Problem URL: Mirror reflection

First, think about the case p = 3 and q = 2. So, this problem can be transformed into finding m * p = n * q, where m = the number of room extension + 1, n = the number of light reflection + 1.

If the number of light reflection is odd (which means n is even), it means the corner is on the left-hand side. The possible corner is 2. Otherwise, the corner is on the right-hand side. The possible corners are 0 and 1.

Given the corner is on the right-hand side. If the number of room extension is even (which means m is odd), it means the corner is 1. Otherwise, the corner is 0.

m is even & n is odd => return 0.
m is odd & n is odd => return 1.
m is odd & n is even => return 2.

The case m is even & n is even is impossible. Because in the equation m * q = n * p, if m and n are even, we can divide both m and n by 2. Then, m or n must be odd.

class Solution:
    def mirrorReflection(self, p: int, q: int) -> int:
        while p % 2 == 0 and q % 2 == 0:
            p, q = p/2, q/2
        return int(1 - (p%2) + (q%2))

Time Complexity: O(n)
Space Complexity: O(n)