LeetCode: Angle Between Hands of a Clock

Angle Between Hands of a Clock

Similar Problems:

• CheatSheet: LeetCode For Code Interview
• CheatSheet: Common Code Problems & Follow-ups
• Tag: #geometry, #math

Given two numbers, hour and minutes. Return the smaller angle (in sexagesimal units) 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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

• Solution:

## https://code.dennyzhang.com/angle-between-hands-of-a-clock
## Basic Ideas: math
##
##  When hour change by one, it has 30 degrees
##  When minute change by one, it has 6 degrees
##  The change of minutes adjust hours' change
##     When hour change by one, it means minutes change by 60
##
##  Change from 00:00 to hour:minute
##
##    d = abs(30*hour + 30*(minutes/60) - minutes*6)
##    min(d, 360-d)
## Complexity: Time O(1), Space O(1)
class Solution:
    def angleClock(self, hour: int, minutes: int) -> float:
        d = abs(30*(hour%12) + 30*(minutes/60) - minutes*6)
        return min(d, 360-d)