Check if Strings Can Be Made Equal With Operations

✅ Day 87 of 100 Days LeetCode Challenge Problem: 🔹 #2840 – Check if Strings Can be Made Equal With Operations II 🔗 https://lnkd.in/gY73RBb5 Learning Journey: 🔹 Today’s problem extended the previous one, allowing swaps between indices where the difference is even. 🔹 I observed that this again partitions the string into two independent groups: • Even indices (0, 2, 4, …) • Odd indices (1, 3, 5, …) 🔹 I extracted characters from even and odd positions separately for both strings. 🔹 Then, I sorted these groups and compared them between s1 and s2. 🔹 If both even-index groups and odd-index groups match, the strings can be made equal. Concepts Used: 🔹 String Manipulation 🔹 Index Grouping (Parity-based) 🔹 Sorting 🔹 Greedy Observation Key Insight: 🔹 Since swaps are allowed only between indices with even distance, characters can only move within their parity group. 🔹 Therefore, the problem reduces to checking if both parity groups have identical character distributions. Complexity: 🔹 Time: O(n log n) 🔹 Space: O(n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

Just proposed 3 new tools for langchain community. The current built ins are great for demos but fall short in production: - Math can't solve equations or do calculus - Web scraping returns raw noisy HTML - Python REPL has zero security controls So I built: CalculatorTool - equations, calculus, unit conversion via sympy + pint WebScraperTool - smart extraction, metadata, retry logic, robots.txt PythonCodeExecutorTool - AST security scanning, stateful sessions, timeout kill 95 tests. no external APIs. ready to merge. here's the issue: https://lnkd.in/gXRYRJpa #OpenSource #LangChain #Python #AIAgents #LLM

Day 10/100 – DSA Challenge Today’s problem: Move Zeroes (LeetCode 283) What I Learned: The goal was to move all zeroes to the end of an array while maintaining the relative order of non-zero elements — and importantly, doing it in-place. Key Idea: Two-Pointer Technique I used a two-pointer approach: One pointer (fast) iterates through the array Another pointer (slow) tracks where the next non-zero element should go Whenever a non-zero element is found, it is swapped with the element at the slow pointer, ensuring all non-zero elements are shifted forward while zeroes naturally move to the end. Why this approach? Maintains order of elements Works in O(n) time complexity Uses O(1) extra space (in-place) Takeaway: This problem reinforced how powerful the two-pointer technique is for array manipulation problems, especially when constraints require in-place operations. Looking forward to tackling more problems and improving consistency! #Day10 #100DaysOfCode #DSA #Python #CodingJourney #LeetCode #ProblemSolving

✅ Day 95 of 100 Days LeetCode Challenge Problem: 🔹 #869 – Reordered Power of 2 🔗 https://lnkd.in/gkNXaSFM Learning Journey: 🔹 Today’s problem focused on checking whether the digits of a number can be rearranged to form a power of 2. 🔹 I created a helper function to extract and store the digits of a number. 🔹 Then I sorted the digits of the input number for comparison. 🔹 Next, I generated powers of 2 iteratively and compared their sorted digit lists with the input. 🔹 If any match was found, I returned True. Otherwise, continued until the digit length exceeded the input. Concepts Used: 🔹 Digit Extraction 🔹 Sorting 🔹 Simulation of Powers of 2 🔹 Brute Force Optimization Key Insight: 🔹 Instead of generating permutations (which is expensive), sorting digits allows quick comparison. 🔹 Any valid rearrangement must have the same digit frequency as some power of 2. Complexity: 🔹 Time: O(log n * d log d) 🔹 Space: O(d) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers

✅ Day 99 of 100 Days LeetCode Challenge Problem: 🔹 #999 – Available Captures for Rook 🔗 https://lnkd.in/gAw_XpbJ Learning Journey: 🔹 Today’s problem focused on simulating rook movement on a chessboard. 🔹 First, I located the position of the rook 'R' on the board. 🔹 Then, I explored all four directions: up, down, left, and right. 🔹 In each direction, I moved step-by-step until: • I hit a bishop 'B' → stop (blocked) • I found a pawn 'p' → increment count and stop • Or reached the board boundary 🔹 Summed all valid captures and returned the result. Concepts Used: 🔹 Matrix Traversal 🔹 Simulation 🔹 Direction Vectors 🔹 Boundary Checking Key Insight: 🔹 The rook’s movement is linear in 4 directions, and each direction is independent. 🔹 Early stopping (on bishop or pawn) is critical for correctness and efficiency. Complexity: 🔹 Time: O(1) (fixed 8×8 board, constant work) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #100DaysOfCode #Python #CodingJourney #ProblemSolving #LearningInPublic

