Java Ice Breakers

How to check whether a number is Disarium or not

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by

anagendrakumar2020
in ,

This is the first solution to check the number for Disarium. This solution uses String class, and a couple of Character functions to evaluate the sum of powers. The Disarium number is say for a given number 351, evaluate the powers from the left side as power (3 * 1) + power (5 * 2) + power (1 *3) , the sum of all powers must be equal to the given number 351.

import java.util.Scanner;

public class Disarium {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner inpInt= new Scanner (System.in);
		int sumCheck=0;
		System.out.println("Enter an integer value");
		int numValue=inpInt.nextInt();

		// Convert the number to String for easy processing and to avoid the second loop
		String numToStr=String.valueOf(numValue);
		int numLen=numToStr.length();

		// Calculate the sum of each digit's power value from high(=length of number) to 1
		for (;numLen>0;numLen--)
			if(Character.isDigit(numToStr.charAt(numLen-1))) 
				sumCheck = (int) (sumCheck + Math.pow(Character.getNumericValue(numToStr.charAt(numLen-1)),numLen));

		//Compare the sumCheck with actual number		
		if (sumCheck == numValue)
			System.out.println("The given number "+numValue+" is Disarium");
		else
			System.out.println("The given number "+numValue+" is not Disarium");

		inpInt.close();
	}
}
Try it >>
Output:
Enter an integer value
135
The given number 135 is Disarium

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