Python Program to Find Number of Occurrences of All Elements in a Linked List

When working with linked lists, it's often useful to find the number of occurrences of elements. This implementation demonstrates how to create a linked list and count occurrences of specific elements using a custom count_elem method.

Below is a demonstration for the same ?

Linked List Implementation

First, we create the basic structure with Node and LinkedList classes:

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList_structure:
    def __init__(self):
        self.head = None
        self.last_node = None

    def add_vals(self, data):
        if self.last_node is None:
            self.head = Node(data)
            self.last_node = self.head
        else:
            self.last_node.next = Node(data)
            self.last_node = self.last_node.next

    def print_it(self):
        curr = self.head
        while curr:
            print(curr.data)
            curr = curr.next

    def count_elem(self, key):
        curr = self.head
        count_val = 0
        while curr:
            if curr.data == key:
                count_val = count_val + 1
            curr = curr.next
        return count_val

Example: Counting Element Occurrences

Here's how to use the linked list to count occurrences of a specific element ?

# Create linked list instance
my_instance = LinkedList_structure()
elements = [56, 78, 98, 12, 34, 55, 0, 12, 56]

# Add elements to the linked list
for elem in elements:
    my_instance.add_vals(elem)

print('The linked list is:')
my_instance.print_it()

# Count occurrences of specific elements
search_element = 12
count_val = my_instance.count_elem(search_element)
print(f'{search_element} occurs {count_val} time(s) in the list.')

# Count another element
search_element = 56
count_val = my_instance.count_elem(search_element)
print(f'{search_element} occurs {count_val} time(s) in the list.')

The linked list is:
56
78
98
12
34
55
0
12
56
12 occurs 2 time(s) in the list.
56 occurs 2 time(s) in the list.

Counting All Elements

To find occurrences of all unique elements, we can extend the functionality ?

class LinkedList_structure:
    def __init__(self):
        self.head = None
        self.last_node = None

    def add_vals(self, data):
        if self.last_node is None:
            self.head = Node(data)
            self.last_node = self.head
        else:
            self.last_node.next = Node(data)
            self.last_node = self.last_node.next

    def count_all_elements(self):
        frequency = {}
        curr = self.head
        while curr:
            if curr.data in frequency:
                frequency[curr.data] += 1
            else:
                frequency[curr.data] = 1
            curr = curr.next
        return frequency

# Example usage
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

my_list = LinkedList_structure()
elements = [1, 2, 3, 1, 2, 4, 1]

for elem in elements:
    my_list.add_vals(elem)

frequencies = my_list.count_all_elements()
print("Element frequencies:")
for element, count in frequencies.items():
    print(f"{element}: {count} time(s)")

Element frequencies:
1: 3 time(s)
2: 2 time(s)
3: 1 time(s)
4: 1 time(s)

Key Components

• The Node class represents individual elements with data and next pointer

• The LinkedList_structure class manages the linked list operations

• The add_vals method appends elements to the end of the list

• The count_elem method counts occurrences of a specific element

• The count_all_elements method returns frequency of all unique elements

Conclusion

This implementation provides an efficient way to count element occurrences in a linked list. The count_elem method handles single element counting, while count_all_elements gives a complete frequency analysis of all elements.