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;
}
}
}
Input
Output
#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;
}
};
Input
Output
#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
Input
Output