Given a linked list and a key in it, the task is to move all occurrences of the given key to the end of the linked list, keeping the order of all other elements the same.
Examples:
Input : 1 -> 2 -> 2 -> 4 -> 3 key = 2 Output : 1 -> 4 -> 3 -> 2 -> 2 Input : 6 -> 6 -> 7 -> 6 -> 3 -> 10 key = 6 Output : 7 -> 3 -> 10 -> 6 -> 6 -> 6
A simple solution is to one by one find all occurrences of a given key in the linked list. For every found occurrence, insert it at the end. We do it till all occurrences of the given key are moved to the end.
Time Complexity: O(n2), as we will be using nested loops to find the node with the key and insert it at the end of the linked list.
Auxiliary Space: O(1), as we are not using any extra space.
Efficient Solution 1: is to keep two pointers:
pCrawl => Pointer to traverse the whole list one by one.
pKey => Pointer to an occurrence of the key if a key is found. Else the same as pCrawl.
We start both of the above pointers from the head of the linked list. We move pKey only when pKey is not pointing to a key. We always move pCrawl. So, when pCrawl and pKey are not the same, we must have found a key that lies before pCrawl, so we swap between pCrawl and pKey, and move pKey to the next location. The loop invariant is, after swapping of data, all elements from pKey to pCrawl are keys.
Below is the implementation of this approach.
// C++ program to move all occurrences of a // given key to end. #include <bits/stdc++.h> // A Linked list Node struct Node {
int data;
struct Node* next;
}; // A utility function to create a new node. struct Node* newNode( int x)
{ Node* temp = new Node;
temp->data = x;
temp->next = NULL;
} // Utility function to print the elements // in Linked list void printList(Node* head)
{ struct Node* temp = head;
while (temp != NULL) {
printf ( "%d " , temp->data);
temp = temp->next;
}
printf ( "\n" );
} // Moves all occurrences of given key to // end of linked list. void moveToEnd(Node* head, int key)
{ // Keeps track of locations where key
// is present.
struct Node* pKey = head;
// Traverse list
struct Node* pCrawl = head;
while (pCrawl != NULL) {
// If current pointer is not same as pointer
// to a key location, then we must have found
// a key in linked list. We swap data of pCrawl
// and pKey and move pKey to next position.
if (pCrawl != pKey && pCrawl->data != key) {
pKey->data = pCrawl->data;
pCrawl->data = key;
pKey = pKey->next;
}
// Find next position where key is present
if (pKey->data != key)
pKey = pKey->next;
// Moving to next Node
pCrawl = pCrawl->next;
}
} // Driver code int main()
{ Node* head = newNode(10);
head->next = newNode(20);
head->next->next = newNode(10);
head->next->next->next = newNode(30);
head->next->next->next->next = newNode(40);
head->next->next->next->next->next = newNode(10);
head->next->next->next->next->next->next = newNode(60);
printf ( "Before moveToEnd(), the Linked list is\n" );
printList(head);
int key = 10;
moveToEnd(head, key);
printf ( "\nAfter moveToEnd(), the Linked list is\n" );
printList(head);
return 0;
} |
// Java program to move all occurrences of a // given key to end. class GFG {
// A Linked list Node
static class Node {
int data;
Node next;
}
// A utility function to create a new node.
static Node newNode( int x)
{
Node temp = new Node();
temp.data = x;
temp.next = null ;
return temp;
}
// Utility function to print the elements
// in Linked list
static void printList(Node head)
{
Node temp = head;
while (temp != null ) {
System.out.printf( "%d " , temp.data);
temp = temp.next;
}
System.out.printf( "\n" );
}
// Moves all occurrences of given key to
// end of linked list.
static void moveToEnd(Node head, int key)
{
// Keeps track of locations where key
// is present.
