The Clock Examiner

A medium Python interview practice problem on DataDriven. Write the Python and run it against test cases, with instant feedback.

Domain
Python
Difficulty
medium
Seniority
L3
Asked in
phData

Problem

You're rendering an analog clock widget and need the angle between its two hands. Given hour (0-23) and minute (0-59), return the smaller angle in degrees between the hour and minute hands on a 12-hour dial, where the hour hand advances smoothly as the minutes pass rather than jumping on the hour.

Two hands on a clock: how wide is the gap?

Example 1

Input:

hour = 3, minute = 0

Output:

90

Example 2

Input:

hour = 12, minute = 30

Output:

165

Worked solution and explanation

What this really is

Strip the clock face away and this is two angular positions on a circle, then the shorter arc between them. The whole problem hinges on one detail almost everyone botches under pressure: the hour hand does not sit on the hour, it creeps forward as the minutes pass. Compute it as hour * 30 and you'll nail 3:00 but be off by 15 degrees at 12:30. The second trap is forgetting a clock is circular, so a raw difference of 240 is really a 120 degree gap.

Break down the requirements

Step 1: Place the hour hand, including its drift

Each hour is 30 degrees apart (360 / 12), and the hour hand drifts another 0.5 degrees for every minute past the hour. Position: (hour % 12) * 30 + minute * 0.5. The % 12 is what folds a 23-hour clock back onto the 12-hour face.

Step 2: Place the minute hand

The minute hand sweeps a full 360 degrees every 60 minutes, so 6 degrees per minute. Position: minute * 6.

Step 3: Reduce to the shorter arc

Take the absolute difference of the two positions, then fold it back across the circle with min(diff, 360 - diff). That guarantees you always report the smaller of the two arcs, never more than 180.

The solution

Continuous angle calculation with circular fold
def clock_angle(hour: int, minute: int) -> float:
    hour_pos = (hour % 12) * 30 + minute * 0.5
    minute_pos = minute * 6
    diff = abs(hour_pos - minute_pos)
    return min(diff, 360 - diff)

Common follow-up questions