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Python Program For Moving All Occurrences Of An Element To End In A Linked List

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  • Last Updated : 18 May, 2022

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)

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 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.  

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 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
")
printList(head)
  
key = 10
head = moveToEnd(head, key)
  
print("
After moveToEnd(), the Linked list is
")
printList(head)
  
# This code is contributed by Arnab Kundu

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 the list.

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

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

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 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. The time complexity of this solution is also O(n) and it also requires only one traversal of the list.

Please refer complete article on Move all occurrences of an element to end in a linked list for more details!


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