Node pKey = head;
// Traverse list
Node pCrawl = head;
while (pCrawl != null ) {
// If current pointer is not same as pointer
// to a key location, then we must have found
// a key in linked list. We swap data of pCrawl
// and pKey and move pKey to next position.
if (pCrawl != pKey && pCrawl.data != key) {
pKey.data = pCrawl.data;
pCrawl.data = key;
pKey = pKey.next;
}
// Find next position where key is present
if (pKey.data != key)
pKey = pKey.next;
// Moving to next Node
pCrawl = pCrawl.next;
}
}
// Driver code
public static void main(String args[])
{
Node head = newNode( 10 );
head.next = newNode( 20 );
head.next.next = newNode( 10 );
head.next.next.next = newNode( 30 );
head.next.next.next.next = newNode( 40 );
head.next.next.next.next.next = newNode( 10 );
head.next.next.next.next.next.next = newNode( 60 );
System.out.printf( "Before moveToEnd(), the Linked list is\n" );
printList(head);
int key = 10 ;
moveToEnd(head, key);
System.out.printf( "\nAfter moveToEnd(), the Linked list is\n" );
printList(head);
}
} // This code is contributed by Arnab Kundu |
# Python3 program to move all occurrences of a # given key to end. # Linked List node class Node:
def __init__( self , data):
self .data = data
self . next = None
# A utility function to create a new node. def newNode(x):
temp = Node( 0 )
temp.data = x
temp. next = None
return temp
# Utility function to print the elements # in Linked list def printList( head):
temp = head
while (temp ! = None ) :
print ( temp.data,end = " " )
temp = temp. next
print ()
# Moves all occurrences of given key to # end of linked list. def moveToEnd(head, key):
# Keeps track of locations where key
# is present.
pKey = head
# Traverse list
pCrawl = head
while (pCrawl ! = None ) :
# If current pointer is not same as pointer
# to a key location, then we must have found
# a key in linked list. We swap data of pCrawl
# and pKey and move pKey to next position.
if (pCrawl ! = pKey and pCrawl.data ! = key) :
pKey.data = pCrawl.data
pCrawl.data = key
pKey = pKey. next
# Find next position where key is present
if (pKey.data ! = key):
pKey = pKey. next
# Moving to next Node
pCrawl = pCrawl. next
return head
# Driver code head = newNode( 10 )
head. next = newNode( 20 )
head. next . next = newNode( 10 )
head. next . next . next = newNode( 30 )
head. next . next . next . next = newNode( 40 )
head. next . next . next . next . next = newNode( 10 )
head. next . next . next . next . next . next = newNode( 60 )
print ( "Before moveToEnd(), the Linked list is\n" )
printList(head) key = 10
head = moveToEnd(head, key)
print ( "\nAfter moveToEnd(), the Linked list is\n" )
printList(head) # This code is contributed by Arnab Kundu |
// C# program to move all occurrences of a // given key to end. using System;
class GFG {
// A Linked list Node
public class Node {
public int data;
public Node next;
}
// A utility function to create a new node.
static Node newNode( int x)
{
Node temp = new Node();
temp.data = x;
temp.next = null ;
return temp;
}
// Utility function to print the elements
// in Linked list
static void printList(Node head)
{
Node temp = head;
while (temp != null ) {
Console.Write( "{0} " , temp.data);
temp = temp.next;
}
Console.Write( "\n" );
}
// Moves all occurrences of given key to
// end of linked list.
static void moveToEnd(Node head, int key)
{
// Keeps track of locations where key
// is present.
Node pKey = head;
// Traverse list
Node pCrawl = head;
while (pCrawl != null ) {
// If current pointer is not same as pointer
// to a key location, then we must have found
// a key in linked list. We swap data of pCrawl
// and pKey and move pKey to next position.
