Linked List Random Node - Practice Coding Problems

Linked List Random Node - Problem

Given a singly linked list, return a random node's value from the linked list. Each node must have the same probability of being chosen.

Implement the Solution class:

Solution(ListNode head) Initializes the object with the head of the singly-linked list head.
int getRandom() Chooses a node randomly from the list and returns its value. All the nodes of the list should be equally likely to be chosen.

Input & Output

Example 1 — Basic Linked List

$ Input: head = [1,2,3]

› Output: 1 (or 2 or 3)

💡 Note: Each call to getRandom() should return one of the values {1, 2, 3} with equal probability 1/3

Example 2 — Single Node

$ Input: head = [5]

› Output: 5

💡 Note: Only one node exists, so getRandom() always returns 5 with probability 1

Example 3 — Larger List

$ Input: head = [10,20,30,40,50]

› Output: 30 (or any value from list)

💡 Note: Each value has equal 1/5 probability of being selected

Constraints

The number of nodes in the linked list will be in the range [1, 10⁴]
-10⁴ ≤ Node.val ≤ 10⁴
At most 10⁴ calls will be made to getRandom

Visualization

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Understanding the Visualization

Input

Singly linked list with unknown length

Process

Apply reservoir sampling or two-pass selection

Output

Return random node value with uniform distribution

Key Takeaway

🎯 Key Insight: Reservoir sampling achieves uniform random selection in one pass without knowing list size

Asked in

G Google 15 f Facebook 12 a Amazon 8 M Microsoft 6

The key insight is using reservoir sampling for uniform random selection in a single pass. The optimal approach uses reservoir sampling with O(n) time and O(1) space, maintaining equal probability without knowing list length upfront.

Common Approaches

Approach	Time	Space	Notes
✓ Two-Pass: Count Then Select	O(n)	O(1)	Count nodes first, then traverse again to find the randomly selected position
Reservoir Sampling	O(n)	O(1)	Single-pass algorithm that maintains uniform probability without knowing list length

Two-Pass: Count Then Select — Algorithm Steps

Step 1: Traverse entire list to count total nodes
Step 2: Generate random index from 0 to count-1
Step 3: Traverse list again to reach the random index

Visualization

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Step-by-Step Walkthrough

Count Pass

Traverse entire list counting nodes

Random Index

Generate random number from 0 to count-1

Select Pass

Traverse again to reach the random position

Code -

solution.c — C

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

struct ListNode {
    int val;
    struct ListNode* next;
};

struct ListNode* createNode(int val) {
    struct ListNode* node = malloc(sizeof(struct ListNode));
    node->val = val;
    node->next = NULL;
    return node;
}

struct ListNode* buildLinkedList(int* arr, int size) {
    if (size == 0) return NULL;
    struct ListNode* head = createNode(arr[0]);
    struct ListNode* current = head;
    for (int i = 1; i < size; i++) {
        current->next = createNode(arr[i]);
        current = current->next;
    }
    return head;
}

int getRandom(struct ListNode* head) {
    // Count total nodes
    int count = 0;
    struct ListNode* current = head;
    while (current) {
        count++;
        current = current->next;
    }
    
    // Generate random index
    int randomIndex = rand() % count;
    
    // Traverse to random position
    current = head;
    for (int i = 0; i < randomIndex; i++) {
        current = current->next;
    }
    
    return current->val;
}

int solution(struct ListNode* head) {
    return getRandom(head);
}

int main() {
    srand(time(NULL));
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int arr[100];
    int size = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL && size < 100) {
        arr[size++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    struct ListNode* head = buildLinkedList(arr, size);
    printf("%d\n", solution(head));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Two passes through the linked list: O(n) + O(n) = O(n)

✓ Linear Growth

Space Complexity

O(1)

Only storing count and current node pointer variables

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Two-Pass: Count Then Select — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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