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

Last Updated : 17 Aug, 2023

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> 
#include<iostream> 
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.

The algorithm implemented in the code is called “Reservoir Sampling”. It is a technique for selecting a random item from a large data stream with equal probability.

Here’s an algorithm to randomly select a node from a singly linked list in C++:

1.Calculate the length of the linked list, let’s call it n.
2.Generate a random number in the range [0, n-1]. This can be done using the rand() function in C++.
3.Start at the head of the linked list and traverse the linked list until the randomly generated index is reached.
4.Return the node at the randomly generated index.
Here’s some sample code to implement the algorithm in C++:

C++

#include <iostream> 
#include <cstdlib> 
#include <ctime> 
  
struct Node { 
  int data; 
  Node* next; 
}; 
  
int getLength(Node* head) { 
  int count = 0; 
  Node* current = head; 
  while (current != nullptr) { 
    count++; 
    current = current->next; 
  } 
  return count; 
} 
  
Node* getRandomNode(Node* head) { 
  int length = getLength(head); 
  int randomIndex = rand() % length; 
  Node* current = head; 
  for (int i = 0; i < randomIndex; i++) { 
    current = current->next; 
  } 
  return current; 
} 
  
int main() { 
  srand(time(0)); 
  
  Node* head = new Node{1, nullptr}; 
  head->next = new Node{2, nullptr}; 
  head->next->next = new Node{3, nullptr}; 
  head->next->next->next = new Node{4, nullptr}; 
  
  Node* randomNode = getRandomNode(head); 
  std::cout << "Random node: " << randomNode->data << std::endl; 
  
  return 0; 
} 

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 3^rd last node and other nodes. Please refer complete article on Select a Random Node from a Singly Linked List for more details!

Suggest improvement

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