TheAlgorithms · debasishbsws · Nov 1, 2022 · Oct 31, 2022 · Oct 31, 2022 · Oct 31, 2022
@@ -6,27 +6,19 @@
  * has divisors 1, 2 and 3 (excluding itself), and 1 + 2 + 3 = 6, so 6 is a
  * perfect number.
  *
- * <p>
  * link:https://en.wikipedia.org/wiki/Perfect_number
  */
 public class PerfectNumber {
 
-    public static void main(String[] args) {
-        assert isPerfectNumber(6);
-        /* 1 + 2 + 3 == 6 */
-        assert !isPerfectNumber(8);
-        /* 1 + 2 + 4 != 8 */
-        assert isPerfectNumber(28);
-        /* 1 + 2 + 4 + 7 + 14 == 28 */
-    }
-
     /**
      * Check if {@code number} is perfect number or not
      *
      * @param number the number
      * @return {@code true} if {@code number} is perfect number, otherwise false
      */
     public static boolean isPerfectNumber(int number) {
+        if (number <= 0)
+            return false;
         int sum = 0;
         /* sum of its positive divisors */
         for (int i = 1; i < number; ++i) {
@@ -36,4 +28,38 @@ public static boolean isPerfectNumber(int number) {
         }
         return sum == number;
     }
+
+    /**
+     * Check if {@code n} is perfect number or not
+     *
+     * @param n the number
+     * @return {@code true} if {@code number} is perfect number, otherwise false
+     */
+    public static boolean isPerfectNumber2(int n) {
+        if (n <= 0)
+            return false;
+        int sum = 1;
+        double root = Math.sqrt(n);
+
+        /*
+         * We can get the factors after the root by dividing number by its factors
+         * before the root.
+         * Ex- Factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50 and 100.
+         * Root of 100 is 10. So factors before 10 are 1, 2, 4 and 5.
+         * Now by dividing 100 by each factor before 10 we get:
+         * 100/1 = 100, 100/2 = 50, 100/4 = 25 and 100/5 = 20
+         * So we get 100, 50, 25 and 20 which are factors of 100 after 10
+         */
+        for (int i = 2; i <= root; i++) {
+            if (n % i == 0) {
+                sum += i + n / i;
+            }
+        }
+
+        // if n is a perfect square then its root was added twice in above loop, so subtracting root from sum
+        if (root == (int) root)
+            sum -= root;
+
+        return sum == n;
+    }
 }
@@ -0,0 +1,21 @@
+package com.thealgorithms.maths;
+
+import static org.junit.jupiter.api.Assertions.*;
+import org.junit.jupiter.api.Test;
+
+class PerfectNumberTest {
+
+    @Test
+    public void perfectNumber() {
+        int trueTestCases[] = { 6, 28, 496, 8128, 33550336 };
+        int falseTestCases[] = { -6, 0, 1, 9, 123 };
+        for (Integer n : trueTestCases) {
+            assertTrue(PerfectNumber.isPerfectNumber(n));
+            assertTrue(PerfectNumber.isPerfectNumber2(n));
+        }
+        for (Integer n : falseTestCases) {
+            assertFalse(PerfectNumber.isPerfectNumber(n));
+            assertFalse(PerfectNumber.isPerfectNumber2(n));
+        }
+    }
+}