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.
C++
/* C++ program to randomly select a node from a singly linked list */#include<stdio.h> #include<stdlib.h> #include <time.h> #include<iostream> using namespace std; /* Link list node */class Node { public: int key; Node* next; void printRandom(Node*); void push(Node**, int); }; // A reservoir sampling based function to print a // random node from a linked list void Node::printRandom(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 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; } cout<<"Randomly selected key is \n"<< result; } /* BELOW FUNCTIONS ARE JUST UTILITY TO TEST */ /* A utility function to create a new node */Node* newNode(int new_key) { // allocate node Node* new_node = (Node*) malloc(sizeof( 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 Node:: push(Node** head_ref, int new_key) { /* allocate node */ 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() { Node n1; Node *head = NULL; n1.push(&head, 5); n1.push(&head, 20); n1.push(&head, 4); n1.push(&head, 3); n1.push(&head, 30); n1.printRandom(head); return 0; } // This code is contributed by SoumikMondal |
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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\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 */ 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; } |
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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 |
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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 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 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) |
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C#
// C# program to select a random node // from singly linked list using System; // Linked List Class public class LinkedList { Node head; // head of list /* Node Class */ public class Node { public int data; public Node next; // Constructor to create a new node public 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 (new Random().Next() % n == 0) { result = current.data; } // Move to next node current = current.next; } Console.WriteLine("Randomly selected key is " + result); } // Driver Code 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(list.head); } } // This code is contributed by 29AjayKumar |
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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 contributed by Rajeev. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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