"Day 66: Binary Search for Minimum in Rotated Array II"

🚀 LeetCode #1625: Lexicographically Smallest String After Applying Operations Today’s problem was about transforming a numeric string using two operations: adding a value to digits at odd indices and rotating the string, to get the lexicographically smallest result possible. The challenge was to handle infinite possibilities smartly. I used a Breadth First Search (BFS) approach to systematically explore all reachable string states while keeping track of visited ones. 💡 Key Takeaways: - Some problems don’t need a direct formula, they need systematic exploration. - BFS is not just for graphs; it’s a powerful tool for exploring state transitions too. - Modular arithmetic and rotation logic often come together in string manipulation problems. This one was a great reminder that clean logic and state tracking can solve even the most “infinite looking” problems efficiently. #LeetCode #ProblemSolving #Python #DSA #CodingChallenge #Algorithms #BFS #StringManipulation #LearningEveryday

🚀 Day 40 of #100DaysOfDSA Solved LeetCode Problem #69 – Sqrt(x) 🧮 💡 Problem Insight: Given a non-negative integer x, return the square root of x rounded down to the nearest integer. For example: Input: x = 8 Output: 2 (since √8 ≈ 2.828, and floor(2.828) = 2) ✨ Key Learnings: Practiced binary search to find results efficiently without using built-in math functions. Learned how to narrow down search space based on mid-square comparisons. Reinforced understanding of integer division and rounding down behavior. Time Complexity: O(log n) — fast and efficient! Space Complexity: O(1) 💬 Lesson: Binary Search isn’t just for sorted arrays — it’s a mindset for narrowing down possibilities quickly 🚀 #LeetCode #Python #DSA #BinarySearch #ProblemSolving #100DaysOfCode #Day40 #CodingJourney

🚀 Day 78 of #100DaysOfDSA 🚀 Solved LeetCode 205 — Isomorphic Strings 🔹 Problem: Check if two strings s and t are isomorphic. Two strings are isomorphic if each character in s can be replaced to get t, with no two characters mapping to the same character (but a character can map to itself). 🔹 Approach: Bidirectional Hash Mapping (One-to-One Mapping Check) If lengths differ → return False immediately. Maintain two dictionaries: mapping1 for s → t mapping2 for t → s Traverse both strings simultaneously: Ensure the mapping is consistent in both directions. If any conflict appears, return False. ✨ Key Insight: Use two-way mapping to ensure one-to-one correspondence — avoiding false positives in character transformations. #LeetCode #Python #HashMap #Strings #DSA #ProblemSolving #100DaysOfCode #CodingJourney 🚀

𝗗𝗮𝘆 𝟯𝟲 𝗼𝗳 #𝟭𝟴𝟬𝗗𝗮𝘆𝘀𝗢𝗳𝗖𝗼𝗱𝗲 Today, I solved 𝗦𝗲𝗮𝗿𝗰𝗵 𝗜𝗻𝘀𝗲𝗿𝘁 𝗣𝗼𝘀𝗶𝘁𝗶𝗼𝗻 — a practical variation of binary search. The goal was to find the index of a target in a sorted array, or the position where it should be inserted to maintain order. Using binary search, I efficiently located the correct spot in O(log n) time. The key was to track the first position where the element is greater than or equal to the target — which is exactly the insert position if the target isn’t found. This is a great example of how small tweaks to a classic algorithm can solve new problems elegantly. A useful technique for search, insertion, and maintaining sorted data dynamically! #Python #Algorithms #BinarySearch #LeetCode #Coding #ProblemSolving

𝗗𝗮𝘆 𝟯𝟵 𝗼𝗳 #𝟭𝟴𝟬𝗗𝗮𝘆𝘀𝗢𝗳𝗖𝗼𝗱𝗲 Today, I built on the first/last occurrence solution to 𝗰𝗼𝘂𝗻𝘁 𝗼𝗰𝗰𝘂𝗿𝗿𝗲𝗻𝗰𝗲𝘀 𝗼𝗳 𝗮 𝘁𝗮𝗿𝗴𝗲𝘁 𝗶𝗻 𝗮 𝘀𝗼𝗿𝘁𝗲𝗱 𝗮𝗿𝗿𝗮𝘆 𝘄𝗶𝘁𝗵 𝗱𝘂𝗽𝗹𝗶𝗰𝗮𝘁𝗲𝘀. Using the same lower_bound and upper_bound helpers: Lower bound → first index where element ≥ target Upper bound → first index where element > target The count is simply: count = upper_bound - lower_bound This gives an O(log n) solution — much faster than scanning the entire array, especially with many duplicates. It’s a great example of how breaking a problem into reusable pieces leads to clean and efficient code. Perfect for analytics, frequency analysis, and search optimizations! 📊 #Python #Algorithms #BinarySearch #FrequencyCount #Coding #ProblemSolving

✅ LeetCode 3446 – Sort Matrix by Diagonals Successfully solved another interesting matrix manipulation problem! 🧩 Problem Summary: Given an n × n matrix, the task is to sort: The bottom-left diagonals (including the main diagonal) in non-increasing order. The top-right diagonals in non-decreasing order. 💡 Approach: Use a dictionary to group elements by diagonal index (j - i). Sort each diagonal individually based on its position (top or bottom triangle). Reconstruct the matrix by placing back the sorted values. ⚙️ Key Concepts: Matrix traversal Diagonal indexing (j - i) Sorting and reconstruction 📊 Result: ✅ Accepted with 0 ms runtime 💪 Optimized, clean, and easy-to-read solution #LeetCode #Python #ProblemSolving #CodingChallenge #DataStructures #Algorithms #Matrix #DeveloperJourney

🔍 Day 44 DSA Challenge – Problem #33: Search in Rotated Sorted Array 📌 Problem Statement: Given a sorted array nums that may have been rotated at an unknown pivot, find the index of a target value. Return -1 if the target doesn’t exist. The solution must run in O(log n) time. ⚙️ How I Solved It: Applied a modified binary search: Identify which half (left or right) is sorted. Narrow the search to the half where the target could exist. Repeat until the target is found or search space is exhausted. 📊 Performance Stats: ⏱ Runtime: 0 ms (⚡ beats 100%) 💾 Memory: 12.65 MB (beats 42.49%) ✅ Testcases Passed: 196 / 196 🧠 Key Takeaway: Understanding array properties like rotation and leveraging binary search ensures optimal search performance in logarithmic time. #LeetCode #BinarySearch #RotatedArray #Problem33 #Python #DSA #100DaysOfCode #CodingJourney