formating guide

functions, classes => camelCase

files, variables=> snake_case

1

Welcome to the world of Python! After your first hour, you’ve likely learned about variables, basic math, user input, and perhaps how to print strings.

Here is a classic "First Hour" challenge that combines all those elements.

---

The Challenge: The "Dinner Bill Splitter"

Write a program that helps a group of friends calculate how much each person owes at a restaurant, including a tip.

Your Goal:

1. Ask the user for the total amount of the bill (e.g., 50.00).
2. Ask the user what percentage they want to tip (e.g., 15).
3. Ask the user how many people are splitting the bill.
4. Calculate the total bill (bill + tip) and then divide it by the number of people.
5. Print the final amount each person should pay.

Things to remember:

• When you use input(), Python treats the data as a string (text). You will need to convert the bill and tip to a float (decimal number) and the number of people to an int (integer) to do math.
• The Math: To get the tip amount, you can use the formula:

---

Example Output:

If the bill is $100, the tip is 18%, and there are 2 people, your code should output something like: Each person should pay: $59.0

---

Great job on the first challenge! If you’ve mastered variables and basic math, your next hour is likely spent learning about If/Else statements (logic) and perhaps your first List or String manipulation.

In "LeetCode" style, the problem focuses less on user interaction and more on a specific logical goal with defined inputs and outputs.

---

2

LeetCode Style Challenge: "The Password Validator"

Problem Statement:

You are building a login system. A valid password must meet two specific criteria:

1. It must be at least 8 characters long.
2. It must not contain any spaces.

Write a script that takes a string variable called password and determines if it is "Valid" or "Invalid."

Constraints:

• Input: A string (e.g., password = "Python123")
• Output: Print the word "Valid" if both conditions are met. Print "Invalid" otherwise.

---

Key Concepts to Use:

• len(): This function counts how many characters are in a string (e.g., len("Hi") is 2).
• in keyword: You can check if a character exists in a string using if " " in password:.
• if / elif / else: To handle the logic flow.

Example Test Cases:

Input	Expected Output
"PythonRocks"	Valid
"12345"	Invalid (Too short)
"Secret Code"	Invalid (Contains a space)

---

3

Excellent progress! If you've just finished your second or third hour, you’ve likely encountered Loops (for or while) and Lists.

This is where Python gets really powerful—handling "collections" of data instead of just one number or string at a time.

---

Challenge: The "Temperature Analyzer"

Problem Statement:

You are given a list of daily high temperatures for a week. You need to write a program that analyzes this data to help a gardener decide if their plants are safe.

Your script should:

1. Loop through the list of temperatures.
2. Count how many days were "Freezing" (below 32°F).
3. Find the highest temperature in the list.
4. Print both the count of freezing days and the max temperature.

Constraints:

• Input: A list of integers, e.g., temps = [35, 28, 40, 31, 30, 45, 50]
• Output: A message like: "There were 3 freezing days. The weekly high was 50."

---

Key Concepts to Use:

• For Loops: To look at each number in the list one by one.
• Counters: A variable (like count = 0) that you increment inside the loop using count = count + 1.
• Comparison Logic: Using > or < inside an if statement to compare the current temperature to your "max" variable.

Tips for Success:

• To find the "Max" without using built-in shortcuts, create a variable called highest and set it to the first number in the list. Every time you find a number bigger than highest, update it!
• Python's indentation is critical—make sure your if statements are indented inside your for loop.

---

4

You are moving fast! After about four hours of Python, you’ve likely reached one of the most important milestones in programming: Functions.

Functions allow you to wrap your code into reusable "tools" so you don't have to write the same logic over and over.

---

Challenge: The "Currency Converter Function"

Problem Statement:

Imagine you are building a travel app. You need a function that converts USD to Euros, but it also needs to subtract a small transaction fee.

Your task: Create a function called convert_currency that:

1. Takes two parameters: amount (USD) and rate (the exchange rate).
2. Calculates the converted amount: .
3. Subtracts a **flat fee of ** from the final result.
4. Returns the final value.

Constraints:

• If the result of the conversion is less than , the function should return 0 (because the fee covers the whole amount).
• You must use the return keyword.

---

Example Test Cases:

# Calling your function should look like this:
print(convert_currency(100, 0.92))  # Expected: 90.0 (100 * 0.92 - 2)
print(convert_currency(1, 0.92))    # Expected: 0    (Conversion is less than 2)

Why this is the "Next Level":

• Scope: Variables created inside a function stay inside the function.
• Return vs. Print: This is a common point of confusion. print() shows a value to the human, but return gives the value back to the computer to use in other calculations.

---

5

You are flying through the fundamentals! By the fifth hour, you’ve likely encountered Dictionaries (key-value pairs) and perhaps how to combine them with Lists. This is where you move from simple variables to managing actual "data."

---

Challenge: The "Inventory Manager"

Problem Statement:

You are managing a small grocery store. You have a Dictionary where the key is the name of the fruit and the value is the quantity currently in stock.

Your task: Write a program that takes an inventory dictionary and a "restock" list. It should update the quantities in the dictionary based on the list.

Constraints:

1. Start with this dictionary: inventory = {"apples": 5, "bananas": 12, "oranges": 0}
2. You are given a list of items that just arrived: shipment = ["apples", "oranges", "apples", "apples"]
3. Loop through the shipment list.
4. For every item in the shipment, increase its count in the inventory dictionary by 1.
5. Print the final dictionary.

---

Key Concepts to Use:

• Dictionary Access: How to get or update a value using its key (e.g., inventory["apples"] += 1).
• Membership Checking: Using if item in inventory: to make sure you don't try to update a fruit that isn't in your store.
• Nested Logic: Using a for loop to go through the list, and an if statement inside that loop to update the dictionary.

Example Output:

After processing the shipment above, your inventory should look like this: {'apples': 8, 'bananas': 12, 'oranges': 1}

---

Why this is a milestone:

In the real world, data rarely comes in a single variable. Most apps (like Instagram or Amazon) work by looping through lists of dictionaries. Mastering this "nesting" is a huge step toward building real apps.

---

6

You are making incredible time! By the sixth hour, you are likely moving beyond "how to write code" and into "how to handle real-world data." This usually involves List Comprehensions (a shorter way to write loops) or handling Nested Data (lists inside of lists).

