Python Program For Selecting A Random Node From A Singly Linked List

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

1. Count the 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 only if the generated number is equal to 0 (or any other fixed number from 0 to N-i).

We get uniform probabilities with the 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.
    (a) Generate a random number from 0 to n-1. 
        Let the generated random number is j.
    (b) If j is equal to 0 (we could choose other fixed numbers 
        between 0 to n-1), then replace result with the current node.
    (c) n = n+1
    (d) current = current->next

Below is the implementation of above algorithm.

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
    # linked list
    def printRandom(self):

        # If list is empty
        if self.head is None:
            return

        if self.head and not self.head.next:
           print "Randomly selected key is %d" %(self.head.data)

        # 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
        # because we iterate from (k+1)th element, or
        # the first node will be picked more easily
        current = self.head.next
        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 code
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)

Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of nodes in the linked list.
Auxiliary Space: O(1), as we are not using any extra space.

Note that the above program is based on the 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 the last node.
The probability that the 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 the 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 3^rd last node and other nodes. Please refer complete article on Select a Random Node from a Singly Linked List for more details!