Javascript 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:
- Count the number of nodes by traversing the list.
- 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, probability 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
result = head->key
(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.
// Javascript program to select a random
// node from singly linked list
// Node Class
class Node {
constructor(d) {
this.data = d;
this.next = null;
}
}
// A reservoir sampling-based function
// to print a random node from a
// linked list
function printrandom(node) {
// If list is empty
if (node == null) {
return;
}
// Use a different seed value so
// that we don't get same result
// each time we run this program
// Math.abs(UUID.randomUUID().
// getMostSignificantBits());
// Initialize result as first node
let result = node.data;
// Iterate from the (k+1)th element
// to nth element
let current = node;
let n;
for (n = 2; current != null; n++) {
// Change result with probability 1/n
if (Math.floor(Math.random() * n) == 0) {
result = current.data;
}
// Move to next node
current = current.next;
}
console.log("Randomly selected key is " + result);
}
// Driver code
head = new Node(5);
head.next = new Node(20);
head.next.next = new Node(4);
head.next.next.next = new Node(3);
head.next.next.next.next = new Node(30);
printrandom(head);
// This code is contributed by rag2127
Output
Randomly selected key is 3
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 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!
