Power of Three

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Problem Highlights

• 🔗 Leetcode Link: Power of Three
• 💡 Problem Difficulty: Easy
• ⏰ Time to complete: 15 mins
• 🛠️ Topics: Recursion
• 🗒️ Similar Questions: Sqrt(x), Super Pow, Pow(x.n), Power of Two, Power of Four

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?

Be sure that you clarify the input and output parameters of the problem:

• How do we stimulate the operation of evaluating exponents?
  • An exponent can easily be evaluated by multiplying the base exponent number of times. So, we can easily simulate this using a loop.
• What time and space complexity are we attempting to achieve?
  • N being the number, O(logN) time and O(1) space.

Run through a set of example cases:

HAPPY CASE
Input: n = 27
Output: true
Explanation: 27 = 3^3

Input: n = -1
Output: false
Explanation: There is no x where 3x = (-1).

EDGE CASE 
Input: n = 0
Output: false
Explanation: There is no x where 3^x = 0.

Input: n = 1
Output: true
Explanation: 2^0 = 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.

For problems outside of the main catagories, we want to consider the input and output:

We are working with a number and checking if it is a power of three. We know for sure any number is a power of three if it can be divided by three until it becomes 1, along the way down it must also be divisible by three.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: We will recursively divide the number by 3 until it either reaches 1 and return true or it is no longer a whole number when divided by 3 and return false. We also need to consider the edge case 0, which is false.

1. Break into base case where n = 1 return True
2. Break into base case where n = 0 or n is no longer a whole number when divided by 3 return False
3. Return the recursive function n/3 

⚠️ Common Mistakes

• This problem looks quite simple at the first glance, however requires considering the edge case 0 and 1.

4: I-mplement

Implement the code to solve the algorithm.

class Solution:
    def isPowerOfThree(self, n: int) -> bool:
        # Break into base case where n = 1 return True
        if n == 1:
            return True

        # Break into base case where n = 0 or n is no longer a whole number when divided by 3 return False
        if n == 0 or n % 3 != 0:
            return False
        
        # Return the recursive function n/3 
        return self.isPowerOfThree(n/3)

class Solution {
  public boolean isPowerOfThree(int n) {
    // Break into base case where n = 1 return True
    if(n == 1) return true;
    
    // Break into base case where n = 0 or n is no longer a whole number when divided by 3 return False
    if(n == 0 || n % 3 != 0) return false;
    
    // Return the recursive function n/3 
    return isPowerOfThree(n/3);    
  }
}

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

6: E-valuate

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

Assume N represents the number given

• Time Complexity: O(logN), because we are dividing n by 3 each time, until it is one
• Space Complexity: O(1), the space complexity is constant if you do not count the recursive stack.