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:

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.

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;
}

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 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!