#38 Medium Time O(n) Space O(1) Animated

Remove Nth Node From End of List

Remove the nth node from the end of a linked list.

LeetCode ↗

Tutor Mode Build the mental model before you memorize the code.

1 Predict
2 Trace
3 Explain
4 Implement

Pattern Pointer rewiring with saved next state

Before changing a link, save the only pointer that would otherwise be lost.

Approach: Two Pointers (Dummy Node)

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Core Insight

Before changing a link, save the only pointer that would otherwise be lost.

Common Pitfall

Always store next before redirecting current.next.

Description

Remove the nth node from the end of a linked list.

Constraints

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

Solution — Two Pointers (Dummy Node)

TimeO(n) SpaceO(1)

Implementation Plan

Translate the invariant into these small, testable moves.

Initialize the pointers.
Save the next node.
Rewire the current node.
Advance both pointers until the list is consumed.

def removeNthFromEnd(head, n):
    dummy = ListNode(0, head)
    left = dummy
    right = head
    
    while n > 0 and right:
        right = right.next
        n -= 1
        
    while right:
        left = left.next
        right = right.next
        
    left.next = left.next.next
    return dummy.next

Reorder List

Copy List with Random Pointer

Remove Nth Node From End of List

Steps

Description

Constraints

Tags

Solution — Two Pointers (Dummy Node)