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Python Program For Removing Middle Points From a Linked List Of Line Segments

  • Last Updated : 27 Dec, 2021

Given a linked list of coordinates where adjacent points either form a vertical line or a horizontal line. Delete points from the linked list which are in the middle of a horizontal or vertical line.

Input:  (0,10)->(1,10)->(5,10)->(7,10)
Output: Linked List should be changed to following
The given linked list represents a horizontal line from (0,10) 
to (7, 10) followed by a vertical line from (7, 10) to (7, 5), 
followed by a horizontal line from (7, 5) to (40, 5).

Input:     (2,3)->(4,3)->(6,3)->(10,3)->(12,3)
Output: Linked List should be changed to following
There is only one vertical line, so all middle points are removed.

Source: Microsoft Interview Experience

The idea is to keep track of the current node, next node, and next-next node. While the next node is the same as the next-next node, keep deleting the next node. In this complete procedure, we need to keep an eye on the shifting of pointers and checking for NULL values.
Following are implementations of the above idea. 


# Python program to remove middle points in a linked list of
# line segments,
class LinkedList(object):
    def __init__(self):
        self.head = None
    # Linked list Node
    class Node(object):
        def __init__(self, x, y):
            self.x = x
            self.y = y
   = None
    # This function deletes middle nodes in a sequence of
    # horizontal and vertical line segments represented
    # by linked list.
    def deleteMiddle(self):
        # If only one node or no node...Return back
        if self.head == None or == None or == None:
            return self.head
        Next =
        NextNext =
        # check if this is vertical or horizontal line
        if self.head.x == Next.x:
            # Find middle nodes with same value as x and
            # delete them.
            while NextNext != None and Next.x == NextNext.x:
       = None
                # Update NextNext for the next iteration
                Next = NextNext
                NextNext =
        elif self.head.y == Next.y:
            # find middle nodes with same value as y and
            # delete them
            while NextNext != None and Next.y == NextNext.y:
       = None
                # Update NextNext for the next iteration
                Next = NextNext
                NextNext =
            # Adjacent points should have same x or same y
            print "Given list is not valid"
            return None
        # recur for other segment
        # temporarily store the head and move head forward.
        temp = self.head
        self.head =
        # call deleteMiddle() for next segment
        # restore head
        self.head = temp
        # return the head
        return self.head
    # Given a reference (pointer to pointer) to the head
    # of a list and an int, push a new node on the front
    # of the list.
    def push(self, x, y):
        # 1 & 2: Allocate the Node &
        # Put in the data
        new_node = self.Node(x, y)
        # 3. Make next of new Node as head = self.head
        # 4. Move the head to point to new Node
        self.head = new_node
    def printList(self):
        temp = self.head
        while temp != None:
            print "(" + str(temp.x) + "," + str(temp.y) + ")->",
            temp =
        print ''
# Driver program
llist = LinkedList()
print "Given list"
if llist.deleteMiddle() != None:
    print "Modified Linked List is"
# This code is contributed by BHAVYA JAIN


Given Linked List:
(0,10)-> (1,10)-> (3,10)-> (10,10)-> (10,8)-> (10,5)-> (20,5)-> (40,5)->
Modified Linked List:
(0,10)-> (10,10)-> (10,5)-> (40,5)-> 

Time Complexity of the above solution is O(n) where n is a number of nodes in the given linked list.
The above code is recursive, write an iterative code for the same problem. Please see below for the solution.
Iterative approach for removing middle points in a linked list of line segments
Please refer complete article on Given a linked list of line segments, remove middle points for more details!

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