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This approach involves calculating the angles made by the hour hand and the minute hand with respect to 12:00 and then finding the smallest angle between them. The minute hand moves 6 degrees per minute (360 degrees / 60 minutes), and the hour hand moves 30 degrees per hour (360 degrees / 12 hours) plus an additional 0.5 degrees per minute. The difference between these angles gives us the desired result.
Time Complexity: O(1) - The computation involves a fixed number of arithmetic operations.
Space Complexity: O(1) - No additional space is required beyond fixed-size variables.
#include <cmath>
class Solution {
public:
double angleClock(int hour, int minutes) {
double minute_angle = minutes * 6;
double hour_angle = (hour % 12) * 30 + minutes * 0.5;
double angle = fabs(minute_angle - hour_angle);
return angle > 180 ? 360 - angle : angle;
}
};
int main() {
Solution sol;
std::cout << std::fixed << sol.angleClock(12, 30) << std::endl; // Output: 165.00000
std::cout << std::fixed << sol.angleClock(3, 30) << std::endl; // Output: 75.00000
std::cout << std::fixed << sol.angleClock(3, 15) << std::endl; // Output: 7.50000
return 0;
}Solve with full IDE support and test cases
The C++ solution employs a similar logic to the C solution, encapsulated in a class structure for encapsulation. It uses fixed point notation for output format. The method calculates angles for the hour and minute hands, determines the absolute difference, and computes the smallest angle.
This alternative approach converts the movement of the clock hands into their equivalent rotations, effectively translating this into angles. The goal is to determine the position of both hands as angles relative to the 12 o'clock position and compute the minimal angular difference.
Time Complexity: O(1) - Fixed-time operations based on input.
Space Complexity: O(1) - Uses a constant number of variables.
1public class Solution {
2 public double angleClock(int hour, int minutes) {
3 int totalMinutes = hour * 60 + minutes;
4 double minuteAngle = (totalMinutes % 60) * 6.0;
5 double hourAngle = (totalMinutes % 720) * 0.5;
6 double angle = Math.abs(minuteAngle - hourAngle);
7 return angle > 180 ? 360 - angle : angle;
8 }
9
10 public static void main(String[] args) {
11 Solution sol = new Solution();
12 System.out.printf("%.5f\n", sol.angleClock(12, 30)); // Output: 165.00000
13 System.out.printf("%.5f\n", sol.angleClock(3, 30)); // Output: 75.00000
14 System.out.printf("%.5f\n", sol.angleClock(3, 15)); // Output: 7.50000
15 }
16}Java variant of the solution follows a reduction of the problem into completed 12-hour period cycles for the hour hand, adjusted by total minutes. Modular arithmetic aids in translating these cyclic movements into angular displacements relative to the top of the hour.