ABC TO XYZ OF AI

Above template is excellent for building a structured, beginner-to-advanced problem-solving flow — and with a small enhancement, it can flexibly handle all types of problems, from simple to complex (e.g. Merge Sort, Backtracking, Dynamic Programming, etc.).

✅ Finalized Python Problem-Solving Template:

Let’s slightly enhance previous version to dynamically adapt to all problem types:

🧠 0. Understand the Problem (Add to every solution)

• Problem Statement: What exactly is asked?
• Input/Output format: Examples + edge cases
• Constraints: Time/space limits, input size
• Keywords: Sort, Search, Max/Min, Count, Path, Decision

✏️ 1. Write Pseudocode / Flow

• Step-by-step breakdown
• Conditions / Decisions
• Use diagrams if applicable (matrix, tree, array, etc.)

🧰 2. Choose Tools

• Which data structures? Why?
• Loop or Recursion?
• Sliding Window / DP / Stack / Queue?

🧪 3. Dry Run with Mini Input

• Manually run on 1–2 inputs
• Track variable changes
• Ensure logic holds

💻 4. Code

def get_input():
    # Can be hardcoded, user, or file
    return "abba"

def process(data):
    # Core logic here
    return data == data[::-1]

def show_output(result):
    print("Is palindrome?", result)

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

main()

✅ 5. Test & Debug

• Add print() at each logic step
• Cover edge cases

🧠 Optional Enhancements:

• Add time/space complexity as docstring
• Write unit tests
• Make reusable function if needed

Now let’s include all previously discussed topics in this enhanced format using small data samples:

🧩 Example 1: Palindrome Check

🧠 Problem

Check if a string is a palindrome (same forwards and backwards).
Input: "abba" → Output: True

✏️ Pseudocode

• Compare start and end characters
• Move inward
• If mismatch → return False

🧰 Data Structure & Pattern

• Use string indexing and loop or slicing

💻 Code:

def get_input():
    return "abba"

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

def show_output(result):
    print("Palindrome?", result)

main()

🧩 Example 2: Merge Sort

🧠 Problem

Sort an array using the Divide & Conquer method.

Input: [4, 2, 1, 3]
Output: [1, 2, 3, 4]

✏️ Pseudocode

1. Split array into halves
2. Recursively sort each half
3. Merge them in sorted order

🧰 Data Structure & Pattern

• Array + recursion
• Use merge technique with two pointers

💻 Code:

def merge_sort(arr):
    if len(arr) <= 1:
        return arr

    mid = len(arr) // 2
    L = merge_sort(arr[:mid])
    R = merge_sort(arr[mid:])
    return merge(L, R)

def merge(left, right):
    i = j = 0
    merged = []

    while i < len(left) and j < len(right):
        if left[i] <= right[j]:
            merged.append(left[i])
            i += 1
        else:
            merged.append(right[j])
            j += 1
    merged.extend(left[i:])
    merged.extend(right[j:])
    return merged

print(merge_sort([4, 2, 1, 3]))

🧩 Example 3: Sliding Window – Max Sum Subarray of Size K

🧠 Problem

Find max sum of subarray of size K
Input: [1, 2, 3, 4, 5], K=2 → Output: 9

✏️ Pseudocode

1. Create initial window of size K
2. Slide forward: add next, remove prev
3. Track max sum

💡 Pattern: Sliding Window

Efficient for continuous segments

💻 Code:

def max_sum_k(arr, k):
    window_sum = sum(arr[:k])
    max_sum = window_sum

    for i in range(k, len(arr)):
        window_sum += arr[i] - arr[i - k]
        max_sum = max(max_sum, window_sum)
    
    return max_sum

print(max_sum_k([1, 2, 3, 4, 5], 2))

🧩 Example 4: Binary Search

🧠 Problem

Find element x in sorted array
Input: arr=[1, 2, 3, 4, 5], x=3 → Output: 2 (index)

✏️ Pseudocode

1. Set low=0, high=len(arr)-1
2. Repeat:
   • mid = (low + high) // 2
   • If mid == x → return
   • Else adjust low/high

💡 Pattern: Binary Search

Use when array is sorted

💻 Code:

def binary_search(arr, x):
    low, high = 0, len(arr) - 1

    while low <= high:
        mid = (low + high) // 2
        if arr[mid] == x:
            return mid
        elif arr[mid] < x:
            low = mid + 1
        else:
            high = mid - 1

    return -1

print(binary_search([1, 2, 3, 4, 5], 3))

Absolutely! Here’s a complete and interactive breakdown of 10 classic Python problems using your preferred template:

✅ Standard Problem-Solving Template

1. Understand the Problem (Statement + Example)
2. Write Pseudocode / Decision Tree / Flow
3. Choose the Right Pattern + Data Structure
4. Code using this structure: def get_input(): ... def process(data): ... def show_result(result): ... def main(): ...

🔟 Core Problems with Template Breakdown

1. 🔙 Backtracking: Subsets Generation

Problem:
Given an array, return all possible subsets (the power set).

Example:
Input: [1,2] → Output: [[], [1], [2], [1,2]]

Pseudocode:

function backtrack(index, path):
    add path to result
    for i from index to len(nums):
        call backtrack(i+1, path + [nums[i]])

Why Backtracking?
We need to explore all combinations (include/exclude).

