# 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
(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``#include``#include ` `// 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 3rd last node and other nodes. Please refer complete article on Select a Random Node from a Singly Linked List for more details!

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