Split Haycorns

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

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 5 mins
• 🛠️ Topics: Divisors, Loops

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?

• The function split_haycorns() should take a positive integer quantity and return a list of all divisors of quantity.

HAPPY CASE
Input: 6
Expected Output: [1, 2, 3, 6]

EDGE CASE
Input: 1
Expected Output: [1]

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.

This problem falls under: Divisors Calculation.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Define a function that iterates through numbers from 1 to quantity and checks if they are divisors of quantity.

1. Define the function `split_haycorns(quantity)`.
2. Initialize an empty list `divisors` to store the divisors.
3. Iterate through all numbers from 1 to `quantity` (inclusive).
4. For each number, check if it is a divisor of `quantity` using the modulo operator (`quantity % i == 0`).
5. If it is a divisor, add it to the `divisors` list.
6. Return the `divisors` list.

⚠️ Common Mistakes

• Not handling the case where quantity is 1, which should correctly return [1].

4: I-mplement

Implement the code to solve the algorithm.

def split_haycorns(quantity):
    # Initialize an empty list to store the divisors
    divisors = []
    
    # Iterate through all numbers from 1 to quantity (inclusive)
    for i in range(1, quantity + 1):
        # If i is a divisor of quantity, add it to the list
        if quantity % i == 0:
            divisors.append(i)
    
    # Return the list of divisors
    return divisors

5: R-eview

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

Call the function with the provided examples:

print(split_haycorns(6))  # Expected Output: [1, 2, 3, 6]
print(split_haycorns(1))  # Expected Output: [1]

Expected outputs:

[1, 2, 3, 6]
[1]

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 value of quantity since we need to check all numbers from 1 to quantity.
• Space Complexity: O(n) for storing the divisors.