12-Hour Clock Multiplication

Given two positive integers a and b, find the product of the two numbers on a 12-hour clock rather than a number line.

Note: Assume the Clock starts from 0 hours to 11 hours.

Examples:

Input: a = 2, b = 3
Output: 6
Explanation: 2*3 = 6. The time in a 12 hour clock is 6.

Input: a = 3, b = 5
Output: 3
Explanation: 3*5 = 15. The time in a 12 hour clock is 3.

[Expected Approach] Modular Multiplication - O(1) Time and O(1) Space

The idea is to reduce both numbers modulo 12 before multiplication. This avoids overflow for large values and gives the same result because (a * b) % 12 = ((a % 12) * (b % 12)) % 12.

#include <iostream>
using namespace std;

int mulClock(int a, int b) {
    
    // Reduce both numbers first to avoid overflow.
    return (a % 12) * (b % 12) % 12;
}

int main() {
    cout << mulClock(2, 3) << endl;
    cout << mulClock(3, 5) << endl;
    return 0;
}

class GFG {
    static int mulClock(int a, int b) {
        
        // Reduce both numbers first to avoid overflow.
        return (a % 12) * (b % 12) % 12;
    }

    public static void main(String[] args) {
        System.out.println(mulClock(2, 3));
        System.out.println(mulClock(3, 5));
    }
}

def mulClock(a, b):
    
    # Reduce both numbers first to avoid overflow.
    return (a % 12) * (b % 12) % 12


if __name__ == "__main__":
    print(mulClock(2, 3))
    print(mulClock(3, 5))

using System;

class GFG {
    static int mulClock(int a, int b) {
        
        // Reduce both numbers first to avoid overflow.
        return (a % 12) * (b % 12) % 12;
    }

    static void Main() {
        Console.WriteLine(mulClock(2, 3));
        Console.WriteLine(mulClock(3, 5));
    }
}

function mulClock(a, b) {
    
    // Reduce both numbers first to avoid overflow.
    return ((a % 12) * (b % 12)) % 12;
}

// Driver Code
console.log(mulClock(2, 3));
console.log(mulClock(3, 5));

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