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Python/mirror-reflection.py

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+# Time:  O(1)
+# Space: O(1)
+
+# 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.
+#
+# Return the number of the receptor that the ray meets first.
+# (It is guaranteed 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.
+#
+# Note:
+# - 1 <= p <= 1000
+# - 0 <= q <= p
+
+
+class Solution(object):
+    def mirrorReflection(self, p, q):
+        """
+        :type p: int
+        :type q: int
+        :rtype: int
+        """
+        # explanation commented in the following solution
+        return 2 if (p & -p) > (q & -q) else 0 if (p & -p) < (q & -q) else 1
+
+
+# Time:  O(log(max(p, q))) = O(1) due to 32-bit integer
+# Space: O(1)
+class Solution2(object):
+    def mirrorReflection(self, p, q):
+        """
+        :type p: int
+        :type q: int
+        :rtype: int
+        """
+        def gcd(a, b):
+            while b:
+                a, b = b, a % b
+            return a
+
+        lcm = p*q // gcd(p, q)
+        # let a = lcm / p, b = lcm / q
+        if lcm // p % 2 == 1:
+            if lcm // q % 2 == 1:
+                return 1  # a is odd, b is odd <=> (p & -p) == (q & -q)
+            return 2  # a is odd, b is even <=> (p & -p) > (q & -q)
+        return 0  # a is even, b is odd <=> (p & -p) < (q & -q)