Recursive Power of 2

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Unit 7 Session 1 (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 15 mins
• 🛠️ Topics: Recursion, Mathematics, Powers of Two

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?

• Q: What should the function return for n = 0?
  • A: The function should return False since 0 is not a power of two.

HAPPY CASE
Input: 8
Output: True
Explanation: (8 = 2^3), hence it is a power of two.

EDGE CASE
Input: 0
Output: False
Explanation: 0 is not a power of any positive integer.

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 aligns with a mathematical and recursive decomposition problem, focusing on:

• Understanding properties of powers of two.
• Reducing the problem size by dividing the number recursively.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use a recursive method to determine if a number is a power of two by continuously dividing the number by two.

1) Base Case: If `n` is 1, return True because \(2^0 = 1\).
2) If `n` is less than 1 or not divisible by 2, return False.
3) Recursive Case: Recursively call the function with \(n/2\).

⚠️ Common Mistakes

• Not correctly identifying non-powers of two that are not immediately obvious (e.g., 0 or negative numbers).

4: I-mplement

Implement the code to solve the algorithm.

def is_power_of_two(n):
    # Base case: if n is 1, it's a power of 2
    if n == 1:
        return True

    # Base case: if n is less than 1 or not divisible by 2, it's not a power of 2
    elif n < 1 or n % 2 != 0:
        return False

    # Recursive case: recursively check if n divided by 2 is a power of 2
    else:
        return is_power_of_two(n // 2)
        

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 of 8 to check for the expected output of True.
• Validate the edge case with input 0 and other non-powers like 3 to ensure the function returns False.

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

• Time Complexity: O(log n) because each recursive call reduces the problem size by half.
• Space Complexity: O(log n) due to the recursion stack depth, also reducing by half each time.