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Select a Random Node from a Singly Linked List

Given a singly linked list, select a random node from linked list (the probability of picking a node should be 1/N if there are N nodes in list). You are given a random number generator.

Below is a Simple Solution
1) Count number of nodes by traversing the list.
2) Traverse the list again and select every node with probability 1/N. The selection can be done by generating a random number from 0 to N-i for i’th node, and selecting the i’th node node only if generated number is equal to 0 (or any other fixed number from 0 to N-i).

We get uniform probabilities with above schemes.

i = 1, probability of selecting first node = 1/N
i = 2, probability of selecting second node =
                   [probability that first node is not selected] * 
                   [probability that second node is selected]
                  = ((N-1)/N)* 1/(N-1)
                  = 1/N  

Similarly, probabilities of other selecting other nodes is 1/N

The above solution requires two traversals of linked list.



How to select a random node with only one traversal allowed?
The idea is to use Reservoir Sampling. Following are the steps. This is a simpler version of Reservoir Sampling as we need to select only one key instead of k keys.

(1) Initialize result as first node
   result = head->key 
(2) Initialize n = 2
(3) Now one by one consider all nodes from 2nd node onward.
    (3.a) Generate a random number from 0 to n-1. 
         Let the generated random number is j.
    (3.b) If j is equal to 0 (we could choose other fixed number 
          between 0 to n-1), then replace result with current node.
    (3.c) n = n+1
    (3.d) current = current->next

Below is the implementation of above algorithm.

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C

/* C program to randomly select a node from a singly
   linked list */
#include<stdio.h>
#include<stdlib.h>
#include <time.h>
  
/* Link list node */
struct Node
{
    int key;
    struct Node* next;
};
  
// A reservoir sampling based function to print a
// random node from a linked list
void printRandom(struct Node *head)
{
    // IF list is empty
    if (head == NULL)
       return;
  
    // Use a different seed value so that we don't get
    // same result each time we run this program
    srand(time(NULL));
  
    // Initialize result as first node
    int result = head->key;
  
    // Iterate from the (k+1)th element to nth element
    struct Node *current = head;
    int n;
    for (n=2; current!=NULL; n++)
    {
        // change result with probability 1/n
        if (rand() % n == 0)
           result = current->key;
  
        // Move to next node
        current = current->next;
    }
  
    printf("Randomly selected key is %d ", result);
}
  
/* BELOW FUNCTIONS ARE JUST UTILITY TO TEST  */
  
/* A utility function to create a new node */
struct Node *newNode(int new_key)
{
    /* allocate node */
    struct Node* new_node =
        (struct Node*) malloc(sizeof(struct Node));
  
    /* put in the key  */
    new_node->key  = new_key;
    new_node->next =  NULL;
  
    return new_node;
}
  
  
/* A utility function to insert a node at the beginning
  of linked list */
void push(struct Node** head_ref, int new_key)
{
    /* allocate node */
    struct Node* new_node = new Node;
  
    /* put in the key  */
    new_node->key  = new_key;
  
    /* link the old list off the new node */
    new_node->next = (*head_ref);
  
    /* move the head to point to the new node */
    (*head_ref)    = new_node;
}
  
  
// Driver program to test above functions
int main()
{
    struct Node *head = NULL;
    push(&head, 5);
    push(&head, 20);
    push(&head, 4);
    push(&head, 3);
    push(&head, 30);
  
    printRandom(head);
  
    return 0;
}

Java

// Java program to select a random node from singly linked list
  
import java.util.*;
  
// Linked List Class
class LinkedList {
  
    static Node head;  // head of list
  
    /* Node Class */
    static class Node {
  
        int data;
        Node next;
  
        // Constructor to create a new node
        Node(int d) {
            data = d;
            next = null;
        }
    }
  
    // A reservoir sampling based function to print a
    // random node from a linked list
    void printrandom(Node node) {
  
        // If list is empty
        if (node == null) {
            return;
        }
  
        // Use a different seed value so that we don't get
        // same result each time we run this program
        Math.abs(UUID.randomUUID().getMostSignificantBits());
  
        // Initialize result as first node
        int result = node.data;
  
        // Iterate from the (k+1)th element to nth element
        Node current = node;
        int n;
        for (n = 2; current != null; n++) {
  
            // change result with probability 1/n
            if (Math.random() % n == 0) {
                result = current.data;
            }
  
            // Move to next node
            current = current.next;
        }
  
        System.out.println("Randomly selected key is " + result);
    }
  
    // Driver program to test above functions
    public static void main(String[] args) {
  
        LinkedList list = new LinkedList();
        list.head = new Node(5);
        list.head.next = new Node(20);
        list.head.next.next = new Node(4);
        list.head.next.next.next = new Node(3);
        list.head.next.next.next.next = new Node(30);
  
        list.printrandom(head);
  
    }
}
  
// This code has been contributed by Mayank Jaiswal

Python

# Python program to randomly select a node from singly
# linked list 
  
import random
  
# Node class 
class Node:
  
    # Constructor to initialize the node object
    def __init__(self, data):
        self.data= data
        self.next = None
  
class LinkedList:
  
    # Function to initialize head
    def __init__(self):
        self.head = None
  
    # A reservoir sampling based function to print a
    # random node from a linkd list
    def printRandom(self):
  
        # If list is empty 
        if self.head is None:
            return 
  
        # Use a different seed value so that we don't get 
        # same result each time we run this program
        random.seed()
  
        # Initialize result as first node
        result = self.head.data
  
        # Iterate from the (k+1)th element nth element
        current = self.head 
        n = 2 
        while(current is not None):
              
            # change result with probability 1/n
            if (random.randrange(n) == 0 ):
                result = current.data 
  
            # Move to next node
            current = current.next
            n += 1
  
        print "Randomly selected key is %d" %(result)
          
    # Function to insert a new node at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node
  
    # Utility function to print the linked LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print temp.data,
            temp = temp.next
  
  
# Driver program to test above function
llist = LinkedList()
llist.push(5)
llist.push(20)
llist.push(4)
llist.push(3)
llist.push(30)
llist.printRandom()
  
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

Note that the above program is based on outcome of a random function and may produce different output.

How does this work?
Let there be total N nodes in list. It is easier to understand from last node.

The probability that last node is result simply 1/N [For last or N’th node, we generate a random number between 0 to N-1 and make last node as result if the generated number is 0 (or any other fixed number]

The probability that second last node is result should also be 1/N.

The probability that the second last node is result 
          = [Probability that the second last node replaces result] X 
            [Probability that the last node doesn't replace the result] 
          = [1 / (N-1)] * [(N-1)/N]
          = 1/N

Similarly we can show probability for 3rd last node and other nodes.



This article is attributed to GeeksforGeeks.org

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