Eeyore's House

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

TIP102 Unit 1 Session 2 Standard (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 10 mins
• 🛠️ Topics: Arrays, Nested Loops, Modulo Operation

U-nderstand

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

• Q: What is the input to the function?

  • A: The input consists of two integer arrays pile1 and pile2, and a positive integer k.

• Q: What is the expected output of the function?

  • A: The function should return the number of good pairs (i, j) where pile1[i] is divisible by pile2[j] * k.

• Q: How is a good pair defined?

  • A: A pair (i, j) is called good if pile1[i] % (pile2[j] * k) == 0.

• Q: What should the function return if there are no good pairs?

  • A: The function should return 0 if no good pairs are found.

• Q: Can the arrays be empty or have elements that are zero?

  • A: The problem assumes the arrays are non-empty and contain positive integers, as stick lengths cannot be zero.

• The function good_pairs() should take two integer arrays pile1 and pile2, and a positive integer k, returning the number of good pairs. A pair (i, j) is good if pile1[i] is divisible by pile2[j] * k.

HAPPY CASE
Input: pile1 = [1, 3, 4], pile2 = [1, 3, 4], k = 1
Expected Output: 5

Input: pile1 = [1, 2, 4, 12], pile2 = [2, 4], k = 3
Expected Output: 2

EDGE CASE
Input: pile1 = [2, 4, 6], pile2 = [1, 1, 1], k = 2
Expected Output: 9

Input: pile1 = [], pile2 = [1, 2, 3], k = 1
Expected Output: 0

P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Iterate through each stick in pile1 and for each stick, iterate through each stick in pile2. Check if pile1[i] is divisible by pile2[j] * k. Count the number of such good pairs.

1. Initialize a counter `count` to 0.
2. Iterate through each stick in `pile1` using index `i`.
3. For each stick in `pile1`, iterate through each stick in `pile2` using index `j`.
4. Check if `pile1[i]` is divisible by `pile2[j] * k`.
5. If the condition is met, increment `count`.
6. Return the total `count` of good pairs.

⚠️ Common Mistakes

• Forgetting to handle empty arrays.
• Incorrectly checking the divisibility condition.

I-mplement

Implement the code to solve the algorithm.

def good_pairs(pile1, pile2, k):
    # Initialize the counter for good pairs
    count = 0
    
    # Iterate through each stick in pile1
    for i in range(len(pile1)):
        # Iterate through each stick in pile2
        for j in range(len(pile2)):
            # Check if pile1[i] is divisible by pile2[j] * k
            if pile1[i] % (pile2[j] * k) == 0:
                # Increment the counter if the condition is met
                count += 1
    
    # Return the total number of good pairs
    return count