🐍 Python Problem 0ne

Problem: Square every digit of a number and concatenate them.
Example:
- 9119 → 811181 (since 9²=81, 1²=1, 1²=1, 9²=81 → "81"+"1"+"1"+"81")
- 765 → 493625 (49-36-25)

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My Solution:

def square_every_digit(num):
    num_str = str(num)
    result = ""
    for digit in num_str:
        squared = int(digit) ** 2
        result += str(squared)
    return int(result)

AI's Solution (via ChatGPT):

def square_every_digit(num):
    squared_digits = ''.join(str(int(digit) ** 2) for digit in str(num))
    return int(squared_digits)

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💡 Key Takeaway:
The AI uses join() with a generator expression to simplify the loop, making the code shorter and more "Pythonic". Both methods work, but join() is cleaner for string concatenation!

🔗 Sources: Codewars, ChatGPT

👉 Try it yourself!
How would YOU approach this problem?

#Python #Codewars #Coding #LearnPython

🚀 Problem 2

Problem: Determine if a string is an isogram (no repeating letters, case-insensitive).
Examples:
- "Dermatoglyphics" → True
- "moOse" → False (duplicate 'o' when lowercased)
- Empty string → True

🧑💻 My Approach:

def is_isogram(string):
    string = string.lower()
    for i in string:  
        if string.count(i) > 1: 
            return False  
    return True

✅ Works but…
- Checks each character's count, leading to O(n²) time complexity.
- Early exit if duplicates found, but inefficient for large strings.

🤖 AI's Optimized Approach:

def is_isogram(string):
    lowered = string.lower()
    return len(set(lowered)) == len(lowered)

🔥 Why Better?
1. O(n) Time: Converts string to a set (auto-removes duplicates).
2. Concise: One-liner using Python's built-in functions.
3. Readable: Leverages set properties for uniqueness check.

💡 Key Takeaways:
- Use Sets for Uniqueness: Perfect for checking duplicates.
- Case Handling: Always lowercase the string first.
- Edge Cases: Empty string is handled gracefully.

🔍 Compare Efficiency:
| Method | Time Complexity | Lines of Code |
|--------------|-----------------|---------------|
| Loop & Count | O(n²) | 5 |
| Set & Length | O(n) | 1 |

*Credits: Codewars problem + ChatGPT optimization.*

Try both methods and share your thoughts! 👇

#Python #CodingTips #Algorithms #Programming

🔷 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

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🧪 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.

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⚡️ 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.

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

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

📘 Day 3: Basic Arithmetic & Input/Output in Python
Let’s build the foundation of your programming skills with math operations and user input handling!

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🔢 Basic Arithmetic Operations

print(5 + 3)    # Addition → 8
print(10 - 4)   # Subtraction → 6
print(7 * 2)    # Multiplication → 14
print(15 / 3)   # Division → 5.0
print(15 // 2)  # Floor Division → 7
print(15 % 4)   # Modulus → 3
print(2 ** 3)   # Exponentiation → 8

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🧮 Order of Operations (PEMDAS)

print(2 + 3 * 4)      # Output: 14
print((2 + 3) * 4)    # Output: 20

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🗣 User Input and Output

name = input("What is your name? ")
print("Hello, " + name + "!")

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### 🔁 Convert Input to Numbers

num1 = int(input("Enter first number: "))
num2 = int(input("Enter second number: "))
print("Sum:", num1 + num2)

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✅ Practice Problems

1️⃣ Area of a Rectangle

length = float(input("Enter length: "))
width = float(input("Enter width: "))
area = length * width
print("Area:", area)

2️⃣ Square of a Number

num = int(input("Enter a number: "))
print("Square:", num ** 2)

3️⃣ Basic Arithmetic Calculator

num1 = float(input("Enter first number: "))
num2 = float(input("Enter second number: "))

print("Addition:", num1 + num2)
print("Subtraction:", num1 - num2)
print("Multiplication:", num1 * num2)
print("Division:", num1 / num2)

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🎯 Goal for Day 3
✅ Use Python’s arithmetic operators confidently
✅ Understand input/output with input() and print()
✅ Practice writing small interactive programs

—

👨‍💻 Keep coding! Day 4 coming soon...
#Python #Day3 #LearnToCode #ProgrammingBasics #BeginnerFriendly

🚶‍♂️ Problem 4: Cartesia City Walk

You live in the city of Cartesia, where roads form a perfect grid.
Your walk app gives you directions like ['n', 's', 'w', 'e'], each taking 1 minute and 1 block.

🕙 You’ve got 10 minutes and must return exactly to where you started.
✅ Return True if the walk takes 10 minutes and ends at the starting point.
❌ Otherwise, return False.

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🧠 Your Idea (Step-by-Step Position Tracking)

def is_valid_walk(walk):
    if len(walk) != 10:
        return False
    x, y = 0, 0
    for drn in walk:
        if drn == 'n':
            y += 1
        elif drn == 's':
            y -= 1
        elif drn == 'e':
            x += 1
        elif drn == 'w':
            x -= 1
    return x == 0 and y == 0

🔎 Why it works:
You simulate the entire walk and check if you’ve come back to the original (0,0) point.

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⚡️ AI’s Optimized Version (Count Matching)

def is_valid_walk(walk):
    if len(walk) != 10:
        return False
    return walk.count('n') == walk.count('s') and walk.count('e') == walk.count('w')

⚠️ Key Note:
This version does not simulate movement. It cleverly checks balance:
- 'n' must equal 's'
- 'e' must equal 'w'
If that’s true and the walk is 10 steps long → ✅ you're back on time and at the start.

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🧪 Example Tests

print(is_valid_walk(['n','s','n','s','n','s','n','s','n','s']))  # ✅ True
print(is_valid_walk(['w','e','w','e','w','e','w','e','w','e']))  # ✅ True
print(is_valid_walk(['n','n','n','s','s','s','e','w','e','w']))  # ✅ True
print(is_valid_walk(['n','n','n','s']))                          # ❌ False

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✅ Key Learning:
- Use coordinate tracking for clarity and understanding
- Use frequency counting for efficiency and elegance
- Always validate input length before logic

—

👨‍💻 Keep walking (and coding)!
#Python #Codewars #GridWalk #LearnToCode #DailyChallenge