Optimizing Range Updates with XOR in Java

🚀 LeetCode Challenge 22/50 💡 Approach: Space-Optimized Dynamic Programming A robot moves only RIGHT or DOWN on an m×n grid. How many unique paths reach the bottom-right corner? Recursion explodes exponentially — DP solves it beautifully, and we can do even better by ditching the full 2D grid! 🔍 Key Insight: → Every cell's paths = paths from ABOVE + paths from LEFT → First row & first column always have exactly 1 path each → We only ever need the CURRENT and PREVIOUS row → So one 1D array updated in-place is enough! 📈 Complexity: ❌ Recursion → O(2^(m+n)) Time ❌ 2D DP → O(m×n) Time, O(m×n) Space ✅ 1D DP → O(m×n) Time, O(n) Space   Same speed, fraction of the memory! DP teaches us to build solutions from the ground up — one step, one row, one decision at a time. 🤖 #LeetCode #DSA #DynamicProgramming #Java #ADA #PBL2 #LeetCodeChallenge #Day22of50 #CodingJourney #ComputerEngineering #AlgorithmDesign #UniquePaths

Day 25 o#SDE Problem on matrix search and in-place array transformation problems. Solved: • Search a 2D Matrix II • Modify an array such that if arr[i] is j, then arr[j] becomes i Key Learning: • Matrix problems can often be optimized by leveraging sorted row/column properties, reducing time complexity significantly. • In-place array modification problems require careful handling to avoid overwriting values — techniques like encoding/decoding values help achieve this in O(1) extra space. #LeetCode #DSA #Arrays #Matrices #Algorithms #Java #SoftwareEngineering

🚀 LeetCode Challenge 8/50 🔍 Problem: Single Number Today’s problem introduced a powerful concept from bit manipulation. 💡 Approach: I used the XOR technique to solve this problem efficiently: XOR of a number with itself becomes 0 XOR of a number with 0 remains the number So, all duplicate elements cancel out, leaving only the unique element ⚡ This eliminates the need for extra space and keeps the solution highly optimized. 📊 Complexity Analysis: Time Complexity: O(n) Space Complexity: O(1) 📚 Key Learning: Bit manipulation can simplify problems significantly when patterns are recognized. The XOR trick is a must-know concept for optimizing array-based problems. Learning something new every day 🚀 #LeetCode #Algorithms #DataStructures #ProblemSolving #CodingJourney #Java #100DaysOfCode #StudentDeveloper #Learning

Day 44 #SDE sliding window techniques and optimization problems. Solved: • Longest Substring Without Repeating Characters • Maximum Points You Can Obtain from Cards Key Learning: • “Longest Substring Without Repeating Characters” is a classic sliding window + hashing problem, where we dynamically adjust the window to maintain unique characters. • “Maximum Points from Cards” uses a clever window optimization, where instead of picking k cards directly, we minimize the sum of the remaining subarray. #LeetCode #DSA #SlidingWindow #Algorithms #Java #SoftwareEngineering

Day 105: Circular Arrays & Shortest Paths 🔄 Problem 2515: Shortest Distance to Target String in a Circular Array Today’s challenge involved finding the minimum steps to reach a target string in a circular array, allowing movement in both directions. The Strategy: • Bidirectional Search: Since the array is circular, the distance can be calculated in two ways: moving forward or moving backward. • Modular Arithmetic: I used (dist + n) % n to handle the wrap-around logic seamlessly, ensuring the index stays within bounds regardless of the direction. • Optimization: By iterating once through the array and comparing the distances for every occurrence of the target, I maintained an O(N) time complexity. Sometimes the most elegant way to handle a "circular" problem is simply embracing the symmetry of the path. 🚀 #LeetCode #Java #Algorithms #ProblemSolving #DailyCode

🚀 Day 556 of #750DaysOfCode 🚀 🧠 Today’s Problem: XOR After Range Multiplication Queries I (LeetCode - Medium) This problem combines array simulation + modular arithmetic + bit manipulation (XOR) — a great mix to strengthen fundamentals. 💡 Key Idea: For each query [l, r, k, v], we don’t update every element — only indices: 👉 l, l+k, l+2k, ... ≤ r Each selected element is multiplied by v and taken modulo 10^9 + 7 After processing all queries → compute XOR of the final array ⚙️ Approach I Used: Direct simulation (since constraints are small) Carefully handled: ✔ Step jumps (k) ✔ Modulo operations to avoid overflow ✔ Final XOR computation 🔥 Learning Takeaways: Not all range problems need segment trees — sometimes smart simulation is enough Always check constraints before optimizing Combining multiple concepts (loops + mod + XOR) builds strong problem-solving intuition 💻 Tech Used: Java, Arrays, Simulation Consistency > Perfection. One problem at a time 💪 #LeetCode #Java #ProblemSolving #DataStructures #Algorithms #CodingJourney #Consistency #100DaysOfCode

Day 48 #SDE two-pointer techniques and cycle detection patterns. Solved: • Container With Most Water • Find the Duplicate Number Key Learning: • “Container With Most Water” is a classic two-pointer problem, where moving the smaller height pointer helps maximize area efficiently. • “Find the Duplicate Number” can be solved using Floyd’s Cycle Detection (Tortoise & Hare) without modifying the array and using O(1) space. #LeetCode #DSA #TwoPointers #Algorithms #Java #SoftwareEngineering

Day 20 of my #30DayCodeChallenge: Efficiency through Pruning! Today's challenge was Word Search II-a complex puzzle that tests how you manage massive search spaces. The Logic: 1. Trie Integration: I stored the dictionary in a Trie to check prefixes in O(1) time. 2. DFS & Backtracking: Explored the grid cell by cell, but with a twist... 3. Intelligent Pruning: If a path doesn't match a Trie prefix, the search stops immediately. This turns an exponential problem into something much more manageable. Coding isn't just about finding the answer; it's about finding it before your timer runs out! #Java #DataStructures #Backtracking #Trie #Algorithms #CleanCode #150DaysOfCode

🚀 Day 55 of #100DaysOfCode – Rotate Image Today I worked on an interesting matrix problem: Rotate Image (90° clockwise) 🔄 💡 Key Learning: Instead of using extra space, the challenge is to rotate the matrix in-place. 🧠 Approach I used: 1️⃣ Transpose the matrix (convert rows → columns) 2️⃣ Reverse each row This combination effectively rotates the matrix by 90° clockwise without using extra memory. 📌 Example: Input: [[1,2,3], [4,5,6], [7,8,9]] Output: [[7,4,1], [8,5,2], [9,6,3]] ⚡ Complexity: Time: O(n²) Space: O(1) (in-place) 💻 Implemented in Java and successfully passed all test cases ✅ This problem really helped me strengthen my understanding of matrix manipulation and in-place algorithms. #LeetCode #DataStructures #Java #CodingPractice #ProblemSolving #Algorithms #100DaysOfCode