---

Challenge: The "Classroom Grader"

Problem Statement:

You have a list of students, where each student is represented by a list containing their name and their test score. Your job is to "filter" this data.

Your task: Create a new list called honors_students that contains only the names of students who scored 90 or higher.

Constraints:

1. Input: A nested list like this: students = [["Alice", 95], ["Bob", 78], ["Charlie", 92], ["David", 85]]
2. Use a For Loop (or a List Comprehension if you’ve learned it!) to check each score.
3. Extract only the name (the first element of the inner list).
4. Print the final honors_students list.

---

Key Concepts to Use:

• Double Indexing: If student is ["Alice", 95], then student[0] is the name and student[1] is the score.
• List Appending: Using .append() to add a name to your new list.
• Filtering Logic: An if statement to check if the score meets the requirement.

Example Output:

Based on the input above, your output should be: ['Alice', 'Charlie']

---

Why this is a milestone:

In data science and web development, data almost always arrives in "nested" formats (like JSON). Learning how to "dig" into a list to pull out a specific piece of information is a superpower.

---

7

You are entering the "intermediate" beginner phase! After about 7 hours of study, you are likely exploring String Methods and basic Data Cleaning. Real-world data is often "dirty" (extra spaces, mixed capitalization, or weird characters), and Python is the king of cleaning it up.

---

Challenge: The "Clean Tweet Hashtags"

Problem Statement:

You are building a social media tool. You are given a list of messy hashtags entered by users. You need to "clean" them so they can be stored in a database.

Your task: Write a program that takes a list of strings and transforms them into a clean format.

The Rules for "Cleaning":

1. Remove any leading or trailing whitespace.
2. Convert everything to lowercase.
3. Ensure every string starts with a #. If it doesn't have one, add it.
4. Skip any strings that are empty after cleaning.

Constraints:

• Input: tags = [" Python ", "coding", "#DataScience", " ", "sql "]
• Output: A new list: ["#python", "#coding", "#datascience", "#sql"]

---

Key Concepts to Use:

• .strip(): Removes spaces from the start and end of a string.
• .lower(): Converts the string to lowercase.
• .startswith("#"): Checks if the first character is a hashtag.
• String Concatenation: Using "# " + tag to add the symbol.

---

Example Test:

raw_tags = ["  LEARN ", "#growth", " teaching", " "]
# Your code should result in:
# ["#learn", "#growth", "#teaching"]

Why this is a milestone:

This is the first time you are mutating data (changing its form) and filtering data (removing the empty space) at the same time. This is the bread and butter of "Data Pre-processing."

---

8

You are absolutely crushing this! After about 8 hours of Python, you’ve likely reached the point where you need to handle Errors and Edge Cases. In the real world, programs crash because users enter the wrong thing or data is missing.

Your next milestone is learning how to make your code "bulletproof" using Try/Except blocks and Type Checking.

---

Challenge: The "Robust Calculator"

Problem Statement:

You are building a division tool. Normally, division is easy, but two things can break a program:

1. A user tries to divide by Zero (which is mathematically impossible).
2. A user enters Text instead of a number.

Your task: Write a function called safe_divide that takes two arguments, a and b.

The Requirements:

• Try to divide a by b and return the result.
• If b is zero, don't let the program crash! Instead, return the string: "Error: Cannot divide by zero."
• If either input is not a number, return the string: "Error: Inputs must be numbers."

---

Constraints:

• Input: Two variables of any type.
• Tools: Use a try...except block or if statements with isinstance(a, (int, float)).

Example Test Cases:

print(safe_divide(10, 2))    # Expected: 5.0
print(safe_divide(10, 0))    # Expected: "Error: Cannot divide by zero."
print(safe_divide(10, "5"))  # Expected: "Error: Inputs must be numbers."

---

Why this is a milestone:

This is the shift from "writing code that works" to "writing professional code." Professional software spends about 20% of its logic on the task and 80% on handling what might go wrong.

---

Learn the difference between a ZeroDivisionError and a TypeError so you can catch them specifically

---

9

You have reached a significant milestone! After about 9 hours of Python, you are ready to tackle Efficiency and Frequency Counting. In the LeetCode world, this is a very common "Easy" level pattern.

This challenge moves away from just "cleaning" data and into algorithmic thinking.

---

LeetCode Style Challenge: "Find the Lone Survivor"

Problem Statement:

You are given a list of integers where every number appears exactly twice, except for one number which appears only once.

Your goal is to find that single, unique number.

Constraints:

• Input: A list of integers (e.g., nums = [2, 4, 6, 2, 4])
• Output: The integer that appears only once (e.g., 6)
• Goal: Solve this using a Dictionary to count occurrences.

---

Key Concepts to Use:

• Frequency Mapping: Create an empty dictionary. Loop through the list. If the number is already in the dictionary, increase its value. If not, add it with a value of 1.
• Dictionary Iteration: After the loop is finished, use .items() to loop through your dictionary and find the key that has a value of 1.

Example Test Cases:

Input	Expected Output
[1, 1, 2, 2, 3]	3
[10, 5, 10]	5
[9, 8, 7, 8, 9]	7

---

Why this is a LeetCode classic:

This is a variation of the "Single Number" problem on LeetCode. It teaches you how to use a Hash Map (which is what a Python Dictionary is) to optimize searching. Searching a dictionary is much faster than searching a list over and over again.

---

learn the "Counter" tool from the Python collections module, or build the frequency dictionary manually

---

10

You are now entering your 10th hour of Python! This is usually the time when learners start exploring Two-Pointer logic or List Reversal. These techniques are the bread and butter of technical interviews.

---

LeetCode Style Challenge: "The Palindrome Checker"

Problem Statement:

A palindrome is a word or phrase that reads the same backward as forward (e.g., "madam" or "racecar").

Your task is to write a function is_palindrome(s) that takes a string s and returns True if it is a palindrome and False otherwise.

Constraints:

• Input: A string (e.g., s = "radar")
• Output: A Boolean (True or False).
• The "Hour 10" Twist: You must ignore capitalization. "Madam" should be considered a palindrome.

---

Key Concepts to Use:

There are two main ways to solve this in Python. I recommend trying both!

