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# Python Program For Selecting A Random Node From A Singly Linked List

• Last Updated : 21 Jul, 2022

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 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
(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.

## Python

 `# Python program to randomly select a ` `# node from singly linked list ` `import` `random`   `# Node class ` `class` `Node:`   `    ``# Constructor to initialize the ` `    ``# node object` `    ``def` `__init__(``self``, data):` `        ``self``.data``=` `data` `        ``self``.``next` `=` `None`   `class` `LinkedList:`   `    ``# Function to initialize head` `    ``def` `__init__(``self``):` `        ``self``.head ``=` `None`   `    ``# A reservoir sampling-based function ` `    ``# to print a random node from a ` `    ``# linked list` `    ``def` `printRandom(``self``):`   `        ``# If list is empty ` `        ``if` `self``.head ``is` `None``:` `            ``return`   `        ``if` `self``.head ``and` `not` `self``.head.``next``:` `           ``print` `"Randomly selected key is %d"` `%``(``self``.head.data)`   `        ``# Use a different seed value so that we don't get ` `        ``# same result each time we run this program` `        ``random.seed()`   `        ``# Initialize result as first node` `        ``result ``=` `self``.head.data`   `        ``# Iterate from the (k+1)th element nth element` `        ``# because we iterate from (k+1)th element, or ` `        ``# the first node will be picked more easily ` `        ``current ``=` `self``.head.``next` `        ``n ``=` `2` `        ``while``(current ``is` `not` `None``):` `            `  `            ``# Change result with probability 1/n` `            ``if` `(random.randrange(n) ``=``=` `0` `):` `                ``result ``=` `current.data `   `            ``# Move to next node` `            ``current ``=` `current.``next` `            ``n ``+``=` `1`   `        ``print` `"Randomly selected key is %d"` `%``(result)` `        `  `    ``# Function to insert a new node at ` `    ``# the beginning` `    ``def` `push(``self``, new_data):` `        ``new_node ``=` `Node(new_data)` `        ``new_node.``next` `=` `self``.head` `        ``self``.head ``=` `new_node`   `    ``# Utility function to print the linked ` `    ``# LinkedList` `    ``def` `printList(``self``):` `        ``temp ``=` `self``.head` `        ``while``(temp):` `            ``print` `temp.data,` `            ``temp ``=` `temp.``next`   `# Driver code` `llist ``=` `LinkedList()` `llist.push(``5``)` `llist.push(``20``)` `llist.push(``4``)` `llist.push(``3``)` `llist.push(``30``)` `llist.printRandom()` `# This code is contributed by Nikhil Kumar Singh(nickzuck_007)`

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 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!

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