# C 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:

- 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.

## C

`/* C program to randomly select a node ` ` ` `from a singly linked list */` `#include<stdio.h> ` `#include<stdlib.h> ` `#include <time.h> ` ` ` `// Linked 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"` `, ` ` ` `result); ` `} ` ` ` `/* 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 code ` `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; ` `} ` |

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 resulting 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 a 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 the 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!