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

• Last Updated : 28 Dec, 2021

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 ` `#include ` `using` `namespace` `std; ` ` `  `// Link list node  ` `class` `Node ` `{ ` `    ``public``: ` `    ``int` `key; ` `    ``Node* next; ` `    ``void` `printRandom(Node*); ` `    ``void` `push(Node**, ``int``); ` `     `  `}; ` ` `  `// A reservoir sampling-based function to  ` `// print a random node from a linked list ` `void` `Node::printRandom(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 ` `    ``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; ` `    ``} ` ` `  `    ``cout << ``"Randomly selected key is "` `<<  ` `             ``result; ` `} ` ` `  `/* A utility function to create  ` `   ``a new node */` `Node* newNode(``int` `new_key) ` `{ ` `    ``// Allocate node  ` `    ``Node* new_node =  ` `          ``(Node*) ``malloc``(``sizeof``( 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` `Node:: push(Node** head_ref,  ` `                 ``int` `new_key) ` `{ ` `    ``// Allocate node  ` `    ``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() ` `{ ` `    ``Node n1; ` `    ``Node *head = NULL; ` `    ``n1.push(&head, 5); ` `    ``n1.push(&head, 20); ` `    ``n1.push(&head, 4); ` `    ``n1.push(&head, 3); ` `    ``n1.push(&head, 30); ` `    ``n1.printRandom(head); ` `    ``return` `0; ` `} ` `// This code is contributed by SoumikMondal `

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 result 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 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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