1. The "Pythonic" Slicing Way: You can reverse a string instantly using the "step" slice: s[::-1].
2. The Algorithmic Way: Compare the first character with the last, the second with the second-to-last, and so on.

Example Test Cases:

Input	Expected Output
"Level"	True
"Python"	False
"stats"	True

---

Why this is a milestone:

On LeetCode, "Palindrome" problems are the gateway to String Manipulation. It forces you to think about how data is indexed. This is often the very first question asked in "screening" interviews for junior developer roles.

---

Pro-Tip:

Before checking if it's a palindrome, remember to use .lower() to make sure "L" matches "l".

Learn how the "step" slicing [::-1] actually works under the hood, or are you ready to write the function

---

11

You have officially hit the double-digit hours mark! By hour 11, you should be comfortable with list indexing, loops, and basic logic. Now it's time to introduce Array Transformation—a common pattern where you modify a list based on specific rules.

---

LeetCode Style Challenge: "FizzBuzz Array"

Problem Statement:

This is perhaps the most famous coding challenge in history. In a LeetCode environment, you wouldn't just print the results; you would return a list of strings.

Write a function fizz_buzz(n) that returns a list of strings from 1 to n, but:

1. For multiples of 3, add "Fizz" instead of the number.
2. For multiples of 5, add "Buzz" instead of the number.
3. For multiples of both 3 and 5, add "FizzBuzz".
4. For everything else, add the number itself as a string.

Constraints:

• Input: An integer n (e.g., n = 5).
• Output: A list of strings (e.g., ["1", "2", "Fizz", "4", "Buzz"]).
• Edge Case: Remember that the number 15 is the first "FizzBuzz".

---

Key Concepts to Use:

• The Modulo Operator (%): This returns the remainder of a division. If x % 3 == 0, then x is a multiple of 3.
• Order of Operations: Think carefully—should you check for 3, 5, or both first?
• String Conversion: Use str(i) to turn a number into a string before adding it to your list.

Example Test Cases:

Input	Expected Output
n = 3	["1", "2", "Fizz"]
n = 15	["1", "2", "Fizz", ... , "14", "FizzBuzz"]

---

Why this is a milestone:

"FizzBuzz" is the ultimate "filter" question. It tests if you understand conditional branching—making sure your code doesn't stop at the first "True" condition if a more specific condition (like 15) exists.

---

Learn why the order of your if and elif statements matters for this specific problem

---

12

You are moving at an impressive pace! By hour 12, you should be comfortable with list manipulation and logic. It’s time to tackle a Search and Optimization problem. This is a "LeetCode Easy" classic that introduces the concept of Efficiency.

---

LeetCode Style Challenge: "The Target Sum" (Two Sum Lite)

Problem Statement:

You are given a list of numbers and a target sum. You need to check if any two distinct numbers in that list add up exactly to the target.

Write a function has_target_sum(nums, target) that returns True if such a pair exists, and False otherwise.

Constraints:

• Input: A list of integers nums and an integer target.
• Output: A Boolean (True or False).
• Note: You cannot use the same element twice (e.g., if the target is 4 and the list is [2], the answer is False).

---

Key Concepts to Use:

1. Nested Loops: You will likely need one loop inside another. The first loop picks a number, and the second loop checks all the other numbers to see if they add up to the target.
2. Indexing: Use range(len(nums)) to make sure you aren't comparing a number to itself.

Example Test Cases:

Input	Target	Expected Output
[1, 4, 6, 10]	10	True (4 + 6)
[3, 2, 4]	6	True (2 + 4)
[1, 2, 3]	7	False

---

Why this is a milestone:

This is your first encounter with Time Complexity.

• A "Nested Loop" solution (looping inside a loop) is called ****.
• As the list gets longer, this approach becomes very slow.

Once you solve this with two loops, the next "hour" of learning will teach you how to do it much faster using a Set or a Dictionary.

---

Learn how to set up the nested range() loops so you don't accidentally count the same number twice

---

13

For your 13th hour, let's look at Sets. A Set in Python is like a dictionary without values; it allows you to check if an item exists almost instantly ().

---

LeetCode Style Challenge: "Contains Duplicate"

Problem Statement:

Given an integer array nums, return True if any value appears at least twice in the array, and return False if every element is distinct.

Constraints:

• Input: A list of integers nums.
• Goal: Solve this without using .count().
• Hint: Think about the properties of a set(). A set can only hold unique items. If you turn a list into a set, what happens to the length?

---

Key Concepts to Use:

1. The set() function: my_set = set(nums) creates a collection of only the unique numbers from the list.
2. Length Comparison: If the number of items in the set is smaller than the number of items in the original list, it means there were duplicates.

Example Test Cases:

Input	Expected Output
[1, 2, 3, 1]	True
[1, 2, 3, 4]	False
[1, 1, 1, 3, 3, 4]	True

---

Why this is a milestone:

This is the "aha!" moment for many programmers. You are learning that choosing the right data structure (a Set instead of a List) can make your code significantly faster and shorter. This is often solved in just 2 lines of code!

---

try writing this using the len() comparison trick, try a "one-by-one" approach where you add numbers to a set as you loop

---

14

You’ve officially unlocked Set Theory! Using set() to handle duplicates is a professional-grade optimization.

For your 14th hour, we are going to look at Array Manipulation and "In-Place" logic. This is a common LeetCode theme: instead of just finding something, you have to "clean up" a list according to a specific rule.

---

LeetCode Style Challenge: "Move Zeroes"

Problem Statement:

Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.

The Catch: On LeetCode, "Easy" problems often ask you to do this in-place, but for your 14th hour, it's perfectly fine to create a new list and return it.

Constraints:

• Input: nums = [0, 1, 0, 3, 12]
• Output: [1, 3, 12, 0, 0]
• Requirement: You cannot just "sort" the list (because sorting would put 1, 3, 12 in the wrong order if the input was [0, 12, 3]).

---

Key Concepts to Use:

1. Filtering: Create a new list (let's call it non_zeroes) that only contains the numbers that are NOT zero.
2. Counting: Keep track of how many zeroes you skipped.
3. Padding: Use the .append() method or list multiplication ([0] * count) to add the zeroes back to the end of your non_zeroes list.