✅ Day 90 of 100 Days LeetCode Challenge Problem: 🔹 #476 – Number Complement 🔗 https://lnkd.in/gzE6gM7d Learning Journey: 🔹 Today’s problem focused on finding the complement of a number by flipping its binary bits. 🔹 I first converted the integer to its binary representation using bin(num)[2:]. 🔹 Then, I created a helper function to flip each bit: • '0' → '1' • '1' → '0' 🔹 After generating the flipped binary string, I converted it back to an integer using int(..., 2). 🔹 Returned the final complemented value. Concepts Used: 🔹 Binary Representation 🔹 Bit Manipulation 🔹 String Traversal 🔹 Base Conversion Key Insight: 🔹 The complement operation is essentially a bitwise NOT, but only within the significant bits of the number (ignoring leading zeros). 🔹 Converting to binary simplifies the flipping logic for beginners. Complexity: 🔹 Time: O(log n) 🔹 Space: O(log n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers

🚀 Day 20 of Consistency | #75DaysLeetCodeChallenge 🧠 Today’s Problem : Min Stack (#155) 💡 Key Learning: This problem teaches how to design a data structure that supports constant time minimum retrieval using an additional stack. ⚡ Approach: Use two stacks → one for values, one for minimums push → add value to main stack & min(current, previous min) to min stack pop → remove from both stacks top → return top of main stack getMin → return top of min stack 🧠 Why this works: Tracks minimum at every step Ensures all operations run in O(1) time Efficient use of auxiliary space 🔥 Result : ✔️ Runtime: 0 ms (Beats 100%) 📈 A classic design problem that strengthens understanding of stacks & optimization. A big thanks to Shivam Singh, Nikhil Yadav & Akshat Tiwari for this amazing challenge 🙌 Consistency is compounding. Keep going. 💪 #Day20 #LeetCode #DSA #CodingJourney #100DaysOfCode #Python #Stack #Consistency

Code that writes code. 🌌 There is a point in every developer's journey where you stop thinking about Logic and start thinking about Patterns. Advanced Python is about: ~Abstraction: Using Protocols and Generics for structural typing. ~Automation: Using advanced Decorators to inject behavior across entire systems. ~Reliability: Understanding the memory manager so you can prevent leaks before they start. We use these "hidden" features not to make the code more complex, but to make the usage of our code more simple for everyone else. Which part of the "Advanced Mindset" was the hardest for you to learn? For me, it was finally mastering asyncio flow control. #CleanCode #Pythonic #ProgrammingPrinciples #SystemDesign #AdvancedCoding

✅ Day 92 of 100 Days LeetCode Challenge Problem: 🔹 #2011 – Final Value of Variable After Performing Operations 🔗 https://lnkd.in/gX-JQNUJ Learning Journey: 🔹 Today’s problem was about evaluating a sequence of increment and decrement operations. 🔹 I initialized a variable ans = 0 to track the value. 🔹 Used a hashmap to map each operation to its effect: • "++X" and "X++" → +1 • "--X" and "X--" → -1 🔹 Iterated through the operations and updated ans accordingly. 🔹 Returned the final computed value. Concepts Used: 🔹 HashMap / Dictionary 🔹 String Matching 🔹 Simple Simulation Key Insight: 🔹 Instead of using multiple condition checks, mapping operations to values simplifies logic and improves readability. Complexity: 🔹 Time: O(n) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers

🚀 Another LeetCode Problem Solved: Palindrome Number! 🔗 Check out my solution: https://lnkd.in/dwDMqXXn 💡 Problem Overview Given an integer x, determine whether it is a palindrome — meaning it reads the same forward and backward. (LeetCode) Examples: ✔ 121 → Palindrome ❌ -121 → Not a palindrome ❌ 10 → Not a palindrome 🧠 My Approach (Digit Reversal) Instead of converting the number to a string, I used a mathematical approach: Extract digits using % 10 Reverse the number step by step Compare reversed number with original ⚙️ Key Learnings ✔ Strong understanding of number manipulation ✔ Importance of handling edge cases (negative numbers, trailing zeros) (leet-solution.com) ✔ Practicing clean and efficient logic ⏱️ Complexity • Time Complexity: O(log n) • Space Complexity: O(1) 🔥 Why this problem matters Even though it’s an “easy” problem, it builds: Logical thinking Problem-solving fundamentals Confidence for bigger challenges #LeetCode #DSA #Python #CodingJourney #ProblemSolving #100DaysOfCode