LeetCode #138 — MEDIUM

Copy List with Random Pointer

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

A linked list of length n is given such that each node contains an additional random pointer, which could point to any node in the list, or null.

Construct a deep copy of the list. The deep copy should consist of exactly n brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next and random pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. None of the pointers in the new list should point to nodes in the original list.

For example, if there are two nodes X and Y in the original list, where X.random --> Y, then for the corresponding two nodes x and y in the copied list, x.random --> y.

Return the head of the copied linked list.

The linked list is represented in the input/output as a list of n nodes. Each node is represented as a pair of [val, random_index] where:

val: an integer representing Node.val
random_index: the index of the node (range from 0 to n-1) that the random pointer points to, or null if it does not point to any node.

Your code will only be given the head of the original linked list.

Example 1:

Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]

Example 2:

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

Example 3:

Input: head = [[3,null],[3,0],[3,null]]
Output: [[3,null],[3,0],[3,null]]

Constraints:

0 <= n <= 1000
-10⁴ <= Node.val <= 10⁴
Node.random is null or is pointing to some node in the linked list.

Step 01

Brute Force Baseline

Problem summary: A linked list of length n is given such that each node contains an additional random pointer, which could point to any node in the list, or null. Construct a deep copy of the list. The deep copy should consist of exactly n brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next and random pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. None of the pointers in the new list should point to nodes in the original list. For example, if there are two nodes X and Y in the original list, where X.random --> Y, then for the corresponding two nodes x and y in the copied list, x.random --> y. Return the head of the copied linked list. The linked list is represented in the input/output as a list of n nodes. Each node is

Baseline thinking

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

Pattern signal: Hash Map · Linked List

Example 1

[[7,null],[13,0],[11,4],[10,2],[1,0]]

Example 2

[[1,1],[2,1]]

Example 3

[[3,null],[3,0],[3,null]]

Core Insight

What unlocks the optimal approach

Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, ensure you are not making multiple copies of the same node.
You may want to use extra space to keep old_node ---> new_node mapping to prevent creating multiple copies of the same node.
We can avoid using extra space for old_node ---> new_node mapping by tweaking the original linked list. Simply interweave the nodes of the old and copied list. For example: Old List: A --> B --> C --> D InterWeaved List: A --> A' --> B --> B' --> C --> C' --> D --> D'
The interweaving is done using next</b> pointers and we can make use of interweaved structure to get the correct reference nodes for random</b> pointers.

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

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #138: Copy List with Random Pointer
/*
// Definition for a Node.
class Node {
    int val;
    Node next;
    Node random;

    public Node(int val) {
        this.val = val;
        this.next = null;
        this.random = null;
    }
}
*/

class Solution {
    public Node copyRandomList(Node head) {
        Map<Node, Node> d = new HashMap<>();
        Node dummy = new Node(0);
        Node tail = dummy;
        for (Node cur = head; cur != null; cur = cur.next) {
            Node node = new Node(cur.val);
            tail.next = node;
            tail = node;
            d.put(cur, node);
        }
        for (Node cur = head; cur != null; cur = cur.next) {
            d.get(cur).random = cur.random == null ? null : d.get(cur.random);
        }
        return dummy.next;
    }
}

// Accepted solution for LeetCode #138: Copy List with Random Pointer
/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Next *Node
 *     Random *Node
 * }
 */

func copyRandomList(head *Node) *Node {
	dummy := &Node{}
	tail := dummy
	d := map[*Node]*Node{}
	for cur := head; cur != nil; cur = cur.Next {
		node := &Node{Val: cur.Val}
		d[cur] = node
		tail.Next = node
		tail = node
	}
	for cur := head; cur != nil; cur = cur.Next {
		if cur.Random != nil {
			d[cur].Random = d[cur.Random]
		}
	}
	return dummy.Next
}

# Accepted solution for LeetCode #138: Copy List with Random Pointer
"""
# Definition for a Node.
class Node:
    def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None):
        self.val = int(x)
        self.next = next
        self.random = random
"""


class Solution:
    def copyRandomList(self, head: "Optional[Node]") -> "Optional[Node]":
        d = {}
        dummy = tail = Node(0)
        cur = head
        while cur:
            node = Node(cur.val)
            tail.next = node
            tail = tail.next
            d[cur] = node
            cur = cur.next
        cur = head
        while cur:
            d[cur].random = d[cur.random] if cur.random else None
            cur = cur.next
        return dummy.next

// Accepted solution for LeetCode #138: Copy List with Random Pointer
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #138: Copy List with Random Pointer
// /*
// // Definition for a Node.
// class Node {
//     int val;
//     Node next;
//     Node random;
// 
//     public Node(int val) {
//         this.val = val;
//         this.next = null;
//         this.random = null;
//     }
// }
// */
// 
// class Solution {
//     public Node copyRandomList(Node head) {
//         Map<Node, Node> d = new HashMap<>();
//         Node dummy = new Node(0);
//         Node tail = dummy;
//         for (Node cur = head; cur != null; cur = cur.next) {
//             Node node = new Node(cur.val);
//             tail.next = node;
//             tail = node;
//             d.put(cur, node);
//         }
//         for (Node cur = head; cur != null; cur = cur.next) {
//             d.get(cur).random = cur.random == null ? null : d.get(cur.random);
//         }
//         return dummy.next;
//     }
// }

// Accepted solution for LeetCode #138: Copy List with Random Pointer
/**
 * Definition for _Node.
 * class _Node {
 *     val: number
 *     next: _Node | null
 *     random: _Node | null
 *
 *     constructor(val?: number, next?: _Node, random?: _Node) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *         this.random = (random===undefined ? null : random)
 *     }
 * }
 */

function copyRandomList(head: _Node | null): _Node | null {
    const d: Map<_Node, _Node> = new Map();
    const dummy = new _Node();
    let tail = dummy;
    for (let cur = head; cur; cur = cur.next) {
        const node = new _Node(cur.val);
        tail.next = node;
        tail = node;
        d.set(cur, node);
    }
    for (let cur = head; cur; cur = cur.next) {
        d.get(cur)!.random = cur.random ? d.get(cur.random)! : null;
    }
    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(n)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.