Copy List with Random Pointer

Copy List with Random Pointer - 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.

Input & Output

Example 1 — Basic List with Random Pointers

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

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

💡 Note: Node 0 (value 7) has no random pointer. Node 1 (value 13) points to node 0. Node 2 (value 11) points to node 4. Node 3 (value 10) points to node 2. Node 4 (value 1) points to node 0.

Example 2 — Simple Two-Node List

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

› Output: [[1,1],[2,1]]

💡 Note: Both nodes point to the second node (index 1). The deep copy preserves these relationships.

Example 3 — Empty List

$ Input: head = []

› Output: []

💡 Note: Empty list returns empty list.

Constraints

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

Visualization

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Asked in

A Amazon 45 M Microsoft 38 G Google 32 F Facebook 28

The key insight is maintaining a mapping between original and copied nodes to preserve pointer relationships. The two-pass hash map approach is most intuitive while the interweaving approach achieves O(1) space. Time: O(n), Space: O(1) for optimal solution.

Common Approaches

✓ Interweaving Approach

⏱️ Time: O(n) Space: O(1)

Insert each new node right after its corresponding original node, set random pointers, then extract the new list.

Two-Pass Hash Map

⏱️ Time: O(n) Space: O(n)

First pass creates all new nodes and maps them. Second pass connects next and random pointers using the mapping.

Interweaving Approach — Algorithm Steps

Insert new nodes after each original node
Set random pointers using the interweaved structure
Extract and separate the new list

Visualization

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Step-by-Step Walkthrough

Interweave

Insert new nodes between original nodes

Connect Random

Set random pointers using interweaved structure

Separate

Extract the new list while restoring original

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct Node {
    int val;
    struct Node* next;
    struct Node* random;
};

struct Node* solution(struct Node* head) {
    if (!head) return NULL;
    
    // Step 1: Create new nodes and interweave with original
    struct Node* current = head;
    while (current) {
        struct Node* newNode = malloc(sizeof(struct Node));
        newNode->val = current->val;
        newNode->next = current->next;
        newNode->random = NULL;
        current->next = newNode;
        current = newNode->next;
    }
    
    // Step 2: Set random pointers for new nodes
    current = head;
    while (current) {
        if (current->random) {
            current->next->random = current->random->next;
        }
        current = current->next->next;
    }
    
    // Step 3: Separate the two lists
    struct Node* newHead = head->next;
    struct Node* currentOld = head;
    struct Node* currentNew = newHead;
    
    while (currentOld) {
        currentOld->next = currentOld->next ? currentOld->next->next : NULL;
        currentNew->next = currentNew->next ? currentNew->next->next : NULL;
        currentOld = currentOld->next;
        currentNew = currentNew->next;
    }
    
    return newHead;
}

struct Node* parseAndCreateList() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing - for production would need robust JSON parser
    if (strcmp(line, "[]\n") == 0) {
        return NULL;
    }
    
    // For demo, create a simple test case
    struct Node* node1 = malloc(sizeof(struct Node));
    struct Node* node2 = malloc(sizeof(struct Node));
    node1->val = 7;
    node1->next = node2;
    node1->random = NULL;
    node2->val = 13;
    node2->next = NULL;
    node2->random = node1;
    
    return node1;
}

void printList(struct Node* head) {
    if (!head) {
        printf("[]\n");
        return;
    }
    
    // Simple output for demo
    printf("[[7,null],[13,0]]\n");
}

int main() {
    struct Node* head = parseAndCreateList();
    struct Node* result = solution(head);
    printList(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Three passes through the list, each taking O(n) time

✓ Linear Growth

Space Complexity

O(1)

Only creates the required new nodes, no extra hash map

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Interweaving Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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