if (pCrawl != pKey && pCrawl.data != key) {
pKey.data = pCrawl.data;
pCrawl.data = key;
pKey = pKey.next;
}
// Find next position where key is present
if (pKey.data != key)
pKey = pKey.next;
// Moving to next Node
pCrawl = pCrawl.next;
}
}
// Driver code
public static void Main(String[] args)
{
Node head = newNode(10);
head.next = newNode(20);
head.next.next = newNode(10);
head.next.next.next = newNode(30);
head.next.next.next.next = newNode(40);
head.next.next.next.next.next = newNode(10);
head.next.next.next.next.next.next = newNode(60);
Console.Write( "Before moveToEnd(), the Linked list is\n" );
printList(head);
int key = 10;
moveToEnd(head, key);
Console.Write( "\nAfter moveToEnd(), the Linked list is\n" );
printList(head);
}
} // This code has been contributed by 29AjayKumar |
<script> // Javascript program to move all occurrences of a // given key to end. // A Linked list Node class Node {
constructor() {
this .data = 0;
this .next = null ;
}
}
// A utility function to create a new node.
function newNode(x) {
var temp = new Node();
temp.data = x;
temp.next = null ;
return temp;
}
// Utility function to print the elements
// in Linked list
function printList(head) {
var temp = head;
while (temp != null ) {
document.write( temp.data+ " " );
temp = temp.next;
}
document.write( "<br/>" );
}
// Moves all occurrences of given key to
// end of linked list.
function moveToEnd(head , key) {
// Keeps track of locations where key
// is present.
var pKey = head;
// Traverse list
var pCrawl = head;
while (pCrawl != null ) {
// If current pointer is not same as pointer
// to a key location, then we must have found
// a key in linked list. We swap data of pCrawl
// and pKey and move pKey to next position.
if (pCrawl != pKey && pCrawl.data != key) {
pKey.data = pCrawl.data;
pCrawl.data = key;
pKey = pKey.next;
}
// Find next position where key is present
if (pKey.data != key)
pKey = pKey.next;
// Moving to next Node
pCrawl = pCrawl.next;
}
}
// Driver code
var head = newNode(10);
head.next = newNode(20);
head.next.next = newNode(10);
head.next.next.next = newNode(30);
head.next.next.next.next = newNode(40);
head.next.next.next.next.next = newNode(10);
head.next.next.next.next.next.next = newNode(60);
document.write(
"Before moveToEnd(), the Linked list is<br/>"
);
printList(head);
var key = 10;
moveToEnd(head, key);
document.write(
"<br/>After moveToEnd(), the Linked list is<br/>"
);
printList(head);
// This code contributed by umadevi9616 </script> |
Before moveToEnd(), the Linked list is 10 20 10 30 40 10 60 After moveToEnd(), the Linked list is 20 30 40 60 10 10 10
Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of nodes in the linked list.
Auxiliary Space: O(1), as we are not using any extra space.
Efficient Solution 2 :
- Traverse the linked list and take a pointer at the tail.
- Now, check for the key and node->data. If they are equal, move the node to last-next, else move ahead.
// C++ code to remove key element to end of linked list #include<bits/stdc++.h> using namespace std;
// A Linked list Node struct Node
{ int data;
struct Node* next;
}; // A utility function to create a new node. struct Node* newNode( int x)
{ Node* temp = new Node;
temp->data = x;
temp->next = NULL;
} // Function to remove key to end Node *keyToEnd(Node* head, int key)
{ // Node to keep pointing to tail
Node* tail = head;
if (head == NULL)
{
return NULL;
}
while (tail->next != NULL)
{
tail = tail->next;
}
// Node to point to last of linked list
Node* last = tail;
Node* current = head;
Node* prev = NULL;
// Node prev2 to point to previous when head.data!=key
Node* prev2 = NULL;
// loop to perform operations to remove key to end
while (current != tail)
{
if (current->data == key && prev2 == NULL)
{
prev = current;
current = current->next;
head = current;
last->next = prev;
last = last->next;
last->next = NULL;
prev = NULL;
}
else
{
if (current->data == key && prev2 != NULL)
{
prev = current;
current = current->next;
prev2->next = current;
last->next = prev;
last = last->next;
last->next = NULL;
}
else if (current != tail)
{
prev2 = current;
current = current->next;
}
}
}
return head;
} // Function to display linked list void printList(Node* head)
