Copy List with Random Pointer

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:

  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:

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✅ Solution:

1. First Solution: Interleaving Copy (Space Optimized)
2. Second Solution: HashMap-based Copy (with Extra Space)

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✨ First Solution: Space Optimized (No Extra Map)

class Solution {
public:
    // Step 1: Interleave copied nodes into original list
    void getCopyInBetween(Node *head){
        Node * temp = head;
        while(temp != NULL){
            Node *nextNode = temp->next;
            Node *copyNode = new Node(temp->val); // copy current node
            temp->next = copyNode;
            copyNode->next = nextNode;
            temp = nextNode;
        }
    }

    // Step 2: Set random pointers for copied nodes
    void addRandomPointer(Node *head){
        Node *temp = head;
        while(temp != NULL){
            Node *copyNode = temp->next;
            if(temp->random)
                copyNode->random = temp->random->next;
            temp = temp->next->next;
        }
    }

    // Step 3: Separate copied list from original list
    Node* getCopyList(Node *head){
        Node *dummyHead = new Node(0);
        Node *copy = dummyHead;
        Node *temp = head;

        while(temp != NULL){
            copy->next = temp->next;         // Add copied node to result list
            copy = copy->next;
            temp->next = temp->next->next;   // Restore original list
            temp = temp->next;
        }

        return dummyHead->next;
    }

    Node* copyRandomList(Node* head) {
        if(head == NULL) return NULL;

        getCopyInBetween(head);
        addRandomPointer(head);
        return getCopyList(head);
    }
};

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✨ Second Solution: Using Hash Map

class Solution {
public:
    Node* copyRandomList(Node* head) {
        if(head == NULL) return NULL;

        unordered_map<Node*, Node*> nodeMap; // Original -> Copy

        Node *temp = head;

        // Step 1: Create copy nodes and store in map
        while(temp != NULL){
            nodeMap[temp] = new Node(temp->val);
            temp = temp->next;
        }

        // Step 2: Assign next and random pointers using map
        temp = head;
        while(temp != NULL){
            nodeMap[temp]->next = nodeMap[temp->next];
            nodeMap[temp]->random = nodeMap[temp->random];
            temp = temp->next;
        }

        return nodeMap[head];
    }
};

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📝 How It Works

Interleaving Copy (1st approach):

1. Interleave copy nodes right after original ones: A → A' → B → B' → ...
2. Assign random pointers: since A.random → C, then A'.random → C' = A.random.next
3. Separate the copy list and restore the original list.

HashMap Copy (2nd approach):

1. Traverse and create new nodes using a hash map to link original → copied.
2. Use the map to set .next and .random of each copy node.

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🧩 Key Ideas

• Interleaving trick avoids extra space by utilizing the original list structure.
• HashMap approach is easier to implement and debug.

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⏱️ Time & Space Complexity

Approach	Time Complexity	Space Complexity
Interleaved Copy	O(N)	O(1)
Hash Map	O(N)	O(N)

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⚠️ Edge Cases

• head == NULL: empty list
• Nodes where .random is NULL
• All .random pointing to themselves
• Cyclic random pointers (not .next)

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💡 Other Approaches

• Recursion + map (less readable, not preferred)
• Serialization-based clone (used when list is serialized into array form)

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🔁 Related Problems

• LC 138: Copy List with Random Pointer
• LC 148: Sort a linked list
• LC 287: Find Duplicate Number (cycle detection)

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🛠️ Other Notes

• Interleaving is a clever space-saving trick.
• HashMap version is easier for interviews, especially when explaining pointer connections.