LeetCode #1344 — MEDIUM

Angle Between Hands of a Clock

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

12
30

Example 2

3
30

Example 3

3
15

Step 02

Core Insight

What unlocks the optimal approach

The tricky part is determining how the minute hand affects the position of the hour hand.
Calculate the angles separately then find the difference.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1344: Angle Between Hands of a Clock
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);
    }
}

// Accepted solution for LeetCode #1344: Angle Between Hands of a Clock
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)
}

# Accepted solution for LeetCode #1344: Angle Between Hands of a Clock
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)

// Accepted solution for LeetCode #1344: Angle Between Hands of a Clock
struct Solution;

impl Solution {
    fn angle_clock(hour: i32, minutes: i32) -> f64 {
        let h = ((hour % 12) as f64 + (minutes as f64 / 60.0)) * 30.0;
        let m = minutes as f64 * 6.0;
        let a = (h - m).abs();
        if a > 180.0 {
            360.0 - a
        } else {
            a
        }
    }
}

#[test]
fn test() {
    use assert_approx_eq::assert_approx_eq;
    let hour = 12;
    let minutes = 30;
    let res = 165.0;
    assert_approx_eq!(Solution::angle_clock(hour, minutes), res);
    let hour = 3;
    let minutes = 30;
    let res = 75.0;
    assert_approx_eq!(Solution::angle_clock(hour, minutes), res);
    let hour = 3;
    let minutes = 15;
    let res = 7.5;
    assert_approx_eq!(Solution::angle_clock(hour, minutes), res);
    let hour = 4;
    let minutes = 50;
    let res = 155.0;
    assert_approx_eq!(Solution::angle_clock(hour, minutes), res);
    let hour = 12;
    let minutes = 0;
    let res = 0.0;
    assert_approx_eq!(Solution::angle_clock(hour, minutes), res);
    let hour = 1;
    let minutes = 57;
    let res = 76.5;
    assert_approx_eq!(Solution::angle_clock(hour, minutes), res);
}

// Accepted solution for LeetCode #1344: Angle Between Hands of a Clock
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);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.