feat(138 day): calculate sum of all divisors with optimized sqrt-based approach

Author: JohnSunny21

FreeCodeCamp/CodingQ/SumOfDivisors.py

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+"""  
+Sum of Divisors
+Given a positive integer, return the sum of all its divisors.
+
+A divisor is any integer that divides the number evenly (the remainder is 0).
+Only count each divisor once.
+For example, given 6, return 12 because the divisors of 6 are 1, 2, 3, and 6, and the sum of those is 12.
+
+"""
+
+import unittest
+
+class SumOfDivisorsTest(unittest.TestCase):
+
+    def test1(self):
+        self.assertEqual(sumDivisors(6), 12)
+
+    def test2(self):
+        self.assertEqual(sumDivisors(13), 14)
+
+    def test3(self):
+        self.assertEqual(sumDivisors(28), 56)
+
+    def test4(self):
+        self.assertEqual(sumDivisors(84), 224)
+
+    def test5(self):
+        self.assertEqual(sumDivisors(549), 806)
+
+    def test6(self):
+        self.assertEqual(sumDivisors(9348), 23520)
+
+
+
+def sumDivisors(n):
+    total = 0
+    for i in range(1, n + 1):
+        if n % i == 0:
+            total += i
+    return total
+
+import math
+def sumDivisors_optimized(n):
+    total = 0
+    for i in range(1, int(math.sqrt(n)) + 1):
+        if n % i == 0:
+            total += i
+            if i != n // i:     # avoid double-counting square roots
+                total += n // i
+    return total 
+
+
+"""  
+
+When finding divisors of a number n, they always come in pairs:
+    => if i divides n, then n / i is also a divisor.
+    => Example: for n = 36:
+        -> i = 2 -> divisor pair (2, 18)
+        -> i = 3 -> divisor pair (3, 12)
+        -> i = 6 -> divisor pair (6, 6) (special case)
+
+    => So instead of looping all the way up to n, you only need to loop up to sqrt(n):
+        -> Because beyond sqrt(n), divisors are just the "other half" of the pairs you already found.
+        -> This reduces complexity from O(n) to O(√n), which is much faster for large numbers.
+
+=> Brute force: loop from 1 to n.
+=> Optimized: loop to sqrt(n) and add divisor pairs.
+=> Both approaches give the correct sum, but the optimized one is much faster for large numbers.
+"""
+
+
+
+if __name__ == "__main__":
+    print(sumDivisors(15))
+    unittest.main()