Remove Linked List Elements Leetcode Problem 203 [Python Solution]

In Remove Linked List Elements Leetcode challenge, we’re given the head of a linked list and an integer val, to remove all conditionally.

This problem falls under the category of Linked List and is classified as an “Easy” level problem.

The task at hand is to remove all nodes from a given linked list that have a specific value, val, and return the updated head of the linked list.

We’ll explore a Python solution to solve this problem efficiently.

Understanding the Problem

Let’s start by understanding the problem with an example:

Example 1:

Suppose we have a linked list as follows:

Input: head = [1, 2, 6, 3, 4, 5, 6]
Value to Remove: val = 6

The linked list initially looks like this: 1 -> 2 -> 6 -> 3 -> 4 -> 5 -> 6

Our goal is to remove all occurrences of the value 6.

After the removal, the modified linked list should look like this: 1 -> 2 -> 3 -> 4 -> 5

Let’s break down the problem and understand how we can approach it.

Problem Overview

The main idea behind solving this problem is to iterate through the linked list, check each node’s value, and remove the nodes with the specified val.

We can efficiently do this by maintaining two pointers: a previous pointer and a current pointer.

Efficient Python Code Solution

Here’s the Python code solution to remove the elements efficiently:

def removeElements(self, head: ListNode, val: int) -> ListNode:
    # Create a dummy node to handle edge cases
    dummy = ListNode(next=head)
    prev, curr = dummy, head

    while curr:
        nxt = curr.next

        if curr.val == val:
            # Remove the node by updating the previous pointer
            prev.next = nxt
        else:
            # If the current node's value doesn't match the target value, update the previous pointer
            prev = curr

        curr = nxt

    # Return the new head
    return dummy.next

Now, let’s dive into the details of this Python solution.

Algorithm Breakdown

1. We start by creating a dummy node.

This dummy node serves as a convenient reference and helps us handle edge cases efficiently.

It doesn’t matter what value the dummy node holds; we are only interested in its next pointer.

2. We initialize two pointers: prev and curr.

The prev pointer is initially set to the dummy node, while the curr pointer is set to the head of the linked list, as provided in the input.

3. We enter a loop that iterates through the linked list.

This loop will continue until the curr pointer reaches the end of the list, i.e., when curr becomes None.

4. Inside the loop, we keep track of the next node using the nxt variable, which holds the value of curr.next.
5. We check whether the value of the current node, curr.val, is equal to the target value, val, that we want to remove from the linked list.
6. If curr.val matches val, we remove the node by updating the prev.next pointer.

This effectively bypasses the current node and points prev.next to the next node in the list, nxt.

7. If the current node’s value does not match the target value, we update the prev pointer to be equal to the curr pointer.

This is because there’s no need to shift prev if we aren’t removing the current node.

8. After processing the current node, we move the curr pointer to the next node by assigning curr to nxt.
9. Once we complete the loop and have removed all nodes with the target value, we return the new head of the modified linked list.

The new head is accessible via dummy.next.

Time and Space Complexity

Before we conclude, let’s analyze the time and space complexity of this solution.

Time Complexity: The time complexity is O(n), where n is the number of nodes in the linked list.

We need to traverse the entire list once to identify and remove the nodes with the target value.

Space Complexity: The space complexity is O(1) since we are using a constant amount of extra memory for the dummy, prev, curr, and nxt pointers.

We do not use additional data structures that scale with the input size.

Reasoning Behind Our Efficient Approach

The key to the efficiency of this solution lies in maintaining two pointers, prev and curr, which allows us to efficiently identify and remove nodes with the target value without the need for extensive node shifting.

The introduction of the dummy node also simplifies edge cases, ensuring that the new head is easily accessible.

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Conclusion

In this blog post, we addressed the LeetCode problem 203, Remove Linked List Elements We provided an efficient Python solution that leverages two pointers and a dummy node to remove nodes with a specified value from the linked list.

This approach offers a time complexity of O(n) and a space complexity of O(1).

We hope this guide has been helpful to you as you explore linked list problems on LeetCode.

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Link to the LeetCode problem: Remove Linked List Elements

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