Unit 6 Session 1 (Click for link to problem statements)
TIP102 Unit 5 Session 2 Advanced (Click for link to problem statements)
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?
False
since there is no node to match val
.HAPPY CASE
Input: head = Node(1, Node(2, Node(3))), val = 2
Output: True
Explanation: The middle node in an odd-numbered list is 2, which matches `val`.
EDGE CASE
Input: head = Node(1, Node(2, Node(3, Node(4)))), val = 3
Output: True
Explanation: In a list with an even number of nodes, the second middle node is 3, which matches `val`.
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.
This problem utilizes the slow-fast pointer technique, commonly used in problems involving linked lists to find middle elements or detect cycles.
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Use the slow-fast pointer technique to locate the middle or the second middle node in the list, and then compare its value to val
.
1) Check if the head is `None` and return `False` immediately, as there's no node to check against `val`.
2) Initialize both `slow` and `fast` pointers at the head.
3) Traverse the list: move `slow` by one step and `fast` by two steps until `fast` reaches the end or the last node.
4) At the end of the traversal, `slow` will point to the middle (or second middle) node.
5) Compare `slow.value` to `val` and return the result.
⚠️ Common Mistakes
Implement the code to solve the algorithm.
class Node:
def __init__(self, value=None, next=None):
self.value = value
self.next = next
def middle_match(head, val):
if not head:
return False # Return False as there's no node to match with val
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow.value == val # Check if the middle node's value matches val
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
O(n/2)
which simplifies to O(n)
because in the worst case, the loop must execute while slow pointer travels through half of the list to reach the middle.O(1)
since no additional data structures are needed beyond the two pointers, slow and fast.