Remove Linked List Elements

To remove all the nodes with a specific value from a linked list, we can use a two-pointer approach: one to keep track of the current node and another to keep track of the previous node. By doing this, we can change the pointers of the nodes to skip over nodes with the value we want to remove.

Here’s the code:

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class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        dummy = ListNode(0)  # Create a dummy node to handle the case where head needs to be removed
        dummy.next = head
        prev = dummy  # Start the previous node at the dummy
        curr = head  # Start the current node at the head

        while curr is not None:
            if curr.val == val:  # If the current node's value equals the target value
                prev.next = curr.next  # Skip over the current node by changing the previous node's next pointer
            else:
                prev = curr  # If the current node's value does not equal the target value, move the previous pointer
            curr = curr.next  # Move to the next node

        return dummy.next  # Return the new head, which is the next node of the dummy

Explanation:

  • We create a dummy node to simplify the handling of the edge case where the head node needs to be removed.
  • We use two pointers, prev and curr, to traverse the list.
  • If the current node’s value equals the target value, we change the previous node’s next pointer to the current node’s next pointer, effectively skipping over the current node and removing it.
  • If the current node’s value does not equal the target value, we move the previous pointer to the current node.
  • Finally, we return the new head, which is the next node of the dummy.

This solution has a time complexity of O(n), where n is the number of nodes in the list, and a space complexity of O(1), as we only use a constant amount of extra space.

  1. Invariant (What does not vary in this problem?)

    • Do not change the order
    • When you find the value remove it from the list and create the link to the element the removed element was pointing to

  2. How to remove the element in the linked list? We cannot accomplish by using just one pointer. In Java yes, we need to make the next of 6 to be null

  3. Two pointers approach Initialize the leader, follower to nil and first element The leader will check if the node is 6, if it is we create the link

    f.next = l.next l.next = nil

    f = f.next l = f.next

  4. Terminating condition Leader.next is not null move leader and follower by one

  5. We should return the head of the linked list If head is 6, we will have a new head

    dummy = Node.new dummy.next = head

    return dummy.next

Time : O(n) Space : O(1)

dummy [1, 1]
f l f

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# Definition for singly-linked list.
# class ListNode
#     attr_accessor :val, :next
#     def initialize(val = 0, _next = nil)
#         @val = val
#         @next = _next
#     end
# end
# @param {ListNode} head
# @param {Integer} val
# @return {ListNode}
def remove_elements(head, val)
    dummy = ListNode.new
    dummy.next = head

    follower = dummy
    leader = dummy.next

    while (leader)
       if leader.val == val
#            remove the element
           follower.next = leader.next   
       else
           follower = leader
       end
       leader = leader.next  
    end

    return dummy.next
end

Identifying Problem Isomorphism

“Remove Linked List Elements” can be approximately mapped to “Move Zeroes”.

In “Remove Linked List Elements”, you’re given a linked list and a value. Your task is to remove all the nodes of the linked list that have the value equal to the given value.

Similarly, in “Move Zeroes”, you’re given an array of integers. Your task is to move all the 0’s to the end of it while maintaining the relative order of the non-zero elements.

The two problems are similar because they both involve a linear data structure (linked list or array) and require the removal or relocation of elements based on their values. However, “Move Zeroes” deals with an array and involves moving the zeros to the end, while “Remove Linked List Elements” deals with a linked list and involves removing nodes.

“Move Zeroes” is simpler because working with arrays tends to be easier than working with linked lists due to their direct access property. The similarity in the required operations makes these problems approximate isomorphs.

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class Solution:
    def removeElements(self, head, val):
        """
        :type head: ListNode
        :type val: int
        :rtype: ListNode
        """

        dummy_head = ListNode(-1)
        dummy_head.next = head

        current_node = dummy_head
        while current_node.next != None:
            if current_node.next.val == val:
                current_node.next = current_node.next.next
            else:
                current_node = current_node.next
                
        return dummy_head.next

Problem Classification

The problem is about “Linked List Manipulation”. It can be classified as “Linked List Deletion” as the task requires removal of certain nodes from the list based on their values.

This problem falls under the category of “Singly-Linked List Operations”, as it involves understanding and manipulating the singly-linked list data structure.

This problem can be classified under “Iterative Methods” as the solution involves iterating through the linked list to achieve the desired output.

Finally, from a practical use-case perspective, this kind of problem could be classified under “Data Cleaning/Preprocessing”, as it involves removing certain data points based on their values.

Language Agnostic Coding Drills

The above code is to remove all elements from a linked list that have a specific value. Here are the concepts and steps we can break the problem into:

  1. Understanding Linked Lists: The primary data structure used here is a linked list. Each node of the list consists of two parts: its value and a pointer to the next node. The last node points to None, indicating the end of the list.

  2. Iteration through a Linked List: The solution needs to traverse the list. Iterating over a linked list involves starting from the head node and following the pointers until reaching a node that points to None.

  3. Dummy Head/Node in Linked Lists: A dummy node is an artificial node that we often add at the front of the linked list to simplify edge cases where the head node might be deleted.

  4. Node Deletion in a Linked List: To remove a node from a linked list, we change the next pointer of the node preceding it to point to the node following it. This essentially “skips over” the node to be deleted, and it’s no longer part of the list.

  5. Conditional Checks: A conditional check is needed to determine whether or not a node’s value matches the target value to be removed.

The steps to solve this problem:

  1. Create a dummy node and make it the new head of the linked list.

  2. Iterate over the linked list, starting from the dummy head. For each node, check if its next node’s value is equal to the target value.

  3. If the next node’s value equals the target value, change the current node’s next pointer to skip the next node (thus deleting it). If not, move to the next node.

  4. Continue until you have checked all nodes in the list.

  5. Return the next node of the dummy head, which now serves as the new head of the processed list.

Targeted Drills in Python

Drill 1: Linked List Creation and Traversal

Before we can manipulate linked lists, we first need to understand how to create and traverse through one.

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class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

# Create linked list: 1 -> 2 -> 3 -> 4
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)

# Traverse and print the linked list
current = head
while current is not None:
    print(current.val)
    current = current.next

Drill 2: Implementing Dummy Head

Create a dummy node and link it to the head of our linked list.

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dummy_head = ListNode(-1)
dummy_head.next = head

print(dummy_head.next.val)  # This should print the value of head node

Drill 3: Linked List Node Deletion

Try to remove the second node from our linked list.

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head.next = head.next.next

# Traverse and print the linked list after deletion
current = head
while current is not None:
    print(current.val)
    current = current.next

Drill 4: Conditional Checks and Node Deletion

Iterate through a linked list and remove any node that has a value of 2.

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dummy_head = ListNode(-1)
dummy_head.next = head

current = dummy_head
while current.next is not None:
    if current.next.val == 2:
        current.next = current.next.next
    else:
        current = current.next

# Traverse and print the linked list after deletion
current = dummy_head.next
while current is not None:
    print(current.val)
    current = current.next

After mastering these individual drills, you should be able to understand and implement the full solution where you can pass any linked list and any value to a function, and it will return the linked list after removing all the nodes with that value.