Bit Manipulation for Subset Generation

#CodeEveryday — My DSA Journey | Day 14 🧩 Problem Solved: Search a 2D Matrix (LeetCode #74) 💭 What I Learned: Used binary search by treating the 2D matrix as a flattened sorted array. Key idea: 👉 Convert 1D index → 2D coordinates using: row = mid / number_of_columns col = mid % number_of_columns At each step: ✔️ Calculated row and column from mid ✔️ Compared matrix value with target ✔️ Adjusted search space accordingly This approach avoids nested loops and gives optimal performance. ⏱ Time Complexity: O(log (m × n)) 🧠 Space Complexity: O(1) ⚡ Key Takeaways: ✔️ 2D problems can often be reduced to 1D using indexing tricks ✔️ Binary search works whenever the data is globally sorted ✔️ Index mapping is a powerful technique in matrix problems 💻 Language Used: Java ☕ 📘 Concepts: Binary Search · Matrix · Index Mapping · Optimization #CodeEveryday #DSA #LeetCode #Java #BinarySearch #Matrix #ProblemSolving #Algorithms #CodingJourney #Consistency 🚀

🔥 DSA Challenge – Day 126/360 🚀 📌 Topic: Array + Backtracking (Recursion) 🧩 Problem: Combination Sum II Problem Statement: Find all unique combinations in an array where numbers sum up to a target. Each number can be used only once, and duplicate combinations are not allowed. 🔍 Example: Input: candidates = [10,1,2,7,6,1,5], target = 8 Output: [[1,1,6], [1,2,5], [1,7], [2,6]] 💡 Approach: Backtracking + Pruning 1️⃣ Step 1 – Sort the array to handle duplicates easily 2️⃣ Step 2 – Use recursion to pick elements and reduce target 3️⃣ Step 3 – Skip duplicates & backtrack after each recursive call 👉 Use condition to skip duplicates: if(i > ind && arr[i] == arr[i-1]) continue; 👉 Stop early if element exceeds target (pruning) ⏱ Complexity: Time: O(2^n) Space: O(k * x) (for storing combinations) 📚 Key Learning: Sorting + duplicate skipping is the key trick to avoid repeated combinations in backtracking problems. #DSA #Java #Coding #InterviewPrep #ProblemSolving #TechJourney #360DaysOfCode #LeetCode

🔥 DSA Challenge – Day 127/360 🚀 📌 Topic: Backtracking 🧩 Problem: Subsets II Problem Statement: Given an integer array that may contain duplicates, return all possible subsets such that the solution set does not contain duplicate subsets. 🔍 Example: Input: nums = [1,2,2] Output: [[], [1], [1,2], [1,2,2], [2], [2,2]] 💡 Approach: Backtracking + Sorting 1️⃣ Step 1 – Sort the array to group duplicate elements together 2️⃣ Step 2 – Use recursion to generate all subsets 3️⃣ Step 3 – Skip duplicate elements using condition (i != ind && nums[i] == nums[i-1]) ⏱ Complexity: Time: O(2^n) Space: O(n) (recursion stack + subset storage) 📚 Key Learning: Sorting + smart skipping of duplicates helps avoid repeated subsets in backtracking problems. #DSA #Java #Coding #InterviewPrep #ProblemSolving #TechJourney #360DaysOfCode #LeetCode #Backtracking

🚀 Starting a New Topic: Recursion 🔥 DSA Challenge – Day 118/360 📌 Topic: String 🧩 Problem: String to Integer (atoi) Problem Statement: Convert a string into a 32-bit signed integer while handling spaces, sign, non-digit characters, and overflow. 🔍 Example: Input: " -42" Output: -42 💡 Approach: Recursion + Parsing 1️⃣ Step 1 – Skip leading spaces and detect sign (+ / -) 2️⃣ Step 2 – Recursively process each digit and build the number 3️⃣ Step 3 – Check overflow at every step and return result ⏱ Complexity: Time: O(n) Space: O(n) (recursive stack) 📚 Key Learning: Always handle overflow during computation, not after building the full number. Recursion simplifies problems by breaking them into smaller subproblems. 💭 New Topic Journey Begins: Excited to dive deep into Recursion and master it step by step! 💪 #DSA #Java #Coding #InterviewPrep #ProblemSolving #TechJourney #118DaysOfCode #LeetCode #Recursion

🔥 DSA Challenge – Day 128/360 🚀 📌 Topic: Recursion 🧩 Problem: Letter Combinations of a Phone Number 💡 Problem Statement: Given a string of digits (2–9), return all possible letter combinations that the number could represent based on the classic phone keypad mapping. 🔍 Example: Input: "23" Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"] 💡 Approach: Backtracking (Recursion) We treat this like a decision tree: Each digit maps to multiple characters For every digit, we try all possible letters Build combinations step-by-step using recursion 👉 Key Idea: Choose → Explore → Backtrack Use StringBuilder for efficient string manipulation ⚙️ Steps: Create a mapping for digits → letters Start recursion from index 0 For each digit: Loop through its mapped letters Add letter to current string Move to next digit Backtrack (remove last character) 🚀 Code Highlights: ✔️ Base case: when index reaches end → add combination ✔️ Efficient backtracking using StringBuilder ✔️ Avoids extra space by modifying same string ⏱️ Time Complexity: O(4^N) 📦 Space Complexity: O(N) (recursion stack) Master this → You master recursion 🔥 #DSA #Java #Backtracking #CodingInterview #LeetCode #128DaysOfCode #Programming #Tech #ProblemSolving

Worked on a challenging problem: “Subarrays with K Different Integers” Key takeaway: Instead of directly solving for exactly K distinct elements, I learned a smarter approach: 👉 count(at most K) − count(at most K−1) 🔹 Concepts I practiced: Sliding Window technique HashMap for frequency tracking Two-pointer approach 🔹 What stood out: The idea of counting all valid subarrays ending at each index using (r - l + 1) was really powerful. It completely changed how I think about subarray problems. Always learning, one problem at a time. #DataStructures #Algorithms #Java #LeetCode #SlidingWindow #LearningJourney #ProblemSolving

🔥 DSA Challenge – Day 131/360 🚀 📌 Topic: Backtracking / Recursion 🧩 Problem: N-Queens Problem Statement: Place N queens on an N×N chessboard such that no two queens attack each other. 🔍 Example: Input: n = 4 Output: [[".Q..","...Q","Q...","..Q."], ["..Q.","Q...","...Q",".Q.."]] 💡 Approach: Optimized Backtracking (Hashing) 1️⃣ Step 1 – Try placing a queen column by column 2️⃣ Step 2 – Use arrays to check if row & diagonals are safe in O(1) 3️⃣ Step 3 – Place queen → recurse → backtrack if needed ⏱ Complexity: Time: O(N!) Space: O(N) + recursion stack 📚 Key Learning: Using hashing for rows & diagonals avoids repeated checks and makes backtracking much faster ⚡ #DSA #Java #Coding #InterviewPrep #ProblemSolving #TechJourney #131DaysOfCode #LeetCode

Day 33 #SDE string algorithms and two-pointer techniques. Solved: • Longest Palindromic Substring • Reverse String Key Learning: • “Longest Palindromic Substring” can be solved using the expand around center approach, checking both odd and even length palindromes efficiently. • “Reverse String” reinforces the two-pointer technique, a simple yet powerful pattern for in-place operations. #LeetCode #DSA #Strings #TwoPointers #Algorithms #Java #SoftwareEngineering