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# Select a Random Node from a Singly Linked List

• Difficulty Level : Hard
• Last Updated : 10 Jan, 2023

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 a probability of 1/N. The selection can be done by generating a random number from 0 to N-i for the 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, the probability of other selecting other nodes is 1/N
The above solution requires two traversals of the 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 the 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 the above algorithm.

## C++

 `/* C++ program to randomly select a node from a singly ` `linked list */` `#include ` `#include ` `#include ` `#include ` `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 of 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 `

## C

 `/* C program to randomly select a node from a singly ` `   ``linked list */` `#include ` `#include ` `#include ` ` `  `/* 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 of 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 `

## Python3

 `# 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 LinkedList ` `    ``def` `printList(``self``): ` `        ``temp ``=` `self``.head ` `        ``while``(temp): ` `            ``print``(temp.data,end``=``" "``) ` `            ``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)`

## 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 `

## Javascript

 ` `

Output

```Randomly selected key is
3```

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 outputs.

How does this work?
Let there be total N nodes in the list. It is easier to understand from the last node.
The probability that the last node is result simply 1/N [For the last or N’th node, we generate a random number between 0 to N-1 and make the last node as the result if the generated number is 0 (or any other fixed number]
The probability that the second last node is the 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 the probability for 3rd last node and other nodes.

My Personal Notes arrow_drop_up
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