Find Minimum in Linked List

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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: Easy
• ⏰ Time to complete: 10-15 mins
• 🛠️ Topics: Linked Lists, Finding Minimum

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 linked list is empty?
  • If the linked list is empty, the function should return None.

HAPPY CASE Input: head = Node(5) -> Node(6) -> Node(7) -> Node(8) Output: 5 Explanation: The smallest value in the linked list is 5. EDGE CASE Input: head = None Output: None Explanation: When the linked list is empty, the function returns None.

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 problems, we want to consider the following approaches:

• Traversal of a linked list
• Comparison of node values

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Traverse the linked list, comparing each node's value to find the minimum.

1) If the head is `None`, return `None`. 2) Initialize `min_value` to the value of the head node. 3) Traverse the linked list starting from the second node. 4) If the current node's value is less than `min_value`, update `min_value`. 5) Continue until all nodes have been visited. 6) Return `min_value`.

⚠️ Common Mistakes

• Forgetting to handle the case where the linked list is empty.
• Not correctly updating the minimum value as you traverse the list.

4: I-mplement

Implement the code to solve the algorithm.

class Node: def __init__(self, value, next=None): self.value = value self.next = next # For testing def print_linked_list(head): current = head while current: print(current.value, end=" -> " if current.next else "\n") current = current.next def find_min(head): if head is None: return None min_value = head.value current = head.next while current: if current.value < min_value: min_value = current.value current = current.next return min_value

5: R-eview

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

• Trace through your code with an input to check for the expected output
• Catch possible edge cases and off-by-one errors

Example:

head1 = Node(5, Node(6, Node(7, Node(8)))) # Linked List: 5 -> 6 -> 7 -> 8 print(find_min(head1)) # Expected Output: 5 head2 = Node(8, Node(5, Node(6, Node(7)))) # Linked List: 8 -> 5 -> 6 -> 7 print(find_min(head2)) # Expected Output: 5

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work. Assume N represents the number of nodes in the linked list.

• Time Complexity: O(N) because we need to traverse all the nodes in the linked list.
• Space Complexity: O(1) because we are only using a constant amount of extra space to store the minimum value.