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Remove Nth Node

Certification: Intermediate Level Accuracy: 0% Submissions: 0 Points: 10

Write a C program to remove the nth node from the end of a linked list and return its head. Given a linked list, remove the n-th node from the end of the list and return the modified list's head.

Example 1
    
  
  • Input: head = [1, 2, 3, 4, 5], n = 2
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  • Output: [1, 2, 3, 5]
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  • Explanation: The linked list is 1->2->3->4->5. We need to remove the 2nd node from the end, which is 4. After removal, the linked list becomes 1->2->3->5.
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Example 2
    
  
  • Input: head = [1], n = 1
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  • Output: []
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  • Explanation: The linked list only has one node: 1. We need to remove the 1st node from the end, which is the only node. After removal, the linked list is empty.
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Constraints
    
  
  • The number of nodes in the list is sz.
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  • 1 <= sz <= 30
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  • 0 <= Node.val <= 100
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  • 1 <= n <= sz
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  • Time Complexity: O(L) where L is the length of the list
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  • Space Complexity: O(1)
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Hint
Solution

Solution Hints

    
  
  • Use the two-pointer technique with a fast and slow pointer
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  • Move the fast pointer n steps ahead first
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  • Then move both pointers together until the fast pointer reaches the end
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  • At this point, the slow pointer will be just before the node to be removed
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  • Update the next pointer of the slow pointer to skip the node to be removed
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Steps to solve by this approach:

 Step 1: Create a dummy node that points to the head to handle edge cases like removing the head node.

 Step 2: Use two pointers, fast and slow, both initially pointing to the dummy node.

 Step 3: Move the fast pointer n+1 steps ahead.

 Step 4: Move both fast and slow pointers one step at a time until the fast pointer reaches the end of the list.

 Step 5: At this point, the slow pointer will be pointing to the node just before the one to be removed.

 Step 6: Update the next pointer of the slow pointer to skip the node to be removed.

 Step 7: Return the next pointer of the dummy node, which is the new head of the list.

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