Python Program For Partitioning A Linked List Around A Given Value And Keeping The Original Order

Given a linked list and a value x, partition it such that all nodes less than x come first, then all nodes with a value equal to x, and finally nodes with a value greater than or equal to x. The original relative order of the nodes in each of the three partitions should be preserved. The partition must work in place.
Examples:

Input: 1->4->3->2->5->2->3, 
        x = 3
Output: 1->2->2->3->3->4->5

Input: 1->4->2->10 
        x = 3
Output: 1->2->4->10

Input: 10->4->20->10->3 
        x = 3
Output: 3->10->4->20->10 

To solve this problem we can use partition method of Quick Sort but this would not preserve the original relative order of the nodes in each of the two partitions.
Below is the algorithm to solve this problem :

1. Initialize first and last nodes of below three linked lists as NULL.
   • Linked list of values smaller than x.
   • Linked list of values equal to x.
   • Linked list of values greater than x.
2. Now iterate through the original linked list. If a node’s value is less than x then append it at the end of the smaller list. If the value is equal to x, then at the end of the equal list. And if a value is greater, then at the end of the greater list.
3. Now concatenate three lists.

Below is the implementation of the above idea. 

Python3

# Python3 program to partition a
# linked list around a given value.

# Link list Node
class Node:
    def __init__(self):
        self.data = 0
        self.next = None

# A utility function to create
# a new node
def newNode(data):
    new_node = Node()
    new_node.data = data
    new_node.next = None
    return new_node

# Function to make two separate lists
# and return head after concatenating
def partition( head, x) :

    # Let us initialize first and last
    # nodes of three linked lists
    # 1) Linked list of values smaller
    #    than x.
    # 2) Linked list of values equal
    #    to x.
    # 3) Linked list of values greater
    #    than x.
    smallerHead = None
    smallerLast = None
    greaterLast = None
    greaterHead = None
    equalHead = None
    equalLast = None

    # Now iterate original list and connect
    # nodes of appropriate linked lists.
    while (head != None) :

        # If current node is equal to x,
        # append it to the list of x values
        if (head.data == x):

            if (equalHead == None):
                equalHead = equalLast = head
            else:

                equalLast.next = head
                equalLast = equalLast.next

        # If current node is less than X,
        # append it to the list of smaller
        # values
        elif (head.data < x):

            if (smallerHead == None):
                smallerLast = smallerHead = head
            else:

                smallerLast.next = head
                smallerLast = head

        else :

        # Append to the list of greater values
            if (greaterHead == None) :
                greaterLast = greaterHead = head
            else:

                greaterLast.next = head
                greaterLast = head

        head = head.next

    # Fix end of greater linked list to None
    # if this list has some nodes
    if (greaterLast != None) :
        greaterLast.next = None

    # Connect three lists

    # If smaller list is empty
    if (smallerHead == None) :

        if (equalHead == None) :
            return greaterHead
        equalLast.next = greaterHead
        return equalHead

    # If smaller list is not empty
    # and equal list is empty
    if (equalHead == None) :

        smallerLast.next = greaterHead
        return smallerHead

    # If both smaller and equal list
    # are non-empty
    smallerLast.next = equalHead
    equalLast.next = greaterHead
    return smallerHead

# Function to print linked list
def printList(head) :

    temp = head
    while (temp != None):

        print(temp.data ,end= " ")
        temp = temp.next

# Driver code
# Start with the empty list
head = newNode(10)
head.next = newNode(4)
head.next.next = newNode(5)
head.next.next.next = newNode(30)
head.next.next.next.next = newNode(2)
head.next.next.next.next.next = newNode(50)

x = 3
head = partition(head, x)
printList(head)
# This code is contributed by Arnab Kundu.

Output:

2 10 4 5 30 50

Time Complexity: O(n) where n is the size of the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please refer complete article on Partitioning a linked list around a given value and keeping the original order for more details!