Decoding Slanted Transposition Cipher in Python

Verify Alien Dictionary: Custom Order Comparison Check if words are lexicographically sorted using custom alphabet ordering. Build char-to-index map from order string. Compare adjacent word pairs character-by-character using custom ordering. Early exit when mismatch found or prefix condition violated. Comparison Logic: HashMap enables O(1) custom order lookup. Prefix check handles "apple" vs "app" case (longer must come after). Break on first mismatch since rest doesn't matter for ordering. Time: O(n × k) where k = avg word length | Space: O(1) — fixed 26 chars #StringComparison #CustomOrdering #HashMap #LexicographicOrder #Python #AlgorithmDesign #SoftwareEngineering

In 𝗙𝗮𝘀𝘁𝗔𝗣𝗜, 𝗿𝗲𝘁𝘂𝗿𝗻 𝘁𝘆𝗽𝗲 and 𝗿𝗲𝘀𝗽𝗼𝗻𝘀𝗲_𝗺𝗼𝗱𝗲𝗹 describe the success body: a single Pydantic model, or a list of that model—whatever the handler returns on a normal 2xx. When a lookup fails or a rule is broken, you raise 𝗛𝗧𝗧𝗣𝗘𝘅𝗰𝗲𝗽𝘁𝗶𝗼𝗻 so the client gets a proper 𝟰𝘅𝘅 𝘀𝘁𝗮𝘁𝘂𝘀 and a 𝗱𝗲𝘁𝗮𝗶𝗹 𝗽𝗮𝘆𝗹𝗼𝗮𝗱. That is separate from the return type, because raise is not return. 𝗿𝗲𝘀𝗽𝗼𝗻𝘀𝗲_𝗺𝗼𝗱𝗲𝗹 defines how the 𝘀𝘂𝗰𝗰𝗲𝘀𝘀 𝗿𝗲𝘀𝗽𝗼𝗻𝘀𝗲 is built and documented; HTTPException is how you answer with an error without faking a 200. 👉 Detailed breakdown here: https://lnkd.in/gcySJqZX 👉 Code Here: https://lnkd.in/gyf4663s #FastAPI #Python #API #BackendDevelopment #SoftwareEngineering #Pydantic #OpenAPI

Day 55/100 – #100DaysOfCode 🚀 Solved LeetCode #205 – Isomorphic Strings (Python). Today I practiced hashmap (dictionary) usage to check whether two strings follow the same pattern. Approach: 1) Create two hashmaps to store character mappings in both directions. 2) Traverse both strings together using zip(). 3) Check if the current mapping is consistent in both maps. 4) If any mismatch is found, return False. 5) Otherwise, update the mappings and continue. 6) If all mappings are valid, return True. Time Complexity: O(n) Space Complexity: O(n) Understanding how bidirectional mapping ensures consistency 💪 #LeetCode #Python #DSA #HashMap #Strings #ProblemSolving #100DaysOfCode

Day 50/100 – #100DaysOfCode 🚀 Solved LeetCode #28 – Find the Index of the First Occurrence in a String (Python). Today I practiced string matching using a brute-force approach to find the first occurrence of a substring. Approach: 1) Traverse the main string (haystack). 2) For each index, try to match the substring (needle). 3) Compare characters one by one. 4) If all characters match, return the starting index. 5) If mismatch occurs, break and move to the next index. 6) If no match is found, return -1. Time Complexity: O(n × m) Space Complexity: O(1) Understanding basic string matching techniques step by step 💪 #LeetCode #Python #DSA #Strings #ProblemSolving #100DaysOfCode

Day 53/100 – #100DaysOfCode 🚀 Solved LeetCode #125 – Valid Palindrome (Python). Today I practiced string cleaning and validation to check whether a given string is a palindrome. Approach: 1) Traverse the string and keep only alphanumeric characters. 2) Convert all characters to lowercase. 3) Build a cleaned string. 4) Compare the string with its reverse. 5) If both are equal, return True; otherwise, return False. Time Complexity: O(n) Space Complexity: O(n) Learning how preprocessing simplifies string problems 💪 #LeetCode #Python #DSA #Strings #ProblemSolving #100DaysOfCode

Day 43/100 – #100DaysOfCode 🚀 Solved LeetCode #2610 – Convert an Array Into a 2D Array With Conditions (Python). Today I practiced hashmap (frequency counting) to construct a 2D array based on given conditions. Approach: 1) Create a frequency map to count occurrences of each element. 2) Initialize an empty result list. 3) While the frequency map is not empty: 4)  Create a new row. 5)  Iterate through keys and add each number once to the row. 6)  Decrease its frequency and remove it if it becomes zero. 7) Add the row to the result. 8) Return the final 2D array. Time Complexity: O(n) Space Complexity: O(n) Learning how frequency maps help in structuring data efficiently 💪 #LeetCode #Python #DSA #HashMap #Arrays #ProblemSolving #100DaysOfCode

🚀 Solved: Find a String (Substring Count) Challenge Just solved another problem on HackerRank under the Python Strings section! ✅ 🧠 Problem Overview: Count how many times a substring appears in a string — including overlapping occurrences. 🔍 Key Learnings: Practiced string traversal techniques Understood why built-in methods like count() may not always work (no overlapping support) Strengthened concepts of slicing and iteration in Python 💡 Example Insight: For string "ABCDCDC" and substring "CDC", the answer is 2 (overlapping counts matter!). ⚡ Approach Used: Iterated through the string Compared substrings using slicing Counted valid matches efficiently 📈 Problems like this help build strong fundamentals in string manipulation, which is crucial for coding interviews and real-world applications. #Python #HackerRank #Coding #Strings #ProblemSolving #DSA #LearningJourney #AI link of #Solution :- https://lnkd.in/gtqcy8fX

Day 41/100 – #100DaysOfCode 🚀 Solved LeetCode #2529 – Maximum Count of Positive Integer and Negative Integer (Python). Today I practiced simple counting logic to determine whether positive or negative numbers are more in the array. Approach: 1) Initialize two counters: neg = 0 and pos = 0. 2) Traverse the array element by element. 3) If the number is negative, increment neg. 4) If the number is positive, increment pos. 5) Return the maximum of neg and pos. Time Complexity: O(n) Space Complexity: O(1) Strengthening fundamentals with simple counting techniques 💪 #LeetCode #Python #DSA #Arrays #ProblemSolving #100DaysOfCode

Group Anagrams: Frequency Array as Hashable Key Sorting each string costs O(k log k) per string. Character frequency array is O(k) and creates identical signature for anagrams. Fixed 26-element array converted to tuple serves as hashable HashMap key — faster, cleaner grouping. Frequency Signature: Character counts uniquely identify anagram groups without sorting. Tuple conversion makes array hashable for dict keys. Pattern applies to document clustering, duplicate detection. Time: O(n × k) vs O(n × k log k) sorting | Space: O(n × k) #FrequencyArray #Anagrams #HashMap #KeyOptimization #Python #AlgorithmOptimization #SoftwareEngineering

Day 48/100 – #100DaysOfCode 🚀 Solved LeetCode #13 – Roman to Integer (Python). Today I practiced string processing and mapping logic to convert Roman numerals into integers. Approach: 1) Create a dictionary to map Roman symbols to their integer values. 2) Traverse the string from left to right. 3) If the current value is less than the next value, subtract it. 4) Otherwise, add it to the total. 5) Return the final result. Time Complexity: O(n) Space Complexity: O(1) Understanding pattern-based problems in strings 💪 #LeetCode #Python #DSA #Strings #ProblemSolving #100DaysOfCode