🔷 Problem 3: A Square of Squares 🔷

👷 You love building blocks—especially square ones! But can you tell if the total number of blocks you have can form a perfect square?

🎯 Task:
Given an integer n, determine whether it's a *perfect square* (i.e., the square of an integer).

🧠 Examples:

-1 => False  
 0 => True  
 3 => False  
 4 => True  
25 => True  
26 => False

---

🧪 My Initial Solution (Not Efficient for Large Numbers)

def is_square(n): 
    def find_factors(n):
        factors = []
        for i in range(1, n + 1):
            if n % i == 0:
                factors.append(i)
        return factors
    factors = find_factors(n)

    for i in factors:
        if i * i == n:
            return True
    if n == 0:
        return True
    return False

⚠️ *Why this fails on Codewars:*
While logically correct, it’s too slow for large values of n, because it checks all factors up to n.

---

⚡️ AI/Optimal Solution (Codewars Accepted ✅)

import math

def is_square(n):
    if n < 0:
        return False
    sqrt_n = math.isqrt(n)  # Efficient integer square root
    return sqrt_n * sqrt_n == n

📌 Key Notes:
- math.isqrt(n) returns the integer square root efficiently (no floating point issues).
- Checking sqrt_n * sqrt_n == n is both fast and accurate.
- Negative numbers can't be perfect squares, so return False early.

✅ Best Practice: Use mathematical functions when available. They are optimized and make your code cleaner and faster.

---

💡 *Follow for more real-world coding challenges and clean solutions!*

#Python #Codewars #Math #PerfectSquare #LearnToCode #ProgrammingTips