How do you solve Remove Linked List Elements on LeetCode?

Remove Linked List Elements (LeetCode #203) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using a dummy node simplifies edge cases like removing the head.

What is the brute force solution for Remove Linked List Elements?

The brute force approach for Remove Linked List Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves iterating through the linked list and collecting all the nodes that do not have the value to be removed. This is simple but inefficient, as we may need to traverse the list multiple times. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Remove Linked List Elements?

Remove Linked List Elements uses the following concepts: Linked List, Recursion. The recommended patterns to study are: Linked List, Two Pointers.

What are the interview tips for Remove Linked List Elements?

Explain your thought process clearly as you code. Consider edge cases, such as an empty list or all nodes being removed. Use a dummy node to simplify your logic and avoid special cases.

What are common mistakes when solving Remove Linked List Elements?

Not handling the case where the head needs to be removed. Forgetting to update the previous pointer when a node is removed.

#203

Remove Linked List Elements

Easy

Linked ListRecursionLinked ListTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution uses a single pass through the linked list, modifying pointers directly to skip over nodes with the value 'val'. This is efficient because it only requires one traversal of the list.

⚙️

Algorithm

5 steps

1Step 1: Create a dummy node that points to the head of the list.
2Step 2: Use a pointer to traverse the list, checking each node's value.
3Step 3: If the current node's value equals 'val', skip it by adjusting the previous node's next pointer.
4Step 4: If it doesn't equal 'val', move the previous pointer to the current node.
5Step 5: Continue until the end of the list, then return the next node of the dummy node.

solution.py18 lines

1# Full working Python code
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7def removeElements(head, val):
8    dummy = ListNode(0)
9    dummy.next = head
10    prev = dummy
11    current = head
12    while current:
13        if current.val == val:
14            prev.next = current.next
15        else:
16            prev = current
17        current = current.next
18    return dummy.next

ℹ

Complexity note: The time complexity is O(n) because we only traverse the list once. The space complexity is O(1) since we are using a constant amount of extra space.

1Using a dummy node simplifies edge cases like removing the head.
2Directly modifying pointers avoids the need for additional data structures.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.