algoadvance

Problem Statement

Given a positive integer n, return the smallest positive integer that is a multiple of both n and 2.

Clarifying Questions

1. Constraints and Edge Cases:
   • What are the minimum and maximum values for n?
   • Should we consider the performance for very large values of n?
2. Inputs and Outputs:
   • Input is a single integer n.
   • Output is a single integer which is the smallest multiple of both n and 2.

Strategy

To find the smallest multiple of both n and 2:

1. We need to consider the properties of even and odd numbers:
   • If n is already even, then n itself is already a multiple of both n and 2.
   • If n is odd, then multiplying n by 2 will make it even.

Thus, the strategy is simple:

1. Check if n is even. If so, return n.
2. If n is odd, return 2 * n.

Code

def smallest_even_multiple(n: int) -> int:
    if n % 2 == 0:
        return n
    else:
        return 2 * n

Time Complexity

• The time complexity of this solution is (O(1)) because it requires only a constant number of operations (a modulus operation and a conditional check).
• The space complexity is also (O(1)) since we are not using any additional data structures.

This solution should be efficient even for very large values of n.

Let’s test the function with a few examples:

# Test cases
print(smallest_even_multiple(5))  # Expected output: 10 (since 5 is odd, 5 * 2 = 10)
print(smallest_even_multiple(6))  # Expected output: 6 (since 6 is even)
print(smallest_even_multiple(1))  # Expected output: 2 (since 1 is odd, 1 * 2 = 2)
print(smallest_even_multiple(8))  # Expected output: 8 (since 8 is even)

This should cover the basic functionality of the solution. Feel free to provide additional test cases or any other specific requirements!

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