## Algorithm

Problem Name: 858. Mirror Reflection

There is a special square room with mirrors on each of the four walls. Except for the southwest corner, there are receptors on each of the remaining corners, numbered 0, 1, and 2.

The square room has walls of length p and a laser ray from the southwest corner first meets the east wall at a distance q from the 0th receptor.

Given the two integers p and q, return the number of the receptor that the ray meets first.

The test cases are guaranteed so that the ray will meet a receptor eventually.

Example 1:

Input: p = 2, q = 1
Output: 2
Explanation: The ray meets receptor 2 the first time it gets reflected back to the left wall.

Example 2:

Input: p = 3, q = 1
Output: 1

Constraints:

• 1 <= q <= p <= 1000

## Code Examples

### #1 Code Example with Java Programming

Code - Java Programming

class Solution {
public int mirrorReflection(int p, int q) {
int extention = q;
int reflection = p;
while (extention % 2 == 0 && reflection % 2 == 0) {
extention /= 2;
reflection /= 2;
}
if (extention % 2 == 0 && reflection % 2 != 0) {
return 0;
} else if (extention % 2 != 0 && reflection % 2 == 0) {
return 2;
} else {
return 1;
}
}
}
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Input

cmd
p = 2, q = 1

Output

cmd
2

### #2 Code Example with Javascript Programming

Code - Javascript Programming

const mirrorReflection = function(p, q) {
while (p % 2 === 0 && q % 2 === 0) {
p /= 2;
q /= 2;
}

if (p % 2 === 0) {
return 2;
} else if (q % 2 === 0) {
return 0;
} else {
return 1;
}
};
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Input

cmd
p = 2, q = 1

Output

cmd
2

### #3 Code Example with Python Programming

Code - Python Programming

class Solution:
def mirrorReflection(self, p, q):
side, up, h = 2, 1, 0
while True:
h += q * up
side = (side + 1) % 2
if side == 0:
side += 2
if h < 0:
h *= -1
up *= -1
elif h > p:
h = p - (h - p)
up *= -1
if h % p == 0:
return h and side or 0
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Input

cmd
p = 3, q = 1

Output

cmd
1