Code:

def process(nums):
    result = []

    def backtrack(index, path):
        result.append(path)
        for i in range(index, len(nums)):
            backtrack(i + 1, path + [nums[i]])

    backtrack(0, [])
    return result

def main():
    print(process([1, 2]))

2. 🧠 DP: Fibonacci with Memoization

Problem:
Return nth Fibonacci number

Pseudocode:

If n in memo: return memo[n]
memo[n] = fib(n-1) + fib(n-2)
return memo[n]

Why DP?
Avoid recomputation of overlapping subproblems.

Code:

def process(n):
    memo = {}
    def fib(n):
        if n <= 1:
            return n
        if n in memo:
            return memo[n]
        memo[n] = fib(n - 1) + fib(n - 2)
        return memo[n]
    return fib(n)

def main():
    print(process(6))  # ➤ 8

3. 🔁 Recursion: Reverse a List

Problem:
Reverse a list using recursion

Pseudocode:

Base: if len(lst) <= 1: return lst
return reverse(rest) + [first]

Code:

def process(lst):
    if len(lst) <= 1:
        return lst
    return process(lst[1:]) + [lst[0]]

def main():
    print(process([1, 2, 3]))  # ➤ [3,2,1]

4. 🔍 Matrix Traversal: DFS

Problem:
Check if a word exists in a grid using DFS

Example:
Search “ABCCED” in a char grid.

Pseudocode:

DFS from every cell
If match, mark visited and search 4 directions
Backtrack if not found

Code:

def process(board, word):
    rows, cols = len(board), len(board[0])
    path = set()

    def dfs(r, c, i):
        if i == len(word): return True
        if (r < 0 or c < 0 or r >= rows or c >= cols or
                word[i] != board[r][c] or (r, c) in path):
            return False
        path.add((r, c))
        res = (dfs(r+1, c, i+1) or dfs(r-1, c, i+1) or
               dfs(r, c+1, i+1) or dfs(r, c-1, i+1))
        path.remove((r, c))
        return res

    for r in range(rows):
        for c in range(cols):
            if dfs(r, c, 0): return True
    return False

def main():
    board = [["A","B","C","E"],
             ["S","F","C","S"],
             ["A","D","E","E"]]
    print(process(board, "ABCCED"))  # ➤ True

5. 💡 Greedy: Interval Scheduling (Max Non-Overlapping)

Problem:
Given intervals, return max number of non-overlapping intervals.

Pseudocode:

Sort intervals by end
Track end; if current start >= end, count++

Code:

def process(intervals):
    intervals.sort(key=lambda x: x[1])
    count = 0
    end = float('-inf')
    for start, stop in intervals:
        if start >= end:
            count += 1
            end = stop
    return count

def main():
    print(process([[1,3],[2,4],[3,5]]))  # ➤ 2

6. 🧩 HashMap: Valid Anagram

Problem:
Check if two strings are anagrams.

Code:

def process(s, t):
    if len(s) != len(t): return False
    from collections import Counter
    return Counter(s) == Counter(t)

def main():
    print(process("listen", "silent"))  # ➤ True

7. 📚 Stack: Valid Parentheses

Problem:
Check if input has valid brackets.

Pseudocode:

For each bracket:
  If open, push to stack
  If close, check top of stack

Code:

def process(s):
    stack = []
    mapping = {')': '(', ']': '[', '}': '{'}
    for ch in s:
        if ch in mapping.values():
            stack.append(ch)
        elif ch in mapping:
            if not stack or stack.pop() != mapping[ch]:
                return False
    return not stack

def main():
    print(process("({[]})"))  # ➤ True

8. 🚍 Queue: Sliding Window Max

Problem:
Return max in each sliding window of size k

Pseudocode (Using Deque):

Use deque to store index
Remove out-of-window + smaller elements

Code:

from collections import deque
def process(nums, k):
    q, res = deque(), []
    for i, n in enumerate(nums):
        while q and q[0] <= i - k:
            q.popleft()
        while q and nums[q[-1]] < n:
            q.pop()
        q.append(i)
        if i >= k - 1:
            res.append(nums[q[0]])
    return res

def main():
    print(process([1,3,-1,-3,5,3,6,7], 3))  # ➤ [3,3,5,5,6,7]

9. ➗ Two Pointers: Sorted Pair Sum

Problem:
Find if two numbers sum to target in sorted array.

Code:

def process(arr, target):
    l, r = 0, len(arr) - 1
    while l < r:
        s = arr[l] + arr[r]
        if s == target:
            return True
        elif s < target:
            l += 1
        else:
            r -= 1
    return False

def main():
    print(process([1,2,4,5,7,11], 9))  # ➤ True

10. 🌲 Trie: Prefix Matching

Problem:
Check if any word in list starts with given prefix.

Pseudocode:

Build Trie → For each char → traverse

Code:

class TrieNode:
    def __init__(self):
        self.children = {}
        self.end = False

class Trie:
    def __init__(self):
        self.root = TrieNode()
    def insert(self, word):
        node = self.root
        for ch in word:
            node = node.children.setdefault(ch, TrieNode())
        node.end = True
    def startsWith(self, prefix):
        node = self.root
        for ch in prefix:
            if ch not in node.children:
                return False
            node = node.children[ch]
        return True

def main():
    t = Trie()
    for word in ["apple", "app", "banana"]:
        t.insert(word)
    print(t.startsWith("ap"))  # ➤ True
    print(t.startsWith("ba"))  # ➤ True
    print(t.startsWith("gr"))  # ➤ False

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