[Medium] 1344. Angle Between Hands of a Clock

Problem Statement

Given two integers hour and minutes, return the smaller angle (in degrees) formed between the hour hand and the minute hand on a clock.

The answer must be in the range ([0, 180]).

Examples

Example 1:

Input:  hour = 12, minutes = 30
Output: 165.0
Explanation:
Minute hand: 30 minutes → 30 * 6 = 180 degrees
Hour hand: (12 % 12 + 30/60) * 30 = 15 degrees
Angle = |180 - 15| = 165

Example 2:

Input:  hour = 3, minutes = 30
Output: 75.0

Example 3:

Input:  hour = 3, minutes = 15
Output: 7.5

Constraints

  • 1 <= hour <= 12
  • 0 <= minutes <= 59
  • Answers are expected as a float with reasonable precision.

Clarification Questions

  1. 12 vs 0 o’clock: Should hour = 12 be treated the same as 0 on the circle?
    Assumption: Yes — use hour % 12.
  2. Return type: Should we return a float or rounded integer?
    Assumption: Float (Python float) with standard precision is acceptable.
  3. Smaller angle: If the two hands form angles x and 360 - x, should we always return the smaller one?
    Assumption: Yes — answer must be in ([0, 180]).
  4. Edge cases: When hands overlap (e.g., 12:00), is 0.0 a valid answer?
    Assumption: Yes.

Interview Deduction Process (20 minutes)

Step 1: Convert positions to degrees (5 min)
On a clock:

  • The minute hand moves 6 degrees per minute (360 / 60).
  • The hour hand moves 30 degrees per hour (360 / 12), plus it moves continuously as minutes pass.

Step 2: Formulas (7 min)
For a given time hour, minutes:

  • Minute hand angle:
minute_angle = minutes * 6
  • Hour hand angle:
hour_angle = (hour % 12 + minutes / 60) * 30

The absolute difference:

diff = |hour_angle - minute_angle|

Step 3: Smaller of the two angles (8 min)
Clock hands split the circle into two angles: diff and 360 - diff. We need the smaller one:

answer = min(diff, 360 - diff)

Return answer.

Solution Approach

  1. Normalize hour with hour % 12 so 12 is treated as 0.
  2. Compute hour_angle using hours plus fractional hours from minutes.
  3. Compute minute_angle as minutes * 6.
  4. Take the absolute difference between these two angles.
  5. Return the smaller angle between diff and 360 - diff.

Key Insights

  1. Continuous hour hand
    The hour hand is not fixed per hour; it moves as minutes progress, which is why we add minutes / 60 to hour.

  2. Normalize 12 to 0
    Using hour % 12 simplifies math and avoids special casing for 12.

  3. Two possible angles
    Since a circle is 360 degrees, the angle between two directions can be taken in two ways; always pick the smaller.

Python Solution

class Solution:
    def angleClock(self, hour: int, minutes: int) -> float:
        # Hour hand: 30 degrees per hour + 0.5 degrees per minute
        hour_angle = (hour % 12) * 30 + minutes * 0.5
        # Minute hand: 6 degrees per minute
        minutes_angle = (minutes % 60) * 6

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

Algorithm Explanation

  • The minute hand angle is straightforward: there are 60 minutes on the clock, so each minute is (360 / 60 = 6) degrees.
  • The hour hand moves 30 degrees per hour; since there are 60 minutes in an hour, it also moves (30 / 60 = 0.5) degrees per minute.
  • By computing both angles and taking their difference, we get one of the angles between the two hands.
  • Taking min(diff, 360 - diff) ensures we return the smaller of the two possible angles.

Complexity Analysis

  • Time Complexity: (O(1)) — just a few arithmetic operations.
  • Space Complexity: (O(1)) — constant extra space.

Edge Cases

  • hour = 12, minutes = 0 → both hands at 0 degrees → angle 0.0.
  • hour = 12, minutes = 30 → hands at 15 and 180 degrees → angle 165.0.
  • hour = 6, minutes = 0 → hands at 180 and 0 degrees → angle 180.0.

Common Mistakes

  • Forgetting to normalize hour with % 12, which can lead to incorrect angles when hour = 12.
  • Ignoring the continuous movement of the hour hand during minutes (using hour * 30 only).
  • Returning diff directly without taking min(diff, 360 - diff).