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.

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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


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Constraints:


	1 &lt;= hour &lt;= 12
	0 &lt;= minutes &lt;= 59


class Solution:
    def angleClock(self, hour: int, minutes: int) -> float:
        
        x = abs(minutes * 6 -(hour * 30 + minutes/2))
        return min(360-x , x)
class Solution {
    public double angleClock(int hour, int minutes) {
        // Position of hour hand in a circle of 0 - 59
        double hrPos = 5 * (hour % 12);

        // Adjust hour hand position according to minute hand
        hrPos += (5 * minutes/60.0);

        double units = Math.abs(minutes - hrPos);

        // Take the min of distance between minute & hour hand and hour & minute hand
        return Math.min(units, 60-units) * 6;
    }
}
class Solution {
public:
    double angleClock(int hour, int minutes) {

        double hourAngle = 30*(double(hour) + double(minutes/60.0));

        double minuteAngle = 6 * (double)minutes;

        return 180 - abs(180 - abs(minuteAngle - hourAngle));

    }
};
var angleClock = function(hour, minutes) {
    const angle = Math.abs((hour * 30) - 5.5 * minutes)
    return angle > 180 ? 360 - angle : angle
};
Angle Between Hands of a Clock