refactor: improving GenericRoot (#6365)

refactor: improving GenericRoot
This commit is contained in:
Oleksandr Klymenko
2025-07-12 09:51:49 +03:00
committed by GitHub
parent 7590d8234f
commit dba2d869f2

View File

@@ -1,30 +1,48 @@
package com.thealgorithms.maths;
/*
* Algorithm explanation:
* https://technotip.com/6774/c-program-to-find-generic-root-of-a-number/#:~:text=Generic%20Root%3A%20of%20a%20number,get%20a%20single%2Ddigit%20output.&text=For%20Example%3A%20If%20user%20input,%2B%204%20%2B%205%20%3D%2015.
/**
* Calculates the generic root (repeated digital sum) of a non-negative integer.
* <p>
* For example, the generic root of 12345 is calculated as:
* 1 + 2 + 3 + 4 + 5 = 15,
* then 1 + 5 = 6, so the generic root is 6.
* <p>
* Reference:
* https://technotip.com/6774/c-program-to-find-generic-root-of-a-number/
*/
public final class GenericRoot {
private static final int BASE = 10;
private GenericRoot() {
}
private static int base = 10;
/**
* Computes the sum of the digits of a non-negative integer in base 10.
*
* @param n non-negative integer
* @return sum of digits of {@code n}
*/
private static int sumOfDigits(final int n) {
assert n >= 0;
if (n < base) {
if (n < BASE) {
return n;
}
return n % base + sumOfDigits(n / base);
return (n % BASE) + sumOfDigits(n / BASE);
}
/**
* Computes the generic root (repeated digital sum) of an integer.
* For negative inputs, the absolute value is used.
*
* @param n integer input
* @return generic root of {@code n}
*/
public static int genericRoot(final int n) {
if (n < 0) {
return genericRoot(-n);
int number = Math.abs(n);
if (number < BASE) {
return number;
}
if (n > base) {
return genericRoot(sumOfDigits(n));
}
return n;
return genericRoot(sumOfDigits(number));
}
}