README_EN.md

1344. Angle Between Hands of a Clock

中文文档

Description

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

Answers within 10-5 of the actual value will be accepted as correct.

 

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

 

Constraints:

• 1 <= hour <= 12
• 0 <= minutes <= 59

Solutions

Solution 1: Mathematics

The hour hand moves 30 degrees every hour and an additional 0.5 degrees for each minute. The minute hand moves 6 degrees every minute. If the angle between the hands is greater than 180 degrees, take its difference from 360 degrees to ensure the smallest angle is obtained.

The time complexity is $O(1)$ and the space complexity is $O(1)$.

Python3

class Solution:
    def angleClock(self, hour: int, minutes: int) -> float:
        h = 30 * hour + 0.5 * minutes
        m = 6 * minutes
        diff = abs(h - m)
        return min(diff, 360 - diff)

Java

class Solution {
    public double angleClock(int hour, int minutes) {
        double h = 30 * hour + 0.5 * minutes;
        double m = 6 * minutes;
        double diff = Math.abs(h - m);
        return Math.min(diff, 360 - diff);
    }
}

C++

class Solution {
public:
    double angleClock(int hour, int minutes) {
        double h = 30 * hour + 0.5 * minutes;
        double m = 6 * minutes;
        double diff = abs(h - m);
        return min(diff, 360 - diff);
    }
};

Go

func angleClock(hour int, minutes int) float64 {
	h := 30*float64(hour) + 0.5*float64(minutes)
	m := 6 * float64(minutes)
	diff := math.Abs(h - m)
	return math.Min(diff, 360-diff)
}

TypeScript

function angleClock(hour: number, minutes: number): number {
    const h = 30 * hour + 0.5 * minutes;
    const m = 6 * minutes;
    const diff = Math.abs(h - m);
    return Math.min(diff, 360 - diff);
}

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