Upasana | July 24, 2020 | 2 min read | 794 views | Java Coding Challenges
We can write a simple Java program to check if the given input is armstrong number or not.
Here is the approach:
Count the number of digits in a number
extract digits from input number one by one using remainder operation
calculate the sum of powers of each digit
if the sum of powers is same as the input itself, then number is armstrong else not.
public class Armstrong {
public boolean check(int input) {
int temp, digit, sumOfPower = 0;
temp = input;
int digits = countDigit(input);
while (temp != 0) {
digit = temp % 10;
System.out.println("Current Digit is " + digit);
sumOfPower = sumOfPower + (int) Math.pow(digit, digits);
System.out.println("Current sumOfPower is " + sumOfPower);
temp /= 10;
}
return sumOfPower == input;
}
static int countDigit(long n) {
return (int) Math.floor(Math.log10(n) + 1);
}
}We know that only 4 armstrong numbers that are of 3 digits (i.e. from 100 to 999), we can write a simple JUnit test to assert the same.
import org.junit.Test;
import static org.hamcrest.core.IsEqual.equalTo;
import static org.junit.Assert.*;
public class ArmstrongTest {
@Test
public void check() {
Armstrong checker = new Armstrong();
assertThat(checker.check(153), equalTo(true));
assertThat(checker.check(370), equalTo(true));
assertThat(checker.check(371), equalTo(true));
assertThat(checker.check(407), equalTo(true));
assertThat(checker.check(8208), equalTo(true));
}
}That’s all.
