Select a Random Node from a Singly Linked List - GeeksforGeeks

Select a Random Node from a Singly Linked List - GeeksforGeeks
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).

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

Reservoir Sampling

(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

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.
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);

}

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