Select 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 a probability of 1/N. The selection can be done by generating a random number from 0 to N-i for the 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, the probability of other selecting other nodes is 1/N
The above solution requires two traversals of the 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 the 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.
 (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

Below is the implementation of the above algorithm. 

Randomly selected key is 
4

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 outputs.

How does this work? 
Let there be total N nodes in the list. It is easier to understand from the last node.
The probability that the last node is result simply 1/N [For the last or N'th node, we generate a random number between 0 to N-1 and make the last node as the result if the generated number is 0 (or any other fixed number]
The probability that the second last node is the 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 the probability for 3^rd last node and other nodes.

Another approach Using rand() Function:

Here in this Approach, we convert linked list to vector by storing every node value and than we apply rand() function on them and return the random node value.

Approach/Intuition:

   here given linked list :

•    5 -> 20 -> 4 -> 3 -> 30.
•    we traverse over linked list and convert it into vector.
•    vector<int>v{5,20,4,3,30};
•    than we use rand() function.
•    int n=v.size() //size of the vector.
•    int RandomIndex=rand() % n;
•    and at the end we will return random node value from singly linked list.

Below is the code to implement the above approach:

20

Complexity Analysis:
Time Complexity: O(n).
Space Complexity:O(n).