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Probability of finding an element K in a Singly Linked List
  • Last Updated : 09 Oct, 2020

Given a Singly Linked List of size N and another key K, we have to find the probability that the key K is present in the Singly Linked List.
Examples: 

Input: Linked list = 2 -> 3 -> 3 -> 3 -> 4 -> 2, Key = 5 
Output:
Explanation: 
Since the value of Key is 5 which is not present in List, the probability of finding the Key in the Linked List is 0.
Input: Linked list = 2 -> 3 -> 5 -> 1 -> 9 -> 8 -> 0 -> 7 -> 6 -> 5, Key = 5 
Output: 0.2 
 

Approach: 
The probability of finding a Key element K in a Singly Linked List is given below:  

Probability = Number of Occurrences of Element K / Size of the Linked List 

In our approach, we will first count the number of Element K present in the Singly Linked List and then the probability will be calculated by dividing the number of occurrences of K by the size of the Singly Linked List.
Below is the implementation of the above approach:

C

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// C code to find the probability
// of finding an Element
// in a Singly Linked List
 
#include <stdio.h>
#include <stdlib.h>
 
// Link list node
struct Node {
    int data;
    struct Node* next;
};
 
/* Given a reference (pointer to pointer)
   to the head of a list and an int,
   push a new node on the front of the list. */
void push(struct Node** head_ref, int new_data)
{
    // allocate node
    struct Node* new_node
        = (struct Node*)malloc(
            sizeof(struct Node));
 
    // put in the data
    new_node->data = new_data;
 
    // 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;
}
 
// Counts nnumber of nodes in linked list
int getCount(struct Node* head)
{
 
    // Initialize count
    int count = 0;
 
    // Initialize current
    struct Node* current = head;
 
    while (current != NULL) {
        count++;
        current = current->next;
    }
    return count;
}
 
float kPresentProbability(
    struct Node* head,
    int n, int k)
{
 
    // Initialize count
    float count = 0;
 
    // Initialize current
    struct Node* current = head;
 
    while (current != NULL) {
        if (current->data == k)
            count++;
        current = current->next;
    }
    return count / n;
}
 
// Driver Code
int main()
{
    // Start with the empty list
    struct Node* head = NULL;
 
    // Use push() to construct below list
    // 1->2->1->3->1
    push(&head, 2);
    push(&head, 3);
    push(&head, 5);
    push(&head, 1);
    push(&head, 9);
    push(&head, 8);
    push(&head, 0);
    push(&head, 7);
    push(&head, 6);
    push(&head, 5);
 
    printf("%.1f",
           kPresentProbability(
               head, getCount(head), 5));
 
    return 0;
}

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Java

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// Java code to find the probability
// of finding an Element
// in a Singly Linked List
class GFG{
 
// Link list node
static class Node
{
  int data;
  Node next;
};
 
// Given a reference (pointer to
// pointer) to the head of a list
// and an int, push a new node
// on the front of the list.
static Node push(Node head_ref,
                 int new_data)
{
  // Allocate node
  Node new_node = new Node();
 
  // Put in the data
  new_node.data = new_data;
 
  // 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;
   
  return head_ref;
}
 
// Counts nnumber of nodes
// in linked list
static int getCount(Node head)
{
  // Initialize count
  int count = 0;
 
  // Initialize current
  Node current = head;
 
  while (current != null)
  {
    count++;
    current = current.next;
  }
  return count;
}
 
static float kPresentProbability(Node head,
                                 int n, int k)
{
  // Initialize count
  float count = 0;
 
  // Initialize current
  Node current = head;
 
  while (current != null)
  {
    if (current.data == k)
      count++;
    current = current.next;
  }
  return count / n;
}
 
// Driver Code
public static void main(String[] args)
{
  // Start with the empty list
  Node head = null;
 
  // Use push() to conbelow list
  // 1.2.1.3.1
  head = push(head, 2);
  head = push(head, 3);
  head = push(head, 5);
  head = push(head, 1);
  head = push(head, 9);
  head = push(head, 8);
  head = push(head, 0);
  head = push(head, 7);
  head = push(head, 6);
  head = push(head, 5);
 
  System.out.printf("%.1f", kPresentProbability(
                     head, getCount(head), 5));
 
}
}
 
// This code is contributed by shikhasingrajput

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Python3

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# Python3 code to find the probability
# of finding an Element
# in a Singly Linked List
 
# Node class
class Node:
     
    def __init__(self, data, next = None):
         
        self.data = data
        self.next = None
 
class LinkedList:
     
    def __init__(self):
         
        self.head = None
     
    def push(self, data):
         
        # Allocate the Node &
        # put the data
        new_node = Node(data)
         
        # Make the next of new Node as head
        new_node.next = self.head
         
        # Move the head to point to new Node
        self.head = new_node
 
    # Counts the number of nodes in linkedlist
    def getCount(self):
         
        # Initialize current
        current = self.head
         
        # Initialize count
        count = 0
         
        while current is not None:
            count += 1
            current = current.next
         
        return count
     
    def kPresentProbability(self, n, k):
         
        # Initialize current
        current = self.head
         
        # Initialize count
        count = 0.0
         
        while current is not None:
            if current.data == k:
                count += 1
                 
            current = current.next
         
        return count / n
     
# Driver Code
if __name__ == "__main__":
     
    # Start with empty list
    llist = LinkedList()
     
    # Use push to construct the linked list
    llist.push(2)
    llist.push(3)
    llist.push(5)
    llist.push(1)
    llist.push(9)
    llist.push(8)
    llist.push(0)
    llist.push(7)
    llist.push(6)
    llist.push(5)
     
    print(llist.kPresentProbability(
          llist.getCount(), 5))
     
# This code is contributed by kevalshah5   

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C#

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// C# code to find the probability
// of finding an Element
// in a Singly Linked List
using System;
class GFG{
 
// Link list node
class Node
{
  public int data;
  public Node next;
};
 
// Given a reference (pointer to
// pointer) to the head of a list
// and an int, push a new node
// on the front of the list.
static Node push(Node head_ref,
                 int new_data)
{
  // Allocate node
  Node new_node = new Node();
 
  // Put in the data
  new_node.data = new_data;
 
  // 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;
   
  return head_ref;
}
 
// Counts nnumber of nodes
// in linked list
static int getCount(Node head)
{
  // Initialize count
  int count = 0;
 
  // Initialize current
  Node current = head;
 
  while (current != null)
  {
    count++;
    current = current.next;
  }
  return count;
}
 
static float kPresentProbability(Node head,
                                 int n, int k)
{
  // Initialize count
  float count = 0;
 
  // Initialize current
  Node current = head;
 
  while (current != null)
  {
    if (current.data == k)
      count++;
    current = current.next;
  }
  return count / n;
}
 
// Driver Code
public static void Main(String[] args)
{
  // Start with the empty list
  Node head = null;
 
  // Use push() to conbelow list
  // 1.2.1.3.1
  head = push(head, 2);
  head = push(head, 3);
  head = push(head, 5);
  head = push(head, 1);
  head = push(head, 9);
  head = push(head, 8);
  head = push(head, 0);
  head = push(head, 7);
  head = push(head, 6);
  head = push(head, 5);
 
  Console.Write("{0:F1}", kPresentProbability(
                 head, getCount(head), 5));
}
}
 
// This code is contributed by Princi Singh

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Output

0.2

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