31_Days_Leetcode_Challenge-July/Day-14_Angle_Between_Hands_Of_a_Clock.py at master · Gangadharbhuvan/31_Days_Leetcode_Challenge-July · GitHub

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'''
	Given two numbers, hour and minutes. Return the smaller angle (in degrees) formed between the hour and the minute hand.


Example 1:


Input: hour = 12, minutes = 30
Output: 165
Example 2:


Input: hour = 3, minutes = 30
Output: 75
Example 3:


Input: hour = 3, minutes = 15
Output: 7.5
Example 4:

Input: hour = 4, minutes = 50
Output: 155
Example 5:

Input: hour = 12, minutes = 0
Output: 0


Constraints:

1 <= hour <= 12
0 <= minutes <= 59
Answers within 10^-5 of the actual value will be accepted as correct.
   Hide Hint #1
The tricky part is determining how the minute hand affects the position of the hour hand.
   Hide Hint #2
Calculate the angles separately then find the difference.

'''
class Solution(object):
    def angleClock(self, hour, minutes):
        t_min = 6*minutes
        t_hour = 30*hour + 30*(minutes/float(60))
        if t_hour>=360:
            t_hour = t_hour-360
        x = abs(t_hour-t_min)
        if x>180:
            return 360-x
        else:
            return x