Print Powers in Python with Time Complexity O(logAB)


Given two non-negative integers A and B, print to the console all numbers less than or equal to B that are powers of A

Powers of a number are: A0, A1, A2, etc.

An = A multiplied by itself n times

A0 = 1
A1 = A
A2 = A * A
A3 = A * A * A
A4 = A * A * A * A
etc.

Example:

Input: A = 3, B = 100

Output:
1
3
9
27
81

Explanation:
30 = 1
31 = 3
32 = 9
33 = 27
34 = 81

Problem Definition

The problem requires us to print all powers of a given number A that are less than or equal to another number B. The powers of A are defined as A0, A1, A2, and so on.

Input:

Output:

Constraints:

Example:

Input: A = 3, B = 100
Output:
1
3
9
27
81

Understanding the Problem

The core challenge is to generate and print all powers of A that do not exceed B. This problem is significant in various applications such as computing exponential growth, generating sequences, and more.

Potential pitfalls include handling the case when A is 0 or 1, as these have unique properties (0n is always 0 for n > 0, and 1n is always 1).

Approach

To solve this problem, we can use a simple iterative approach:

  1. Start with the initial power of 1 (i.e., A0).
  2. Use a loop to multiply the current power by A until the power exceeds B.
  3. Print each power during the iteration.

Naive Solution

A naive solution would involve calculating each power of A and checking if it is less than or equal to B. However, this is not optimal as it involves unnecessary calculations.

Optimized Solution

An optimized solution involves using a loop to multiply the current power by A and checking if it is less than or equal to B. This approach is efficient and has a time complexity of O(logAB).

Algorithm

Here is a step-by-step breakdown of the optimized algorithm:

  1. Initialize a variable power to 1 (representing A0).
  2. Use a while loop to check if power is less than or equal to B.
  3. Print the current value of power.
  4. Multiply power by A to get the next power.
  5. Repeat steps 2-4 until power exceeds B.

Code Implementation

def print_powers(A, B):
    # Initialize the first power of A
    power = 1
    
    # Loop until power exceeds B
    while power <= B:
        # Print the current power
        print(power)
        
        # Calculate the next power
        power *= A

# Example usage
A = 3
B = 100
print_powers(A, B)

Complexity Analysis

The time complexity of the optimized solution is O(logAB) because we are repeatedly multiplying by A until the power exceeds B. The space complexity is O(1) as we are using a constant amount of extra space.

Edge Cases

Potential edge cases include:

These cases can be handled by adding specific checks in the code.

Testing

To test the solution comprehensively, consider the following test cases:

Use a testing framework like unittest or pytest to automate the testing process.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Conclusion

In this blog post, we discussed how to print all powers of a given number A that are less than or equal to another number B. We explored the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.

Additional Resources

For further reading and practice, consider the following resources: