ABC TO XYZ OF AI

Perfect — that’s exactly the depth we want. For each problem, I’ll now include:

🧠 Updated Problem-Solving Framework

1. ✅ Problem Statement
2. 📄 Pseudocode / Flow / Logic
3. 🎯 Why this Pattern / Data Structure
4. 🔁 Looping & Decision Points
5. 🔽 Input → Logic → Output (Structured Code)
6. 🧪 Dry Run / Output Example

Let’s go all-in with this new format! Below are 5 full examples in this format. Once you’re happy with this structure, I’ll send 10 more in batches.

✅ Example 1: Check if a String is a Palindrome

1. ✅ Problem Statement

Check if the given string reads the same backward as forward.
Example: "madam" is a palindrome; "hello" is not.

2. 📄 Pseudocode:

- Input a string
- Reverse the string
- Compare reversed with original
- If equal → Palindrome

3. 🎯 Why This Pattern / Data Structure?

• We use string slicing ([::-1]) — no need for additional memory
• Efficient O(n) time, O(1) extra space

4. 🔁 Looping / Decisions

• No explicit loop, slicing handles reversal.
• One comparison: s == s[::-1]

5. 🔽 Code (Input → Logic → Output):

def get_input():
    return input("Enter a string: ")

def process(data):
    return data == data[::-1]

def show_result(output):
    print("Palindrome?", output)

def main():
    data = get_input()
    result = process(data)
    show_result(result)

main()

6. 🧪 Dry Run:

Input: madam
data[::-1] = madam
madam == madam → True

✅ Example 2: Count Frequency of Characters

1. ✅ Problem Statement

Given a string, count the frequency of each character.

2. 📄 Pseudocode:

- Create empty dict
- For each char in string:
    - If already in dict: increment
    - Else: set count = 1
- Return dict

3. 🎯 Why This Pattern / DS?

• HashMap (dict) gives O(1) lookup/update for frequency
• Best for counting problems

4. 🔁 Looping / Decisions

• Loop over each character
• if char in dict decides update or initialize

5. 🔽 Code:

def get_input():
    return input("Enter a word: ")

def process(word):
    freq = {}
    for ch in word:
        freq[ch] = freq.get(ch, 0) + 1
    return freq

def show_result(output):
    print("Character Frequency:", output)

def main():
    data = get_input()
    result = process(data)
    show_result(result)

main()

6. 🧪 Dry Run:

Input: apple
freq = {'a':1, 'p':2, 'l':1, 'e':1}

✅ Example 3: Find the Minimum in a List

1. ✅ Problem Statement

Given a list of numbers, find the minimum element.

2. 📄 Pseudocode:

- Initialize min_val = first element
- Loop from index 1 → n:
    - if arr[i] < min_val → update min_val
- Return min_val

3. 🎯 Why This Pattern / DS?

• Simple iteration, no need for sorting
• Optimal in one pass → O(n)

4. 🔁 Looping / Decisions

• Loop: index 1 to end
• Compare each with current min_val

5. 🔽 Code:

def get_input():
    return list(map(int, input("Enter numbers separated by space: ").split()))

def process(arr):
    min_val = arr[0]
    for num in arr[1:]:
        if num < min_val:
            min_val = num
    return min_val

def show_result(output):
    print("Minimum value:", output)

def main():
    data = get_input()
    result = process(data)
    show_result(result)

main()

6. 🧪 Dry Run:

Input: [4, 9, 1, 2]
min_val starts at 4 → becomes 1 → stays 1

✅ Example 4: Check if Two Strings are Anagrams

1. ✅ Problem Statement

An anagram has the same characters in different order.
Check if two strings are anagrams.

2. 📄 Pseudocode:

- Count characters in both strings using dict
- Compare both dicts

3. 🎯 Why This Pattern / DS?

• HashMap allows fast comparison
• O(n) for each string

4. 🔁 Looping / Decisions

• 2 loops for 2 words → 2 dicts
• 1 comparison → dict1 == dict2

5. 🔽 Code:

def get_input():
    a = input("Enter first string: ")
    b = input("Enter second string: ")
    return a, b

def count_chars(word):
    freq = {}
    for ch in word:
        freq[ch] = freq.get(ch, 0) + 1
    return freq

def process(a, b):
    return count_chars(a) == count_chars(b)

def show_result(output):
    print("Anagram?", output)

def main():
    a, b = get_input()
    result = process(a, b)
    show_result(result)

main()

6. 🧪 Dry Run:

a = "listen", b = "silent"
Both produce {'l':1,'i':1,'s':1,'t':1,'e':1,'n':1} → True

✅ Example 5: FizzBuzz (Classic Logic)

1. ✅ Problem Statement

Print numbers from 1 to n with these rules:

• If divisible by 3 → “Fizz”
• If divisible by 5 → “Buzz”
• If both → “FizzBuzz”
• Else → number

2. 📄 Pseudocode:

For i from 1 to n:
    if i % 3 == 0 and i % 5 == 0 → FizzBuzz
    elif i % 3 == 0 → Fizz
    elif i % 5 == 0 → Buzz
    else → i

3. 🎯 Why This Pattern?

• Simple modulo logic
• Sequence generation + condition branching

4. 🔁 Looping / Decision

• Loop from 1 to n
• Decision tree with 3 levels

5. 🔽 Code:

def get_input():
    return int(input("Enter a number: "))

def process(n):
    result = []
    for i in range(1, n+1):
        if i % 15 == 0:
            result.append("FizzBuzz")
        elif i % 3 == 0:
            result.append("Fizz")
        elif i % 5 == 0:
            result.append("Buzz")
        else:
            result.append(str(i))
    return result

def show_result(output):
    print("FizzBuzz Sequence:")
    print(", ".join(output))

def main():
    n = get_input()
    result = process(n)
    show_result(result)

main()

6. 🧪 Dry Run:

Input: 5 → Output: 1, 2, Fizz, 4, Buzz

✅ Ready for the Next 10+?

I’ll follow the exact same format for:

• Prime Number Check
• Reverse Words in Sentence
• Matrix Diagonal Sum
• First Non-Repeating Character
• Longest Unique Substring
• Valid Parentheses
• Count Words in String
• Roman to Integer
• Binary Search
• Bubble Sort

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