Remove Linked List Elements Solution for Capgemini

Company: Capgemini · Difficulty: Easy · Topics: Linked List, Recursion

Problem Statement

Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.

The linked list is defined as:

• Each node contains a value and a reference to the next node
• The head is the first node of the linked list

Examples

Example 1:

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

Output: [1,2,3,4,5]

Explanation: Remove all nodes with value 6.

Example 2:

Input: head = [], val = 1

Output: []

Explanation: Empty list remains empty.

Example 3:

Input: head = [7,7,7,7], val = 7

Output: []

Explanation: Remove all nodes with value 7.

Intuition

We need to traverse the linked list and remove all nodes that have the target value. We can use a dummy node to handle the case where the head node needs to be removed.

Approach

Approach 1: Dummy Node (Optimal)

1. Create a dummy node pointing to head

2. Use a pointer to traverse the list

3. When current node's next has target value, skip it

4. Return dummy.next as new head

Approach 2: Recursion

1. Base case: if head is null, return null

2. If head.val == val, return removeElements(head.next, val)

3. Otherwise, head.next = removeElements(head.next, val)

4. Return head

Approach 3: Iterative without Dummy

1. Handle head removal separately

2. Use two pointers: prev and current

3. When current.val == val, update prev.next

4. Update pointers accordingly

Dry Run

Let's trace through Example 1: head = [1,2,6,3,4,5,6], val = 6

Dummy Node Approach:

• Create dummy → [1,2,6,3,4,5,6]
• curr = dummy
• curr.next = 1, 1 != 6 → move curr
• curr.next = 2, 2 != 6 → move curr
• curr.next = 6, 6 == 6 → curr.next = curr.next.next
• curr.next = 3, 3 != 6 → move curr
• curr.next = 4, 4 != 6 → move curr
• curr.next = 5, 5 != 6 → move curr
• curr.next = 6, 6 == 6 → curr.next = curr.next.next
• curr.next = null
• Return dummy.next = [1,2,3,4,5]

Key Insights

1. Dummy Node: Simplifies handling head removal

2. Two Pointers: Use prev and curr for efficient removal

3. Memory Management: Update pointers to avoid memory leaks

4. Edge Cases: Handle empty list and all nodes being removed

Edge Cases

1. Empty list: Return null

2. Head removal: First node has target value

3. All nodes removed: Return empty list

4. Consecutive removals: Multiple target values in a row

5. Last node removal: Target value at end

Time and Space Complexity

• Time Complexity: O(n) - Single pass through list
• Space Complexity: O(1) - Only constant extra space used

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