
      return;
    }
    while (head != null) {
      Console.Write(head.data + " ");
      if (head.next != null)
        Console.Write("-> ");
      head = head.next;
    }
    Console.WriteLine();
  }
 
  /*
     * function to arrange consonants and vowels nodes
     */
 
  static Node arrange(Node head) {
    Node vowel = null, consonant = null, start = null, end = null;
    while (head != null) {
      char x = head.data;
 
      // Checking the the current node data is vowel or
      // not
      if (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u') {
        if (vowel == null) {
          vowel = new Node(x);
          start = vowel;
        } else {
          vowel.next = new Node(x);
          vowel = vowel.next;
        }
      } else {
        if (consonant == null) {
          consonant = new Node(x);
          end = consonant;
        } else {
          consonant.next = new Node(x);
          consonant = consonant.next;
        }
      }
      head = head.next;
    }
 
    // In case when there is no vowel in the incoming LL
    // then we have to return the head of the consonant LL
    if (start == null)
      return end;
 
    // Connecting the vowel and consonant LL
    vowel.next = end;
    return start;
  }
 
  // Driver code
  public static void Main(String[] args) {
    Node head = null;
    head = append(head, 'a');
    head = append(head, 'b');
    head = append(head, 'c');
    head = append(head, 'e');
    head = append(head, 'd');
    head = append(head, 'o');
    head = append(head, 'x');
    head = append(head, 'i');
 
    Console.Write("Linked list before :\n");
    printlist(head);
 
    head = arrange(head);
 
    Console.Write("Linked list after :\n");
    printlist(head);
 
  }
}
 
// This code is contributed by Rajput-Ji

TIME COMPLEXITY:- O(n)

SPACE COMPLEXITY:- O(n)
 


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