# 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
(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.

## C

```/* C program to randomly select a node from a singly
#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\n", 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 */

/* move the head to point to the new node */
}

// Driver program to test above functions
int main()
{
struct Node *head = NULL;

return 0;
}
```

## Java

```// Java program to select a random node from singly linked list

import java.util.*;

// Linked List Class

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) {

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);

}
}

// This code has been contributed by Mayank Jaiswal

```

## Python

```
# Python program to randomly select a node from singly

import random

# Node class
class Node:

# Constructor to initialize the node object
def __init__(self, data):
self.data= data
self.next = None

# Function to initialize head
def __init__(self):

# 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

# Iterate from the (k+1)th element nth element
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)

# Utility function to print the linked LinkedList
def printList(self):
while(temp):
print temp.data,
temp = temp.next

# Driver program to test above function
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.

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