Example Test Cases:

Input	Expected Output
[0, 1, 0, 3, 12]	[1, 3, 12, 0, 0]
[0, 0, 1]	[1, 0, 0]
[4, 5, 0, 1]	[4, 5, 1, 0]

---

Why this is a milestone:

This problem introduces Stable Filtering. You are learning how to extract specific data while preserving the "history" (the order) of the rest of the data. In data processing, maintaining the sequence of events while removing "noise" (the zeroes) is a fundamental task.

Learn how to use "List Multiplication" to create a list of zeroes instantly

---

15

You are now at Hour 15, and you’ve mastered basic filtering and set-based optimization. Now, it's time to learn how to search for a specific pattern within a collection. This is a "LeetCode Easy" classic that introduces Two-Pointer Logic.

---

LeetCode Style Challenge: "Is Subsequence"

Problem Statement:

Given two strings s and t, return True if s is a subsequence of t, or False otherwise.

A subsequence of a string is a new string formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of "abcde" while "aec" is not).

Constraints:

• Input: Two strings, s and t.
• Output: A Boolean (True or False).
• Goal: Use a loop to "walk" through string t and see if you can find the characters of s in order.

---

Key Concepts to Use:

1. The Pointer Method: Create a variable i = 0 to track your position in string s.
2. The Main Loop: Loop through every character in string t.
3. The Match: If the current character in t matches s[i], move your pointer i forward by 1.
4. The Finish Line: If i ever becomes equal to the length of s, it means you found all the characters in order!

Example Test Cases:

Input s	Input t	Expected Output
"abc"	"ahbgdc"	True
"axc"	"ahbgdc"	False (Missing 'x')
"ace"	"abcde"	True

---

Why this is a milestone:

This is your first Two-Pointer problem. Instead of comparing two items in the same list, you are coordinating movement across two different strings. This logic is used in everything from DNA sequencing to search engine "auto-complete" features.

---

Learn how to handle the "IndexError" that might happen if you move your pointer past the end of string s, or are you ready to code

---

16

You’ve reached Hour 16! You’ve shown you can handle two-pointer logic and array transformations. Now, we are going to test your ability to manage state and handle logic shifts within a single loop.

This is a classic "String Processing" problem that mimics how computers read documents.

---

LeetCode Style Challenge: "Length of Last Word"

Problem Statement:

Given a string s consisting of words and spaces, return the length of the last word in the string.

A word is defined as a maximal substring consisting of non-space characters only.

Constraints:

• Input: A string s (e.g., s = "Hello World ")
• Output: An integer (e.g., 5)
• The Twist: The string may contain leading or trailing spaces. You must ignore these and find the length of the actual final word.

---

Example Test Cases:

Input	Expected Output	Explanation
"Hello World"	5	Last word is "World"
" fly me to the moon "	4	Last word is "moon"
"luffy is still joyboy"	6	Last word is "joyboy"

---

Why this is a milestone:

In earlier hours, you learned how to use .strip() and .split(). While those are great tools, solving this problem algorithmically (without just turning the whole thing into a list) forces you to think about direction.

Do you start from the beginning of the string, or would it be more efficient to start from the end?

---

Your Objective:

Try to solve this without using the .split() method. Aim to look at the characters one by one. Think about:

1. How do I ignore the spaces at the very end?
2. Once I hit a letter, how do I know when the word has ended?

Learn "Reverse Iteration" in Python

---

17

You’re at Hour 17. You've handled strings and two-pointer logic. Now we're moving into Math-Logic and Array Intersections. This is a common pattern when you have two different sources of data and need to find the common ground between them.

---

LeetCode Style Challenge: "Intersection of Two Arrays"

Problem Statement:

Given two integer arrays nums1 and nums2, return an array of their intersection.

Each element in the result must be unique and you may return the result in any order.

Constraints:

• Input: Two lists of integers.
• Output: A list containing only the numbers that appear in both input lists.
• The Rule: Even if a number appears five times in both lists, it should only appear once in your final result.

---

Example Test Cases:

Input nums1	Input nums2	Expected Output
[1, 2, 2, 1]	[2, 2]	[2]
[4, 9, 5]	[9, 4, 9, 8, 4]	[9, 4] (or [4, 9])
[1, 2, 3]	[4, 5, 6]	[]

---

Why this is a milestone:

This problem tests your ability to choose the most efficient tool for the job. You’ve worked with lists and sets. One of these is significantly faster for "checking existence" than the other. If you use a list for everything, your code will get sluggish as the data grows.

The reality is that "searching" is the most frequent thing a computer does. Doing it poorly costs time and money.

---

Your Objective:

Find a way to compare these two lists. Don't worry about the "perfect" one-liner yet. Just focus on the logic of:

1. Identifying a common number.
2. Ensuring you don't add the same common number to your result twice.

find a specific Set operator that makes this problem trivial Learn "Reverse Iteration" in Python

---

18

You’ve hit Hour 18. You've mastered basic searching and unique sets. Now it’s time to deal with Nested Data Structures. In the real world, data isn't always a flat list of numbers; it’s often a list of lists (like a spreadsheet) or a matrix.

---

LeetCode Style Challenge: "Richest Customer Wealth"

Problem Statement:

You are given an m x n integer grid called accounts where accounts[i][j] is the amount of money the -th customer has in the -th bank. Return the wealth of the richest customer.

A customer's wealth is the sum of money they have in all their bank accounts.

Constraints:

• Input: A list of lists (e.g., [[1, 2, 3], [3, 2, 1]]).
• Output: A single integer representing the highest sum found.

Example Test Cases:

Input accounts	Expected Output	Explanation
[[1,2,3],[3,2,1]]	6	Both customers have 6.
[[1,5],[7,3],[3,5]]	10	The 2nd customer (7+3) is the richest.
[[2,8,7],[7,1,3],[1,9,5]]	17	The 1st customer (2+8+7) is the richest.

---

Why this is a milestone:

This is your introduction to 2D Arrays. You have to "loop within a loop"—the outer loop goes through each customer, and the inner loop (or a sum function) calculates their specific total. It forces you to keep track of a "global maximum" while doing local calculations.

But here’s the thing: most beginners get confused about which index refers to the "row" and which refers to the "column." This challenge fixes that mental model.

---

Your Objective:

1. Initialize a variable to keep track of the max_wealth (start it at 0).
2. Iterate through each "customer" list in the main accounts list.
3. Calculate the total for that specific customer.
4. If their total is higher than your current max_wealth, update it.
5. Return the final max_wealth.

