Step by Step solution about How to resolve the algorithm First power of 2 that has leading decimal digits of 12 step by step in the Java programming language

Explanation:

1. Introduction:

   • The provided Java code calculates the first power of two (to the base 10) that equals a given integer l when multiplied by n. It's designed to find the smallest test value (representing the power) that satisfies the condition val = l.

2. Main Method:

   • In the main method:
     • It calls runTest with different pairs of l and n values to demonstrate the functionality of the p method.

3. runTest Method:

   • Prints the input parameters l and n along with the result of calling p(l, n).

4. p Method:

   • This method takes two parameters: l (the integer to match) and n (the number of times to multiply by a power of two).

5. Initialization:

   • test is initialized to 0 (representing the starting power).
   • Various constants and variables are calculated, including log (log base 10 of 2), factor (used for scaling numbers), and loop (used for determining the appropriate scaling factor).

6. Loop to Calculate Power:

   • The loop continues as long as n is greater than 0.
   • It increments test by 1 in each iteration.
   • Calculates val as factor * 10^(test * log % 1) to get a power of two based on the current test value.
   • If val equals l, it means a suitable power of two is found, and n is decremented by 1.

7. Returning the Result:

   • After the loop completes, test represents the smallest power of two that satisfies the condition for the given parameters. It is returned as the result.

Example: If we call p(12, 2), it calculates the smallest power of two (in base 10) that, when multiplied by 2, equals 12. In this case, 8 * 2 = 16, which is the first power of two that matches the condition. Therefore, p(12, 2) returns 8.

public class FirstPowerOfTwo {

    public static void main(String[] args) {
        runTest(12, 1);
        runTest(12, 2);
        runTest(123, 45);
        runTest(123, 12345);
        runTest(123, 678910);
    }
    
    private static void runTest(int l, int n) {
        System.out.printf("p(%d, %d) = %,d%n", l, n, p(l, n));
    }
    
    public static int p(int l, int n) {
        int test = 0;
        double log = Math.log(2) / Math.log(10);
        int factor = 1;
        int loop = l;
        while ( loop > 10 ) {
            factor *= 10;
            loop /= 10;
        }
        while ( n > 0) {
            test++;
            int val = (int) (factor * Math.pow(10, test * log % 1));
            if ( val == l ) {
                n--;
            }
        }
        return test;
    }
    
}