Merge pull request #127 from BrianLusina/feat/math-perfect-squares

feat(math, perfect squares): least number pefect squares that sum to n

Author: BrianLusina

algorithms/search/binary_search/divide_chocolate/test_divide_chocolate.py

@@ -1,7 +1,10 @@
 import unittest
 from typing import List
 from parameterized import parameterized
-from algorithms.search.binary_search.divide_chocolate import maximize_sweetness, maximize_sweetness_2
+from algorithms.search.binary_search.divide_chocolate import (
+    maximize_sweetness,
+    maximize_sweetness_2,
+)
 
 TEST_CASES = [
     ([1, 2, 3, 4, 5, 6, 7, 8, 9], 5, 6),

pymath/perfect_square/README.md

@@ -0,0 +1,51 @@
+# Perfect Squares
+
+Given an integer, n, return the least number of perfect square numbers that sum to n.
+
+> A perfect square is an integer that is the square of an integer. In other words, it is an integer that is the result of 
+> multiplying a whole integer by itself. For example, 1, 4, 9 and 16 are perfect squares, but 3, 5, and 11 are not.
+
+## Constraints 
+
+- 1 <= `n` < 10^3
+
+## Solution
+
+One of the first solutions that comes to mind is to keep subtracting perfect squares (like 9, 16, 25, …) until we reach 
+zero, counting how many times it takes. But that greedy idea doesn’t always find the smallest number of squares. For 
+example, trying the biggest square each time may miss a better combination. Instead, this problem has a clever 
+mathematical shortcut that utilizes some deep results from number theory, specifically the Four-Square theorem and the 
+Three-Square theorem.
+
+The Four-Square theorem says every number can be written as the sum of at most four perfect squares. So, the answer will
+always be one of 1, 2, 3, or 4. The Three-Square theorem tells us that some numbers can’t be expressed as the sum of 
+three squares, and these are exactly the numbers that look like 4^a(8b+7) That means, if a number (after dividing out 
+all factors of 4) is of the form 8b+7, then it needs four squares. Using these ideas, we can build a simple check-and-decide
+algorithm instead of trying all combinations. First, we remove all factors of 4 from the number, because multiplying or 
+dividing by 4 doesn’t change how many squares are needed; it just scales them. Then, we check the remainder when divided 
+by 8. If it’s 7, the number must have four squares. Otherwise, we check if it’s already a perfect square (then the answer
+is 1). If not, we test if it can be written as the sum of two perfect squares (then the answer is 2). If none of those 
+conditions are true, we know from the theorems that it must be 3. So, rather than testing every combination, this 
+approach uses mathematical reasoning to narrow the answer step by step, making it very fast and elegant.
+
+Let’s look at the algorithm steps:
+
+- Keep dividing n by 4 while it is divisible by 4. This simplifies the number without changing the answer. If a number 
+  is built from perfect squares, then four times that number is built from the same squares, just doubled. So, dividing 
+  by 4 doesn’t affect how many squares we need; it only makes the number smaller to work with.
+- If the reduced number has a remainder of 7 when divided by 8 (n % 8 == 7), return 4 immediately, because it must need
+  four squares.
+- Check if n is a perfect square itself. If yes, return 1.
+  - Try to write n as a sum of two perfect squares. Iterate over all possible i from 1 to √n, and check if n - i² is also 
+  a perfect square. If such a pair exists, return 2.
+- If none of the above conditions are true, return 3. By elimination, the number can be expressed as the sum of three 
+  squares.
+
+### Time Complexity
+
+We check if the number can be decomposed into the sum of two squares, which takes O(sqrt(n)) iterations. In the remaining 
+cases, we perform the check in constant time.
+
+### Space Complexity
+
+The algorithm consumes a constant space, regardless of the size of the input number, so O(1).

pymath/perfect_square/__init__.py

@@ -1,9 +1,87 @@
 from math import sqrt
 
 
-# function to check if a number is a perfect square
-def is_square(n):
+def is_square(n: int) -> bool:
+    """
+    Checks if a number is a perfect square.
+    Args:
+        n (int): The number to check.
+    Returns:
+        bool: True if n is a perfect square, False otherwise.
+    """
     if n < 0:
         return False
     else:
         return sqrt(n).is_integer()
+
+
+def is_perfect_square(n: int) -> bool:
+    """
+    Checks if a number is a perfect square.
+    Args:
+        n (int): The number to check.
+    Returns:
+        bool: True if n is a perfect square, False otherwise.
+    """
+    if n < 0:
+        return False
+    sqrt_num = int(sqrt(n))
+    return sqrt_num * sqrt_num == n
+
+
+def num_squares(n: int) -> int:
+    """
+    Finds the least number of perfect square numbers that sum to n.
+    Args:
+        n (int): The target number to find the least number of perfect square numbers that sum to it.
+    Returns:
+        int: The least number of perfect square numbers that sum to n.
+    """
+    if n < 0:
+        raise ValueError("n must be non-negative")
+    if n == 0:
+        return 0
+
+    dp = [float("inf")] * (n + 1)
+    dp[0] = 0
+
+    for i in range(1, n + 1):
+        j = 1
+        while j * j <= i:
+            dp[i] = min(dp[i], dp[i - j * j] + 1)
+            j += 1
+
+    return dp[n]
+
+
+def num_squares_2(n: int) -> int:
+    """
+    Finds the least number of perfect square numbers that sum to n.
+    Args:
+        n (int): The target number to find the least number of perfect square numbers that sum to it.
+    Returns:
+        int: The least number of perfect square numbers that sum to n.
+    """
+    if n < 0:
+        raise ValueError("n must be non-negative")
+    if n == 0:
+        return 0
+
+    # Apply reduction by removing factors of 4
+    while n % 4 == 0:
+        n = n // 4
+
+    # Check if n is of form (8k + 7)
+    if n % 8 == 7:
+        return 4
+
+    # Check if n itself is a perfect square
+    if is_perfect_square(n):
+        return 1
+
+    # Check if n is the sum of two perfect squares
+    for value in range(1, int(sqrt(n)) + 1):
+        if is_perfect_square(n - value * value):
+            return 2
+
+    return 3

pymath/perfect_square/test_perfect_squares.py

@@ -0,0 +1,27 @@
+import unittest
+from parameterized import parameterized
+from pymath.perfect_square import num_squares, num_squares_2
+
+
+TEST_CASES = [
+    (1, 1),
+    (12, 3),
+    (13, 2),
+    (23, 4),
+    (997, 2),
+]
+
+
+class NumOfPerfectSquaresTestCases(unittest.TestCase):
+    @parameterized.expand(TEST_CASES)
+    def test_num_of_perfect_squares(self, n: int, expected: int):
+        actual = num_squares(n)
+        (self.assertEqual(expected, actual) @ parameterized.expand(TEST_CASES))
+
+    def test_num_of_perfect_squares_2(self, n: int, expected: int):
+        actual = num_squares_2(n)
+        self.assertEqual(expected, actual)
+
+
+if __name__ == "__main__":
+    unittest.main()