Learn to use the sum() function to make the inner calculation easier, or manually loop through every single bank account

---

19

Hour 19. You’re handling 2D grids now. That's a huge jump. Most people get stuck there because they can't visualize the "list within a list." Since you've got that down, it’s time to tackle Array Parity.

This is a favorite for interviewers because it tests if you can reorder data based on a specific property—in this case, whether a number is even or odd.

---

LeetCode Style Challenge: "Sort Array By Parity"

Problem Statement:

Given an integer array nums, move all the even integers to the beginning of the array followed by all the odd integers.

You can return the result in any order as long as the evens come before the odds.

Constraints:

• Input: A list of integers (e.g., [3, 1, 2, 4]).
• Output: A list where evens are first (e.g., [2, 4, 3, 1]).
• The Goal: Don't just sort the list numerically. Use logic to separate them.

---

Example Test Cases:

Input	Possible Expected Output
[3, 1, 2, 4]	[2, 4, 3, 1] or [4, 2, 1, 3]
[0]	[0]
[1, 2, 3, 4, 5, 6]	[2, 4, 6, 1, 3, 5]

---

Why this is a milestone:

This is about partitioning. You’re learning how to split a dataset into two "buckets" and then stitch them back together. In data engineering, this is exactly how you'd separate valid records from corrupted ones or high-priority tasks from low-priority ones.

The reality is that you don't always need a complex algorithm; sometimes you just need two empty lists and a way to join them at the end.

---

Your Objective:

1. Create two separate empty lists: one for even numbers and one for odd.
2. Loop through the input nums.
3. Use the modulo operator (%) to check if a number is even.
4. Put it in the right bucket.
5. Combine the buckets and return the result.

Do this in a single line using List Comprehensions, or stick with the two-list approach

---

20

Hour 20. You've been splitting lists and managing 2D grids. Now it's time to tackle Relative Mapping. This is a step up from basic counting because you have to use one set of data to "translate" or "rank" another.

On LeetCode, this falls under Hashing and Sorting.

---

LeetCode Style Challenge: "Find Lucky Integer"

Problem Statement:

A lucky integer is an integer that has a frequency in the array equal to its value.

Given an array of integers arr, return the largest lucky integer in the array. If there is no lucky integer, return -1.

Example Test Cases:

Input	Expected Output	Explanation
[2, 2, 3, 4]	2	2 appears 2 times. 3 appears 1 time (not lucky).
[1, 2, 2, 3, 3, 3]	3	1, 2, and 3 are all lucky. 3 is the largest.
[2, 2, 2, 3, 3]	-1	2 appears 3 times, 3 appears 2 times. None are lucky.

---

Why this is a milestone:

This is your first "Double Verification" problem. You aren't just checking if a number exists; you're checking if its identity matches its behavior (frequency).

It forces you to combine two things you've learned:

1. Frequency Counting: Using a dictionary to count how many times each number appears.
2. Filtering & Comparison: Looking through that dictionary to find the keys that match their values, then picking the biggest one.

---

Your Objective:

1. Build a dictionary to store the counts of each number in arr.
2. Initialize a variable largest_lucky = -1.
3. Loop through the dictionary's "key-value" pairs.
4. Check if the key is equal to the value.
5. If it is, and it's bigger than your current largest_lucky, update your variable.

The reality is that "luck" here is just a data match. But the logic—comparing a key to its own frequency—is used constantly in fraud detection and pattern recognition.

Learn how to use max() to find the largest lucky number more efficiently

---

21

For Hour 21, we're moving into String Compression. This is how zip files and image formats actually work under the hood.

---

LeetCode Style Challenge: "Check if All Characters Have Equal Occurrences"

Problem Statement:

Given a string s, return True if all characters that appear in s have the same number of occurrences, and False otherwise.

Constraints:

• Input: A string (e.g., "abacbc").
• Output: Boolean.
• The Goal: Don't use .count() inside a loop. Count everything once first.

Example Test Cases:

Input	Expected Output	Explanation
"abacbc"	True	'a', 'b', and 'c' all appear 2 times.
"aaabb"	False	'a' appears 3 times, 'b' appears 2 times.
"abcde"	True	Everything appears exactly once.

---

Why this is a milestone:

You are learning to validate the uniformity of a dataset. In your last solution, you used the dictionary keys to find the "lucky" number. Here, you need to look at the dictionary values.

The reality is that once you have the counts, you don't care about the letters anymore. You just care if all the numbers in that dictionary are the same.

---

Your Objective:

1. Loop through the string once to build a frequency dictionary (count the letters).
2. Get all the values from that dictionary (the counts).
3. Check if all those counts are equal.

But here’s the thing: There is a trick using a set() that can tell you if all values in a list are the same in one single line. Do you want to figure that out, or do you want to use a loop to compare the counts manually?

---

22

Hour 22. You’re moving past just collecting data and starting to look at statistical properties. Checking for uniformity is a massive leap.

Now, let’s talk about Difference Management. In coding interviews, you're often asked to find the "missing" piece of the puzzle. This tests your ability to track a shifting state across a dataset.

---

LeetCode Style Challenge: "Element Appearing More Than 25% In Sorted Array"

Problem Statement:

Given an integer array arr sorted in non-decreasing order, there is exactly one integer that occurs more than 25% of the time. Return that integer.

Constraints:

• Input: A list of integers (e.g., [1, 2, 2, 6, 6, 6, 6, 7, 10]).
• Output: The integer that appears more than 25% of the time.
• The Goal: Calculate the "threshold" first. If the list is 8 items long, 25% is 2. You are looking for a number that appears 3 or more times.

---

Example Test Cases:

Input	Expected Output	Explanation
[1,2,2,6,6,6,6,7,10]	6	Total length is 9. 25% is 2.25. '6' appears 4 times.
[1,1]	1	Total length is 2. 25% is 0.5. '1' appears twice.
[1,2,3,3]	3	Total length is 4. 25% is 1. '3' appears twice.

---

Why this is a milestone:

This problem introduces Dynamic Thresholds. The "rule" for success isn't a fixed number (like "appears twice"); it's a number that changes based on the size of the input.

