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, 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
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.
C
/* C program to randomly select a node from a singly linked list */#include<stdio.h>#include<stdlib.h>#include <time.h>// Linked list node struct Node{ int key; struct Node* next;};// A reservoir sampling-based function // to print a random node from a // linked listvoid 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 codeint 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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