Fishing Probability

Edit PagePage History

TIP102 Unit 5 Session 1 Advanced (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 10 mins
• 🛠️ Topics: Linked List, Probability, String Manipulation

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?

• How do we calculate the probability?
  • By dividing the number of nodes with the given fish name by the total number of nodes in the list.
• What should the function return if the list is empty?
  • The function should return 0.00.

HAPPY CASE
Input: A linked list with nodes "Carp" -> "Dace" -> "Cherry Salmon", fish_name = "Dace"
Output: 0.33
Explanation: The probability of catching "Dace" is 1/3, which is 0.33 when rounded down to the nearest hundredth.

EDGE CASE
Input: An empty linked list, fish_name = "Dace"
Output: 0.00
Explanation: There are no fish to catch, so the probability is 0.00.

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:

• Traversing the list
• Counting occurrences
• Calculating probability

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Traverse the linked list, count the occurrences of the given fish name, and calculate the probability by dividing the count by the total number of nodes.

1) Initialize a variable `total_count` to 0.
2) Initialize a variable `fish_count` to 0.
3) Set the current node to the head of the list.
4) While the current node is not None:
    a) Increment `total_count`.
    b) If the current node's fish name matches the given fish name, increment `fish_count`.
    c) Move to the next node.
5) If `total_count` is 0, return 0.00.
6) Calculate the probability by dividing `fish_count` by `total_count`.
7) Round the probability to the nearest hundredth and return it.

⚠️ Common Mistakes

• Forgetting to handle the case where the linked list is empty.
• Not properly rounding the probability to the nearest hundredth.

4: I-mplement

Implement the code to solve the algorithm.

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

# For testing
def print_linked_list(head):
    current = head
    while current:
        print(current.fish_name, end=" -> " if current.next else "\n")
        current = current.next

def fish_chances(head, fish_name):
    current = head
    total_count = 0
    fish_count = 0
    
    while current:
        total_count += 1
        if current.fish_name == fish_name:
            fish_count += 1
        current = current.next
    
    if total_count == 0:
        return 0.00
    
    probability = fish_count / total_count
    return round(probability, 2)

# Example Usage:
fish_list = Node("Carp", Node("Dace", Node("Cherry Salmon")))
print(fish_chances(fish_list, "Dace"))          # Output: 0.33
print(fish_chances(fish_list, "Rainbow Trout")) # Output: 0.00

empty_list = None
print(fish_chances(empty_list, "Dace"))         # Output: 0.00

5: R-eview

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

• Initial list: "Carp" -> "Dace" -> "Cherry Salmon"
• Probability of "Dace": 1/3 = 0.33
• Probability of "Rainbow Trout": 0/3 = 0.00
• When the list is empty: 0.00

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 using a fixed amount of additional space.