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Move all occurrences of an element to end in a linked list
  • Difficulty Level : Medium
  • Last Updated : 31 Dec, 2020

Given a linked list and a key in it, the task is to move all occurrences of given key to end of linked list, keeping order of all other elements 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 given key in linked list. For every found occurrence, insert it at the end. We do it till all occurrences of given key are moved to end.

Time Complexity : O(n2)

Efficient Solution 1 : is to keep two pointers: 
pCrawl => Pointer to traverse the whole list one by one. 
pKey => Pointer to an occurrence of key if a key is found. Else same as pCrawl.
We start both of the above pointers from head of 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 same, we must have found a key which lies before pCrawl, so we swap data of pCrawl and pKey, and move pKey to 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++




// 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 urility 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




// 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 urility 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




# 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 urility 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#




// 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 urility 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


Output: 

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) requires only one traversal of list.

Efficient Solution 2 : 
1. Traverse the linked list and take a pointer at tail. 
2. Now, check for the key and node->data, if they are equal, move the node to last-next, else move 
ahead.

C++




// 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 urility 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




// 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




# 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
             
            elif (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#




// 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


Output: 

Linked List before operations :
5 2 2 7 2 2 2 
Linked List after operations :
5 7 2 2 2 2 2

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 given list. For every key found, we remove it from the original list and insert into the separate list of keys. We finally link list of keys at the end of remaining given list. Time complexity of this solution is also O(n) and it also requires only one traversal of list.

This article is contributed by MAZHAR IMAM KHAN. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@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.
 

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