Find Length of Doubly Linked List from Any Node

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Unit 5 Session 2 (Click for link to problem statements)

TIP102 Unit 5 Session 2 Standard (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 15-20 mins
• 🛠️ Topics: Linked Lists, Traversal

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• What happens if the node is None?
  • The function should return 0.
• What happens if the doubly linked list contains only one node?
  • The function should return 1.

HAPPY CASE
Input: node is the node with value 6 in the list 3 <-> 5 <-> 6 <-> 7
Output: 4
Explanation: The function calculates the length of the entire doubly linked list.


EDGE CASE
Input: node = None
Output: 0
Explanation: When the node is `None`, the function returns `0`.

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For Linked List length problems, we want to consider the following approaches:

• Traversal in both directions to find the head and tail
• Counting nodes while traversing the list

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Traverse the list from the given node to find the head, then traverse the entire list from head to tail to count the number of nodes.

Traverse back to the head via prev.
Traverse forward via next to count nodes.

⚠️ Common Mistakes

• Forgetting to handle the case where the node is None.
• Not correctly traversing to the start of the list before counting.

4: I-mplement

Implement the code to solve the algorithm.

def get_length(node):
    if node is None:
        return 0

    # Find the start of the list
    start = node
    while start.prev:
        start = start.prev

    # Traverse from start to end, counting nodes
    length = 0
    current = start
    while current:
        length += 1
        current = current.next

    return length

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Verify the length returned matches the expected length of the doubly linked list for different starting nodes.

Example:

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

# Build: 3 <-> 5 <-> 6 <-> 7
n1 = Node(3)
n2 = Node(5)
n3 = Node(6)
n4 = Node(7)

n1.next, n2.prev = n2, n1
n2.next, n3.prev = n3, n2
n3.next, n4.prev = n4, n3

# Test from any node:
print(get_length(n3))   # Expected: 4 (start in the middle, value 6)
print(get_length(n1))   # Expected: 4 (head)
print(get_length(n4))   # Expected: 4 (tail)
print(get_length(None)) # Expected: 0

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

• Time Complexity: O(N) where N is the number of nodes in the list, as we need to traverse the entire list.
• Space Complexity: O(1) because we are only using a constant amount of extra space for the pointers.