# C Programming – Select a Random Node from a Singly Linked List

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.

Given a singly linked list, select a random node from linked list (the probability of picking a node should be 1/N if there are N nodes in list). You are given a random number generator.

Below is a Simple Solution
1) Count 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 node only if generated number is equal to 0 (or any other fixed number from 0 to N-i).

We get uniform probabilities with 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.
(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 above algorithm.

C Program
``````/* C program to randomly select a node from a singly
#include<stdio.h>
#include<stdlib.h>
#include <time.h>

struct node
{
int key;
struct node* next;
};

// A reservoir sampling based function to print a
// random node from a linked list
{
// IF list is empty
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

// Iterate from the (k+1)th element to nth element
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\n", result);
}

/* BELOW FUNCTIONS ARE JUST UTILITY TO TEST  */

/* 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
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 */

/* move the head to point to the new node */
}

// Driver program to test above functions
int main()
{

return 0;
}``````

Note that the above program is based on outcome of a random function and may produce different output.

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How does this work?
Let there be total N nodes in list. It is easier to understand from last node.

The probability that 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 last node as 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. #### Wikitechy Editor

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