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:
- Count the number of nodes by traversing the list.
- 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 = Noneclass 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 codellist = 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 3rd last node and other nodes. Please refer complete article on Select a Random Node from a Singly Linked List for more details!
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