LeetCode 38. Copy List with Random Pointer — Python Programming Solution

Blind 75 — Programming & Technical Interview Questions — Explanation Series

The Problem:

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]]

The Constraints:

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

The Explanation:

This problem boils down to solving one problem: how do you keep track of random pointers when you have to full reconstruct the list from scratch? You can’t just map values to their nodes because we are not guaranteed each node has a different value. So we know we have go through the list at list twice: once to construct the list, another to assign random pointers. Now the problem is how to keep track of what nodes from the original list correspond to the new list, and like I said we can’t map values since those aren’t all distinct. The answer to this is just map the pointer from the original node to the pointer of the new node. This creates a distinct connection between each node. When we first build the array create a dictionary that stores old pointers as the key and new pointers as the value. Now when we go through the second time we can get the random pointer from the original list and find out what node that corresponds to in our new list.

The Python Programming Solution:

Nothing fancy here. Create the dictionary, construct the new list without random pointers, go through a second time and use our map to assing random pointers in our new list.

"""
# 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: "Node") -> "Node":
    oldToCopy = {None: None}

    cur = head
    while cur:
      copy = Node(cur.val)
      oldToCopy[cur] = copy
      cur = cur.next

    cur = head
    while cur:
      copy = oldToCopy[cur]
      copy.next = oldToCopy[cur.next]
      copy.random = oldToCopy[cur.random]
      cur = cur.next
      
    return oldToCopy[head]

LeetCode 38. Copy List with Random Pointer — Python Programming Solution

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