Skip to main content

Mirror Reflection

Problem Description​

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.

Constraints​

  • 1 <= q <= p <= 1000

Approach​

This Java function mirrorReflection determines where a laser ray, starting from the southwest corner of a square room with mirrored walls, will reflect and reach a receptor. It continuously reduces the room dimensions p and q by dividing them by 2 while both are even, simulating the path of the laser until it reaches a receptor. Depending on whether p or q becomes odd first, it returns 0, 1, or 2 indicating the receptor the laser reaches first (based on their positions in the room).

Java​

class Solution {

    // taken help
    public int mirrorReflection(int p, int q) {
        while(p%2==0 && q%2==0){
            p=p/2;
            q=q/2;
        }

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

  • Time Complexity The time complexity is o(n)o(n).

  • Space Complexity The space complexity is O(1)O(1).