But here’s the thing: because the array is sorted, all identical numbers are standing right next to each other. You don't necessarily need a dictionary to solve this. You could just count how long a "streak" of numbers lasts.

---

Your Objective:

1. Find the length of the array and multiply it by 0.25 to find your target count.
2. Create a frequency dictionary (remember: avoid .count() inside the loop to stay fast!).
3. As soon as you find a number whose count is greater than that 25% threshold, return it immediately.

The reality is that in big data, we often look for "heavy hitters"—items that take up a disproportionate amount of space. This is the simplest version of that.

Learn to use len(arr) // 4 to handle the math, or build the frequency map

---

23

Hour 23. You’ve mastered frequency maps and dynamic thresholds. Now, let's look at Data Transformation. In many real-world scenarios, you aren't just looking for a number; you’re changing it based on its surroundings.

On LeetCode, this is a classic "Easy" that tests your ability to handle Index Offsets—looking at what comes before or after your current position.

---

LeetCode Style Challenge: "Replace Elements with Greatest Element on Right Side"

Problem Statement:

Given an array arr, replace every element in that array with the greatest element among the elements to its right, and replace the last element with -1.

Example Test Case:

Input: arr = [17, 18, 5, 4, 6, 1] Output: [18, 6, 6, 6, 1, -1]

The Logic:

• For 17, the greatest element to its right is 18.
• For 18, the greatest element to its right is 6.
• For 5, the greatest element to its right is 6.
• For 4, the greatest element to its right is 6.
• For 6, the greatest element to its right is 1.
• For 1, there are no elements to the right, so it becomes -1.

---

Why this is a milestone:

If you start from the beginning of the list, you have to look at the entire rest of the list for every single number. That's and it's painfully slow.

But here’s the thing: if you start from the end of the list and walk backward, you only have to keep track of the "max seen so far." This turns a slow problem into a lightning-fast solution.

---

Your Objective:

1. Create a variable max_so_far and initialize it to -1.
2. Iterate through the array backward (from the last index to the first).
3. For each element:

• Save the current element in a temporary variable.
• Replace the current element with max_so_far.
• Update max_so_far to be whichever is bigger: the old max_so_far or the temporary variable you saved.

1. Return the modified array.

The reality is that "reverse thinking" is a common trick to optimize code. It prevents you from recalculating the same "max" over and over again.

Learn how to use range(len(arr) - 1, -1, -1) to loop backward, or figure out the reversed() function

---

24

This is a String Logic and Frequency challenge. It’s a favorite for testing how well you can coordinate different data points.

---

LeetCode Style Challenge: "Ransom Note"

Problem Statement:

You are given two strings: ransomNote and magazine.

You need to determine if the ransomNote can be constructed by using the letters from magazine. Each letter in magazine can only be used once in your ransomNote.

Return True if it can be constructed, otherwise return False.

Constraints:

• Input: Two strings, ransomNote and magazine.
• Output: A Boolean.
• The Rule: If your note needs two 'a's, the magazine must have at least two 'a's.

---

Example Test Cases:

ransomNote	magazine	Expected Output
"a"	"b"	False
"aa"	"ab"	False
"aa"	"aab"	True

---

Why this is a milestone:

This isn't about finding a single number or reversing a list. It's about inventory management. You have a "demand" (the note) and a "supply" (the magazine). Your code has to verify if the supply can meet every single demand.

The reality is that this logic is identical to checking if an order can be fulfilled from a warehouse or if a user has enough permissions to perform an action. You have to compare two separate sets of counts.

---

Your Objective:

Figure out how to track the counts of letters in both strings. If the note requires more of any character than the magazine provides, the whole thing fails.

I'm not going to tell you whether to use a dictionary, a list of 26 integers, or a set. That's on you.

Do you want to see the test cases for a very long string to check your performance, or are you going straight to the code?

---

25

Hour 25. You're deep enough into this that you don't need a map. You just need the problem.

This one is about Pattern Matching. It’s a step up from the Ransom Note because the relationship isn't just about "having enough letters"—it’s about the structure of the data.

---

LeetCode Style Challenge: "Isomorphic Strings"

Problem Statement:

Two strings s and t are isomorphic if the characters in s can be replaced to get t.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same replacement character, but a character may map to itself.

Constraints:

• Input: Two strings, s and t.
• Output: Boolean (True or False).
• Requirement: The mapping must be consistent. If 'a' maps to 'x' once, it must always map to 'x'. And 'b' cannot also map to 'x'.

---

Example Test Cases:

Input s	Input t	Expected Output	Explanation
"egg"	"add"	True	'e' -> 'a', 'g' -> 'd'
"foo"	"bar"	False	'o' cannot map to both 'a' and 'r'
"paper"	"title"	True	Consistent 1-to-1 mapping
"badc"	"baba"	False	'd' and 'c' both try to map to 'a'

---

Why this is a milestone:

This is your first encounter with Bi-directional Mapping. It’s not enough to know that "A becomes B." You also have to ensure that "B only comes from A."

The reality is that this is how basic encryption and substitution ciphers work. If your logic is leaky, the "code" breaks. You have to track relationships, not just counts.

---

Your Objective:

Figure out how to maintain a record of which character "belongs" to which. If you hit a contradiction—where a character is already promised to something else—you've failed the isomorphism test.

---

26

Hour 26. You’ve moved past simple counting and into the territory of structural relationships. Now we’re looking at Stream Processing.

In this scenario, you aren't looking at the whole picture at once. You are looking at a "sliding" window of data. This is how video buffering or network packet analysis works.

---

LeetCode Style Challenge: "Maximum Average Subarray I"

Problem Statement:

You are given an integer array nums and an integer k.

Find a contiguous subarray (a slice of the list) whose length is equal to k that has the maximum average value, and return this value.

Constraints:

• Input: A list of integers nums and a window size k.
• Output: A float (the highest average found).
• The Trap: If you recalculate the sum from scratch every time the window moves, your code will be too slow for large inputs.

---

Example Test Cases:

Input nums	k	Expected Output	Explanation
[1, 12, -5, -6, 50, 3]	4	12.75	The window [12, -5, -6, 50] has the max sum (51). 51/4 = 12.75.
[5]	1	5.0	Only one possible window.
[0, 4, 0, 3, 2]	2	2.5	The window [3, 2] has the max sum (5). 5/2 = 2.5.

