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LESSON 438 OF 673 · LINKED LISTS

Copy List with Random Pointer II in Python (O(n) Time Complexity)

Practice O(n) Read the full explanation

You get the head of a linked list. Build a deep copy of it and return the copy's head.

Each node has a value, a next pointer and one more pointer, random. This random pointer can lead to any node of the list, even the node itself. It can also be None.

The copy is made only of new nodes, one for each original node, with the same values. In the copy, each next and random leads to the matching new node. Nothing points back into the original list.

Say an original node's random leads to the third node. Then its copy's random leads to the copy of the third node.

The examples describe the list with two arrays of the same length. The first, values, holds the values from the head on. The second, random, holds one entry per node. For node i, random[i] is the index of the node its random leads to. Indexes start at 0, and None means no node.

Your function gets only head.

 

Example 1:

4 9 1 6 null next random
Input: values = [4,9,1,6], random = [None,3,0,2]
Output: values = [4,9,1,6], random = [None,3,0,2]
Explanation: The dashed blue arrows are the random pointers.
The copy has four new nodes with the same links:
the new 4's random is None,
the new 9's random leads to the new 6,
the new 1's random leads to the new 4,
the new 6's random leads to the new 1.

Example 2:

8 8 null next random
Input: values = [8,8], random = [None,0]
Output: values = [8,8], random = [None,0]
Explanation: Both nodes hold 8.
The second 8's random leads to the first 8.
In the copy, it must lead to the new first 8, not the old one.

Example 3:

7 next random

Input: values = [7], random = [0]
Output: values = [7], random = [0]
Explanation: The only node's random leads to itself.
In the copy, the new node's random leads to the new node.

 

Constraints:

  • 0 to 1,000 nodes.
  • For an empty list (head is None), return None.
  • Each value is from -10,000 to 10,000.
  • Each random is None or leads to a node of the same list.

Note:

Your algorithm should run in O(n) time and use O(1) extra space.


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    PYTHON · LINKED LISTS

    Copy List Random Pointer in Python, explained

    Understanding the Problem

    The core challenge of this problem is to create a deep copy of a linked list where each node has an additional random pointer. The deep copy should be a completely new list with no shared nodes with the original list. This problem is significant in scenarios where data integrity and isolation are crucial, such as in undo operations, cloning complex data structures, etc.

    Approach

    To solve this problem, we can use a three-step approach:

    1. Interweaving the original list with copied nodes: For each node in the original list, create a new node and insert it right after the original node.
    2. Assigning random pointers: Traverse the interwoven list and assign the random pointers for the copied nodes.
    3. Restoring the original list and extracting the copied list: Separate the interwoven list into the original list and the copied list.

    Algorithm

    Let's break down the algorithm step-by-step:

    1. Traverse the original list and create new nodes. Insert each new node right after its corresponding original node.
    2. Traverse the interwoven list and set the random pointers for the new nodes. If the original node's random pointer is not null, set the new node's random pointer to the new node corresponding to the original node's random pointer.
    3. Separate the interwoven list into the original list and the copied list. Restore the original list by skipping the new nodes and extract the copied list by skipping the original nodes.

    Code Implementation

    class Node:
        def __init__(self, val: int, next: 'Node' = None, random: 'Node' = None):
            self.val = val
            self.next = next
            self.random = random
    
    def copyRandomList(head: 'Node') -> 'Node':
        if not head:
            return None
    
        # Step 1: Create new nodes and interweave them with the original nodes
        current = head
        while current:
            new_node = Node(current.val, current.next)
            current.next = new_node
            current = new_node.next
    
        # Step 2: Assign random pointers for the new nodes
        current = head
        while current:
            if current.random:
                current.next.random = current.random.next
            current = current.next.next
    
        # Step 3: Restore the original list and extract the copied list
        current = head
        new_head = head.next
        while current:
            new_node = current.next
            current.next = new_node.next
            if new_node.next:
                new_node.next = new_node.next.next
            current = current.next
    
        return new_head
    

    Complexity Analysis

    The time complexity of this algorithm is O(n) because we traverse the list a constant number of times. The space complexity is O(1) extra space because we are not using any additional data structures that grow with the input size.

    Edge Cases

    Consider the following edge cases:

    • An empty list (head is null): The function should return null.
    • A list with only one node: Ensure the random pointer is correctly assigned.
    • Random pointers that point to null: Ensure these are correctly handled.

    Testing

    To test the solution comprehensively, consider the following test cases:

    def print_list(head):
        result = []
        while head:
            random_index = None if not head.random else head.random.val
            result.append([head.val, random_index])
            head = head.next
        return result
    
    # Test case 1
    head = Node(7)
    node1 = Node(13)
    node2 = Node(11)
    node3 = Node(10)
    node4 = Node(1)
    head.next = node1
    node1.next = node2
    node2.next = node3
    node3.next = node4
    node1.random = head
    node2.random = node4
    node3.random = node2
    node4.random = head
    
    copied_head = copyRandomList(head)
    print(print_list(copied_head))  # Expected: [[7, None], [13, 7], [11, 1], [10, 11], [1, 7]]
    
    # Test case 2
    head = Node(1)
    node1 = Node(2)
    head.next = node1
    head.random = node1
    node1.random = node1
    
    copied_head = copyRandomList(head)
    print(print_list(copied_head))  # Expected: [[1, 2], [2, 2]]
    
    # Test case 3
    head = Node(3)
    node1 = Node(3)
    node2 = Node(3)
    head.next = node1
    node1.next = node2
    node1.random = head
    
    copied_head = copyRandomList(head)
    print(print_list(copied_head))  # Expected: [[3, None], [3, 3], [3, None]]
    
    # Test case 4
    head = None
    copied_head = copyRandomList(head)
    print(print_list(copied_head))  # Expected: []
    

    Thinking and Problem-Solving Tips

    When approaching such problems, consider the following tips:

    • Break down the problem into smaller, manageable steps.
    • Think about edge cases and how to handle them.
    • Write clean, readable code with comments to explain your thought process.
    • Practice similar problems to improve your problem-solving skills.

    Conclusion

    In this blog post, we discussed how to solve the problem of copying a linked list with random pointers. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong problem-solving skills and improving your coding abilities.

    Additional Resources

    For further reading and practice, consider the following resources:

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