{ struct Node* temp = head;
while (temp != NULL)
{
printf ( "%d " , temp->data);
temp = temp->next;
}
printf ( "\n" );
} // Driver Code int main()
{ Node* root = newNode(5);
root->next = newNode(2);
root->next->next = newNode(2);
root->next->next->next = newNode(7);
root->next->next->next->next = newNode(2);
root->next->next->next->next->next = newNode(2);
root->next->next->next->next->next->next = newNode(2);
int key = 2;
cout << "Linked List before operations :" ;
printList(root);
cout << "\nLinked List after operations :" ;
root = keyToEnd(root, key);
printList(root);
return 0;
} // This code is contributed by Rajout-Ji |
// Java code to remove key element to end of linked list import java.util.*;
// Node class class Node {
int data;
Node next;
public Node( int data)
{
this .data = data;
this .next = null ;
}
} class gfg {
static Node root;
// Function to remove key to end
public static Node keyToEnd(Node head, int key)
{
// Node to keep pointing to tail
Node tail = head;
if (head == null ) {
return null ;
}
while (tail.next != null ) {
tail = tail.next;
}
// Node to point to last of linked list
Node last = tail;
Node current = head;
Node prev = null ;
// Node prev2 to point to previous when head.data!=key
Node prev2 = null ;
// loop to perform operations to remove key to end
while (current != tail) {
if (current.data == key && prev2 == null ) {
prev = current;
current = current.next;
head = current;
last.next = prev;
last = last.next;
last.next = null ;
prev = null ;
}
else {
if (current.data == key && prev2 != null ) {
prev = current;
current = current.next;
prev2.next = current;
last.next = prev;
last = last.next;
last.next = null ;
}
else if (current != tail) {
prev2 = current;
current = current.next;
}
}
}
return head;
}
// Function to display linked list
public static void display(Node root)
{
while (root != null ) {
System.out.print(root.data + " " );
root = root.next;
}
}
// Driver Code
public static void main(String args[])
{
root = new Node( 5 );
root.next = new Node( 2 );
root.next.next = new Node( 2 );
root.next.next.next = new Node( 7 );
root.next.next.next.next = new Node( 2 );
root.next.next.next.next.next = new Node( 2 );
root.next.next.next.next.next.next = new Node( 2 );
int key = 2 ;
System.out.println( "Linked List before operations :" );
display(root);
System.out.println( "\nLinked List after operations :" );
root = keyToEnd(root, key);
display(root);
}
} |
# Python3 code to remove key element to # end of linked list # A Linked list Node class Node:
def __init__( self , data):
self .data = data
self . next = None
# A utility function to create a new node. def newNode(x):
temp = Node(x)
return temp
# Function to remove key to end def keyToEnd(head, key):
# Node to keep pointing to tail
tail = head
if (head = = None ):
return None
while (tail. next ! = None ):
tail = tail. next
# Node to point to last of linked list
last = tail
current = head
prev = None
# Node prev2 to point to previous
# when head.data!=key
prev2 = None
# Loop to perform operations to
# remove key to end
while (current ! = tail):
if (current.data = = key and prev2 = = None ):
prev = current
current = current. next
head = current
last. next = prev
last = last. next
last. next = None
prev = None
else :
if (current.data = = key and prev2 ! = None ):
prev = current
current = current. next
prev2. next = current
last. next = prev
last = last. next
last. next = None
else if (current ! = tail):
prev2 = current
current = current. next
return head
# Function to display linked list def printList(head):
temp = head
while (temp ! = None ):
print (temp.data, end = ' ' )
temp = temp. next
print ()
# Driver Code if __name__ = = '__main__' :
root = newNode( 5 )
root. next = newNode( 2 )
root. next . next = newNode( 2 )
root. next . next . next = newNode( 7 )
root. next . next . next . next = newNode( 2 )
root. next . next . next . next . next = newNode( 2 )
root. next . next . next . next . next . next = newNode( 2 )
key = 2
print ( "Linked List before operations :" )
printList(root)
print ( "Linked List after operations :" )
root = keyToEnd(root, key)
printList(root)
# This code is contributed by rutvik_56 |
// C# code to remove key // element to end of linked list using System;