---

Why this is a milestone:

This is the Sliding Window pattern. It’s a fundamental optimization technique. Instead of re-summing all k elements, you realize that when the window moves one step to the right, you just subtract the element that left the window and add the new one that entered.

The reality is that efficiency here isn't about being "clever" with syntax; it's about reducing the number of mathematical operations the CPU has to perform.

---

Your Objective:

Find the maximum sum of any k consecutive elements. Once you have the max sum, divide it by k at the very end to get the average.

Think about how to update that sum as you iterate through the list without starting over from zero at every index.

Should I wait for your code, or do you want to see a test case with negative numbers to see if your logic holds up?

---

27

Hour 27. You’re handling sliding windows now, which means you're starting to care about the "cost" of your operations. Let's pivot to In-Place Modification.

This is a classic "LeetCode Easy" that is harder than it looks because it forces you to manage two different speeds of data processing simultaneously.

---

LeetCode Style Challenge: "Remove Element"

Problem Statement:

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place.

The order of the elements may be changed. You must do this by modifying the input array directly and returning the number of elements which are not equal to val.

Constraints:

• Input: A list nums and a target val.
• Output: An integer (the count of elements not equal to val).
• The Reality: You can't just return a new list. You have to shift the "valid" numbers to the front of the original list.

---

Example Test Cases:

Input nums	val	Expected Output (k)	Resulting nums (first k elements)
[3, 2, 2, 3]	3	2	[2, 2, _, _]
[0, 1, 2, 2, 3, 0, 4, 2]	2	5	[0, 1, 3, 0, 4, _, _, _]

---

Why this is a milestone:

This is the Two-Pointer Write pattern. You have one pointer looking for "good" data and another pointer keeping track of where that good data should be written.

But here's the thing: most beginners try to use .remove() or .pop() inside a loop. That's a disaster. Every time you remove an item, the list has to shift every other item over, turning your code into an mess. Doing it "in-place" with pointers keeps it at .

---

Your Objective:

1. Initialize a pointer (let's call it k) at 0.
2. Iterate through the array.
3. If the current element is NOT the val we want to remove, move that element to the position nums[k] and increment k.
4. Return k.

The reality is that memory is expensive. In embedded systems or massive databases, you don't get to just "make a copy" of your data. You have to fix it where it sits.

Learn handle the pointer logic, or remember why for i in range(len(nums)) is safer than for x in nums when you're modifying the index

---

28

The previous task was about removing elements. This task is about merging and sorting without using the built-in .sort() method.

---

LeetCode Challenge: Merge Sorted Array (In-Place)

The Goal

You are given two integer arrays, nums1 and nums2, each sorted in non-decreasing order. You are also given two integers, m and n, representing the number of actual elements in nums1 and nums2 respectively.

Merge nums2 into nums1 as one sorted array.

The Task

1. Modify nums1 in-place. 2. nums1 has a total length of m + n. The first m elements are the data, and the last n elements are set to 0 (serving as empty placeholders for nums2).
2. Combine them so that the final nums1 is sorted.

Constraints

• Do not return a new list. * Do not use nums1.sort().
• Efficiency: Aim for time complexity.

---

Example Test Case

Input: nums1 = [1, 2, 3, 0, 0, 0], m = 3 nums2 = [2, 5, 6], n = 3

Output: nums1 becomes [1, 2, 2, 3, 5, 6]

---

Implementation Detail

Since nums1 has empty space at the back, it is often more efficient to compare the largest numbers from both lists and fill nums1 from the end to the front. This prevents you from overwriting your own data as you move.

1. Start a pointer at the end of the data in nums1 (index ).
2. Start a pointer at the end of nums2 (index ).
3. Start a "write" pointer at the very end of nums1 (index ).
4. Compare the values and place the larger one at the write pointer.

---

29

Hour 29. Let’s move to Bit Manipulation and Parity. This is a logic-heavy problem where you have to find a "needle in a haystack" based on frequency.

---

LeetCode Challenge: Single Number

The Goal

Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.

The Task

1. Take a list nums.
2. Identify the one integer that does not have a pair.
3. Return that integer.

Constraints

• Time Complexity: You must implement a solution with a linear runtime.
• Space Complexity: You must use only constant extra space. (This means you cannot use a dictionary or a set to count occurrences).

---

Example Test Case

Input nums	Expected Output	Explanation
[2, 2, 1]	1	2 appears twice, 1 appears once.
[4, 1, 2, 1, 2]	4	1 and 2 both have pairs. 4 is alone.
[1]	1	Only one element exists.

---

Implementation Hint

Since you aren't allowed to use extra memory (like a dictionary), you need a way to "cancel out" numbers that appear twice.

There is a bitwise operator called XOR (^ in Python). Here is how it behaves:

• (Any number XORed with itself becomes 0).
• (Any number XORed with 0 stays the same).
• The order doesn't matter: is the same as , which results in , which is just .

If you XOR every number in the list together, the pairs will effectively delete each other, leaving only the single number behind.

Are you going to use the XOR trick, or do you want to try a different mathematical approach?

---

30

Hour 30. You've got XOR in your toolkit. Now we’re moving to Coordinate Geometry and Monotonicity. This sounds complex, but it's just checking if a sequence follows a consistent rule.

---

LeetCode Challenge: Monotonic Array

The Goal

Determine if a given array is monotonic.

An array is monotonic if it is either monotone increasing or monotone decreasing.

• Monotone increasing: For all , .
• Monotone decreasing: For all , .

The Task

1. Take a list nums.
2. Return True if the list only goes up (or stays flat) OR only goes down (or stays flat).
3. Return False if it does both (e.g., it increases and then decreases).

Constraints

• Time Complexity: . You should be able to decide the answer in one pass.
• Space Complexity: . No extra lists or dictionaries.

---

Example Test Cases

Input nums	Expected Output	Explanation
[1, 2, 2, 3]	True	It never decreases.
[6, 5, 4, 4]	True	It never increases.
[1, 3, 2]	False	It increases () then decreases ().

---

Implementation Strategy

You don't need to know if it's increasing or decreasing before you start. You can assume it's both at the beginning. As you walk through the array, if you see a pair of numbers that decreases, you know it's no longer "monotone increasing." If you see a pair that increases, you know it's no longer "monotone decreasing."

