1344. Angle Between Hands of a Clock

Intuition

Each hand has a fixed angular speed, so we can compute the exact angle of each hand from the 12 o'clock position and take the difference. The only subtlety is that the hour hand also moves as the minutes pass, it is not fixed at the hour mark.

Approach

1. The minute hand sweeps a full 360° in 60 minutes, so it sits at 6 * minutes degrees.
2. The hour hand sweeps 360° in 12 hours = 30° per hour, plus 0.5° per minute. Using hour % 12 (so 12 maps to 0), it sits at 30 * (hour % 12) + 0.5 * minutes degrees.
3. Take the absolute difference of the two angles.
4. The hands form two angles summing to 360°; return the smaller one: min(diff, 360 - diff).

Complexity

• Time: O(1)
• Space: O(1)

Edge Cases

• hour = 12 -> treat as 0 via hour % 12
• minutes = 0 -> hour hand exactly on the hour mark
• Difference exactly 180° -> both diff and 360 - diff are 180, returns 180
• Hour hand past the mark (e.g. 3:30) -> the 0.5 * minutes term accounts for it

class Solution:
    def angleClock(self, hour: int, minutes: int) -> float:
        """
        1344. Angle Between Hands of a Clock
        Time: O(1)
        Space: O(1)
        """
        minute_angle = 6 * minutes
        hour_angle = 30 * (hour % 12) + 0.5 * minutes

        diff = abs(hour_angle - minute_angle)
        return min(diff, 360 - diff)

/**
 * 1344. Angle Between Hands of a Clock
 * Time: O(1)
 * Space: O(1)
 */
class Solution {
    public double angleClock(int hour, int minutes) {
        double minuteAngle = 6.0 * minutes;
        double hourAngle = 30.0 * (hour % 12) + 0.5 * minutes;

        double diff = Math.abs(hourAngle - minuteAngle);
        return Math.min(diff, 360.0 - diff);
    }
}