// Node class public class Node {
public int data;
public Node next;
public Node( int data)
{
this .data = data;
this .next = null ;
}
} class GFG {
static Node root;
// Function to remove key to end
public static Node keyToEnd(Node head, int key)
{
// Node to keep pointing to tail
Node tail = head;
if (head == null ) {
return null ;
}
while (tail.next != null ) {
tail = tail.next;
}
// Node to point to last of linked list
Node last = tail;
Node current = head;
Node prev = null ;
// Node prev2 to point to
// previous when head.data!=key
Node prev2 = null ;
// loop to perform operations
// to remove key to end
while (current != tail) {
if (current.data == key && prev2 == null ) {
prev = current;
current = current.next;
head = current;
last.next = prev;
last = last.next;
last.next = null ;
prev = null ;
}
else {
if (current.data == key && prev2 != null ) {
prev = current;
current = current.next;
prev2.next = current;
last.next = prev;
last = last.next;
last.next = null ;
}
else if (current != tail) {
prev2 = current;
current = current.next;
}
}
}
return head;
}
// Function to display linked list
public static void display(Node root)
{
while (root != null ) {
Console.Write(root.data + " " );
root = root.next;
}
}
// Driver Code
public static void Main()
{
root = new Node(5);
root.next = new Node(2);
root.next.next = new Node(2);
root.next.next.next = new Node(7);
root.next.next.next.next = new Node(2);
root.next.next.next.next.next = new Node(2);
root.next.next.next.next.next.next = new Node(2);
int key = 2;
Console.WriteLine( "Linked List before operations :" );
display(root);
Console.WriteLine( "\nLinked List after operations :" );
root = keyToEnd(root, key);
display(root);
}
} // This code is contributed by PrinciRaj1992 |
<script> // javascript code to remove key element to end of linked list// Node class class Node {
constructor(val) {
this .data = val;
this .next = null ;
}
}
var root;
// Function to remove key to end
function keyToEnd(head , key) {
// Node to keep pointing to tail
var tail = head;
if (head == null ) {
return null ;
}
while (tail.next != null ) {
tail = tail.next;
}
// Node to point to last of linked list
var last = tail;
var current = head;
var prev = null ;
// Node prev2 to point to previous when head.data!=key
var prev2 = null ;
// loop to perform operations to remove key to end
while (current != tail) {
if (current.data == key && prev2 == null ) {
prev = current;
current = current.next;
head = current;
last.next = prev;
last = last.next;
last.next = null ;
prev = null ;
} else {
if (current.data == key && prev2 != null ) {
prev = current;
current = current.next;
prev2.next = current;
last.next = prev;
last = last.next;
last.next = null ;
} else if (current != tail) {
prev2 = current;
current = current.next;
}
}
}
return head;
}
// Function to display linked list
function display(root) {
while (root != null ) {
document.write(root.data + " " );
root = root.next;
}
}
// Driver Code
root = new Node(5);
root.next = new Node(2);
root.next.next = new Node(2);
root.next.next.next = new Node(7);
root.next.next.next.next = new Node(2);
root.next.next.next.next.next = new Node(2);
root.next.next.next.next.next.next = new Node(2);
var key = 2;
document.write( "Linked List before operations :<br/>" );
display(root);
document.write( "<br/>Linked List after operations :<br/>" );
root = keyToEnd(root, key);
display(root);
// This code contributed by aashish1995 </script> |
Linked List before operations :5 2 2 7 2 2 2 Linked List after operations :5 7 2 2 2 2 2
Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of nodes in the linked list.
Auxiliary Space: O(1), as we are not using any extra space.
Thanks to Ravinder Kumar for suggesting this method.
Efficient Solution 3: is to maintain a separate list of keys. We initialize this list of keys as empty. We traverse the given list. For every key found, we remove it from the original list and insert it into a separate list of keys. We finally link the list of keys at the end of the remaining given list.
Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of nodes in the linked list.
Auxiliary Space: O(1), as we are not using any extra space.