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1344. Angle Between Hands of a Clock

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

Thinking

The smaller angle between the hour and minute hands. The hour hand moves \(30^\circ\) per hour and an extra \(0.5^\circ\) per minute; the minute hand moves \(6^\circ\) per minute. The answer is the minimum of the absolute difference and \(360^\circ\) minus that difference.

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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)

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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);
    }
}

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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);
    }
};

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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)
}

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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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