Leetcode #138: Copy List with Random Pointer
In this guide, we solve Leetcode #138 Copy List with Random Pointer in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Hash Table, Linked List
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]
Python Solution
"""
# 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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.