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Probability of finding an element K in a Singly Linked List

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  • Last Updated : 21 Feb, 2022
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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




// 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 number 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;
}

Java




// 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 number 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 construct below 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

Python3




# 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    

C#




// 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 number 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 construct below 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

Javascript




<script>
      // JavaScript code to find the probability
      // of finding an Element
      // in a Singly Linked List
      // Link list node
      class Node {
        constructor() {
          this.data = 0;
          this.next = null;
        }
      }
  
      // 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.
      function push(head_ref, new_data) {
        // Allocate node
        var 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 number of nodes
      // in linked list
      function getCount(head) {
        // Initialize count
        var count = 0;
  
        // Initialize current
        var current = head;
  
        while (current != null) {
          count++;
          current = current.next;
        }
        return count;
      }
  
      function kPresentProbability(head, n, k) {
        // Initialize count
        var count = 0;
  
        // Initialize current
        var current = head;
  
        while (current != null) {
          if (current.data == k) count++;
          current = current.next;
        }
        return count / n;
      }
  
      // Driver Code
      // Start with the empty list
      var head = null;
  
      // Use push() to construct below 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);
  
      document.write(kPresentProbability(head, getCount(head), 5).toFixed(1));
        
      // This code is contributed by rdtank.
    </script>

Output

0.2

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