Delete Nth Node from End of Linked List

hardcodingAlgorithmsLinked ListsTwo Pointers~20 min

Asked at Meta, Google

Overview

Given the head of a linked list, remove the nth node from the end of the list and return the modified list's head.

Use a one-pass algorithm with O(1) extra space.

Constraints

The number of nodes is in range [1, 30].
1 <= n <= length of list.
Node values are integers.

Examples

// List: 1->2->3->4->5, n=2
removeNthFromEnd(head, 2);
// => 1->2->3->5  (removed node 4)

// List: 1, n=1
removeNthFromEnd(head, 1);
// => null  (removed the only node)

// List: 1->2, n=1
removeNthFromEnd(head, 1);
// => 1  (removed node 2)

Solution

Reveal solution

function removeNthFromEnd(head, n) {
  const dummy = { val: 0, next: head };
  let fast = dummy;
  let slow = dummy;

  // Advance fast by n+1 steps
  for (let i = 0; i <= n; i++) {
    fast = fast.next;
  }

  // Move both until fast reaches end
  while (fast) {
    fast = fast.next;
    slow = slow.next;
  }

  // Skip the target node
  slow.next = slow.next.next;
  return dummy.next;
}

Two-pointer technique: fast pointer leads by n+1 nodes. When fast reaches null, slow is right before the target.

Resources

delete-nth-node-from-end.js

Delete Nth Node from End of Linked List

hardcodingAlgorithmsLinked Lists

Overview

Given the head of a linked list, remove the nth node from the end of the list and return the modified list's head.

Use a one-pass algorithm with O(1) extra space.

Constraints

The number of nodes is in range [1, 30].
1 <= n <= length of list.
Node values are integers.

Examples

// List: 1->2->3->4->5, n=2
removeNthFromEnd(head, 2);
// => 1->2->3->5  (removed node 4)

// List: 1, n=1
removeNthFromEnd(head, 1);
// => null  (removed the only node)

// List: 1->2, n=1
removeNthFromEnd(head, 1);
// => 1  (removed node 2)

Solution

Reveal solution

function removeNthFromEnd(head, n) {
  const dummy = { val: 0, next: head };
  let fast = dummy;
  let slow = dummy;

  // Advance fast by n+1 steps
  for (let i = 0; i <= n; i++) {
    fast = fast.next;
  }

  // Move both until fast reaches end
  while (fast) {
    fast = fast.next;
    slow = slow.next;
  }

  // Skip the target node
  slow.next = slow.next.next;
  return dummy.next;
}

Two-pointer technique: fast pointer leads by n+1 nodes. When fast reaches null, slow is right before the target.

Resources

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