LeetCode #82 — MEDIUM

Remove Duplicates from Sorted List II

Move from brute-force thinking to an efficient approach using linked list strategy.

The Problem

Problem Statement

Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.

Example 1:

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

Example 2:

Input: head = [1,1,1,2,3]
Output: [2,3]

Constraints:

The number of nodes in the list is in the range [0, 300].
-100 <= Node.val <= 100
The list is guaranteed to be sorted in ascending order.

Step 01

Brute Force Baseline

Problem summary: Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Linked List · Two Pointers

Example 1

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

Example 2

[1,1,1,2,3]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode dummy = new ListNode(0, head);
        ListNode prev = dummy;

        while (head != null) {
            if (head.next != null && head.val == head.next.val) {
                int duplicateVal = head.val;
                while (head != null && head.val == duplicateVal) head = head.next;
                prev.next = head;
            } else {
                prev = prev.next;
                head = head.next;
            }
        }

        return dummy.next;
    }
}

/**
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func deleteDuplicates(head *ListNode) *ListNode {
    dummy := &ListNode{Val: 0, Next: head}
    prev := dummy

    for head != nil {
        if head.Next != nil && head.Val == head.Next.Val {
            dup := head.Val
            for head != nil && head.Val == dup {
                head = head.Next
            }
            prev.Next = head
        } else {
            prev = prev.Next
            head = head.Next
        }
    }

    return dummy.Next
}

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy = ListNode(0, head)
        prev = dummy

        while head:
            if head.next and head.val == head.next.val:
                dup = head.val
                while head and head.val == dup:
                    head = head.next
                prev.next = head
            else:
                prev = prev.next
                head = head.next

        return dummy.next

/**
 * Definition for singly-linked list.
 * #[derive(PartialEq, Eq, Clone, Debug)]
 * pub struct ListNode {
 *   pub val: i32,
 *   pub next: Option<Box<ListNode>>
 * }
 * impl ListNode {
 *   #[inline]
 *   fn new(val: i32) -> Self {
 *     ListNode { next: None, val }
 *   }
 * }
 */
impl Solution {
    pub fn delete_duplicates(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut vals = Vec::new();
        let mut cur = head;
        while let Some(mut node) = cur {
            vals.push(node.val);
            cur = node.next.take();
        }

        let mut kept = Vec::new();
        let mut i = 0usize;
        while i < vals.len() {
            let mut j = i + 1;
            while j < vals.len() && vals[j] == vals[i] {
                j += 1;
            }
            if j == i + 1 {
                kept.push(vals[i]);
            }
            i = j;
        }

        let mut out: Option<Box<ListNode>> = None;
        for &v in kept.iter().rev() {
            out = Some(Box::new(ListNode { val: v, next: out }));
        }
        out
    }
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */
function deleteDuplicates(head: ListNode | null): ListNode | null {
  const dummy = new ListNode(0, head);
  let prev: ListNode = dummy;

  while (head) {
    if (head.next && head.val === head.next.val) {
      const dup = head.val;
      while (head && head.val === dup) head = head.next;
      prev.next = head;
    } else {
      prev = prev.next!;
      head = head.next;
    }
  }

  return dummy.next;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

COPY TO ARRAY

O(n) time

O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS

O(n) time

O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.