Probability of finding an element K in a Singly Linked List

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: 0
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 with 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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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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Output:

0.2

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