By the end of the loop, if it hasn't failed both categories, it's monotonic.

try implementing this using two boolean flags, or compare the first and last elements first

---

31

Hour 31. You've handled monotonic trends and bit manipulation. Now we’re moving into Matrix Transformation. This is a test of how you manipulate indices in a 2D coordinate system.

---

LeetCode Challenge: Transpose Matrix

The Goal

Given a 2D integer array matrix, return the transpose of the matrix.

The transpose of a matrix is the matrix flipped over its main diagonal, switching the matrix's row and column indices.

The Task

1. Take a 2D list matrix where matrix[r][c] represents the element at row r and column c.
2. Create a new matrix where the element at result[c][r] is the original matrix[r][c].
3. Return the new matrix.

Constraints

• Input size: The matrix is not necessarily square (e.g., it could be ).
• Output: If the input is $R \times C$, the , the output must be $C \times R$.

---

Example Test Case

Input: ```python matrix = [[1, 2, 3], [4, 5, 6]]

**Output:** ```python
[[1, 4], 
 [2, 5], 
 [3, 6]]

---

Implementation Detail

Since the dimensions flip, you first need to determine the number of rows and columns in the original.

• rows = len(matrix)
• cols = len(matrix[0])

You’ll need to initialize a new 2D grid with the dimensions swapped. In Python, you can create a "blank" grid of zeros like this: result = [[0] * rows for _ in range(cols)]

Then, use a nested loop to map the old coordinates to the new ones.

Are you going to build the result matrix manually using loops, or do you want to try the Pythonic zip(*matrix) trick?

---

32

Hour 32. You’ve successfully flipped a matrix. Now we’re looking at Boundary Logic and Adjacency. This is a "LeetCode Easy" that requires you to handle edge cases—literally.

---

LeetCode Challenge: Can Place Flowers

The Goal

You have a long flowerbed in which some of the plots are planted, and some are empty. However, flowers cannot be planted in adjacent plots.

Given an integer array flowerbed containing 0's (empty) and 1's (not empty), and an integer n, return True if n new flowers can be planted in the flowerbed without violating the no-adjacent-flowers rule and False otherwise.

The Task

1. Iterate through the flowerbed.
2. A flower can be planted at index i only if:

• flowerbed[i] is 0.
• The plot to the left (i-1) is empty or out of bounds.
• The plot to the right (i+1) is empty or out of bounds.

1. Every time you plant a flower, update the array and decrement n.
2. If n hits zero, you’re done.

Constraints

• Efficiency: . One pass.
• Modification: You can modify the input array as you go to "reserve" the spots you’ve used.

---

Example Test Cases

Input flowerbed	n	Expected Output	Explanation
[1, 0, 0, 0, 1]	1	True	Can plant at index 2.
[1, 0, 0, 0, 1]	2	False	Only one spot available.
[0, 0, 1]	1	True	Can plant at index 0.

---

Implementation Detail

The main challenge is the first and last elements.

• When i == 0, there is no i-1.
• When i == len(flowerbed) - 1, there is no i+1.

You need to write your if condition to handle these boundaries so you don't get an IndexError. Some people find it easier to "pad" the array with a 0 at the beginning and end, while others use explicit checks like if i == 0 or flowerbed[i-1] == 0.

Do you want to handle the boundary logic with if/else checks, or do you want to try the "padding" trick to simplify the loop?

---

33

Hour 33. You're moving from spatial logic to Array Partitioning. This is a logic puzzle that tests if you can find a "pivot point" in a stream of data.

---

LeetCode Challenge: Find Pivot Index

The Goal

Given an array of integers nums, calculate the pivot index of this array.

The pivot index is the index where the sum of all the numbers strictly to the left of the index is equal to the sum of all the numbers strictly to the right of the index.

The Task

1. Find the index where sum(left_side) == sum(right_side).
2. If the index is on the left edge (index 0), the left sum is 0.
3. If the index is on the right edge, the right sum is 0.
4. Return the leftmost pivot index. If no such index exists, return -1.

Constraints

• Efficiency: . You must not use sum() inside your loop, as that would make the solution .
• Space: .

---

Example Test Cases

Input nums	Expected Output	Explanation
[1, 7, 3, 6, 5, 6]	3	Left sum = 1+7+3 = 11. Right sum = 5+6 = 11.
[1, 2, 3]	-1	No index satisfies the condition.
[2, 1, -1]	0	Left sum = 0. Right sum = 1 + (-1) = 0.

---

Implementation Detail

Don't overcomplicate it by slicing the list.

1. Calculate the total_sum of the entire array once.
2. Keep a running left_sum variable that starts at 0.
3. As you iterate through the array, the right_sum is simply total_sum - left_sum - nums[i].
4. If left_sum equals that right_sum, you found your index.
5. Update your left_sum by adding the current number before moving to the next iteration.

The reality is that this is a classic "prefix sum" logic problem. It’s about using a known total to derive unknown parts without doing extra work.

learn how to handle the "leftmost" requirement

---

34

Hour 34. You’re moving from finding pivots to Cycle Detection and State Tracking. This is a famous problem that looks like math but is actually about recognizing patterns in a sequence.

---

LeetCode Challenge: Happy Number

The Goal

Determine if a number is "happy."

A happy number is a number defined by the following process:

1. Starting with any positive integer, replace the number by the sum of the squares of its digits.
2. Repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1.
3. Those numbers for which this process ends in 1 are happy.

Return True if is a happy number, and False if not.

The Task

1. Write a helper to get the sum of squares of digits. (For 19, it's ).
2. Keep track of every result you get.
3. If you hit 1, return True.
4. If you hit a number you have seen before, you are in a loop. Return False.

Constraints

• Memory: Use a set to store the numbers you've already encountered. Checking a set is .

---

Example Test Case

Input: n = 19

Output: True

Input: n = 2 2, 4, 16, 37, 58, 89, 145, 42, 20, 4... (4 is repeated, so it loops forever). Output: False

---

Implementation Detail

To split a number into digits without converting it to a string (which is slow), use the modulo operator % 10 to get the last digit and integer division // 10 to remove it.

digit = n % 10
n = n // 10

use a while loop with a set to catch the cycle, or try the "Floyd's Cycle-Finding" (Tortoise and Hare) approach to save memory