Count frequency of given node in doubly linked list
The problem involves counting the frequency of a given node in a doubly linked list. A doubly linked list is a data structure in which each node contains a data element and two pointers, one pointing to the next node and another pointing to the previous node. The task is to determine how many times a specific data value appears in the list.
Problem Statement
Given a doubly linked list and a target data value, the objective is to count how many times the target value appears in the linked list.
Example
Consider the following doubly linked list:
7 <> 2 <> 3 <> 4 <> 5 <> 3 <> 7 <> 3
If we want to count the frequency of the value 3, the result should be 3, since the value 3 appears three times in the list.
Idea to Solve
To solve this problem, you need to traverse the entire doubly linked list while maintaining a count of how many times the target value appears. During traversal, check if the data value of each node matches the target value. If it does, increment the count. After traversing the entire list, the count will represent the frequency of the target value in the linked list.
Pseudocode
function frequency(key):
result = 0
temp = head
while temp is not null:
if temp.data equals key:
increment result by 1
move temp to the next node
print "Frequency of node " + key + " is : " + result
Algorithm Explanation
 Initialize a variable
result
to store the frequency count and set it to 0.  Start traversing the doubly linked list from the
head
node using atemp
pointer.  While the
temp
pointer is not null: Check if the data value of the current node (
temp.data
) equals the target value (key
).  If they are equal, increment the
result
by 1.  Move the
temp
pointer to the next node in the linked list.
 Check if the data value of the current node (
 After traversing the entire linked list, print the frequency count along with the target value.
Program Solution

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Time Complexity
The time complexity of this algorithm is O(n), where n is the number of nodes in the doubly linked list. This is because, in the worst case, you need to traverse through all the nodes once to count the frequency of the given value.
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