 Open in App
Not now

# C Program For Selecting A Random Node From A Singly Linked List

• Last Updated : 21 Jul, 2022

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!

My Personal Notes arrow_drop_up
Related Articles