Dileep Kumar Maurya’s Post

🔥 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

🔥 DSA Challenge – Day 125/360 🚀 📌 Topic: Recursion 🧩 Problem: Combination Sum Problem Statement: Given an array of distinct integers and a target, return all unique combinations where numbers sum up to the target. Same element can be used multiple times. 🔍 Example: Input: candidates = [2,3,6,7], target = 7 Output: [[2,2,3],[7]] 💡 Approach: Backtracking 1️⃣ Step 1 – Start from index 0 and try picking each element 2️⃣ Step 2 – If element ≤ target, include it and reduce target 3️⃣ Step 3 – Backtrack (remove element) and move to next index ✔ Use recursion to explore all possibilities ✔ Reuse same element (stay on same index) ✔ Stop when target becomes 0 (valid answer) ✔ Skip when index reaches end ⏱ Complexity: Time: O(2^n * k) (k = avg length of combination) Space: O(k * x) (x = number of combinations) 📚 Key Learning: Backtracking is all about making choices, exploring, and undoing them efficiently. #DSA #Java #Coding #InterviewPrep #ProblemSolving #TechJourney #360DaysOfCode #LeetCode #Backtracking 🚀

🚀 Day 82 – DSA Journey | Same Tree using Recursion Continuing my daily DSA practice, today I worked on a problem that strengthened my understanding of tree comparison and recursion. 📌 Problem Practiced: Same Tree (LeetCode 100) 🔍 Problem Idea: Given two binary trees, determine whether they are identical in both structure and node values. 💡 Key Insight: To check if two trees are the same, we need to compare nodes at every level — both their values and their left & right subtrees. Recursion makes this process clean and intuitive. 📌 Approach Used: • If both nodes are null → trees match at this point • If one is null or values differ → trees are not the same • Recursively compare left subtrees • Recursively compare right subtrees • Both sides must match for the trees to be identical 📌 Concepts Strengthened: • Tree traversal • Recursion • Structural comparison • Base case handling ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(h) (recursion stack) 🔥 Today’s takeaway: Tree problems often become simple when broken down recursively — solve smaller subtrees to solve the whole tree. On to Day 83! 🚀 #Day82 #DSAJourney #LeetCode #BinaryTree #Recursion #Java #ProblemSolving #Coding #LearningInPublic #Consistency

🔥 DSA Challenge – Day 123/360 🚀 📌 Topic: Recursion 🧩 Problem: Generate Parentheses Problem Statement: Given n pairs of parentheses, generate all combinations of well-formed (valid) parentheses. 🔍 Example: Input: n = 3 Output: ["((()))","(()())","(())()","()(())","()()()"] 💡 Approach: Backtracking 1️⃣ Step 1 – Start with an empty string and track open (oc) and close (cc) brackets 2️⃣ Step 2 – Add "(" if oc < n (we still have opening brackets left) 3️⃣ Step 3 – Add ")" only if cc < oc (to maintain valid structure) 👉 This ensures we only generate valid combinations (no extra filtering needed) ⏱ Complexity: Time: O(2ⁿ) (Catalan growth) Space: O(n) recursion stack 📚 Key Learning: Backtracking works best when we avoid invalid choices early, instead of fixing them later. #DSA #Java #Coding #InterviewPrep #ProblemSolving #TechJourney #123DaysOfCode #LeetCode #Backtracking

🚀 Day 19/100 — #100DaysOfLeetCode Back to mastering one of the most powerful patterns in DSA — Sliding Window 💻🔥 ✅ Problem Solved: 🔹 LeetCode 1358 — Number of Substrings Containing All Three Characters 💡 Concept Used: Sliding Window + Frequency Tracking 🧠 Key Learning: The goal was to count all substrings containing 'a', 'b', and 'c'. Instead of checking every possible substring, I learned how: Expand the window until all required characters are present. Once valid, every further extension also forms valid substrings. Count substrings efficiently while shrinking the window. This converts a brute-force O(n²) approach into an optimized O(n) solution. ⚡ Big Insight: Sliding Window isn’t just about finding length — it can also be used for counting valid substrings efficiently. Consistency is slowly turning patterns into instincts 🚀 #100DaysOfLeetCode #LeetCode #DSA #SlidingWindow #Algorithms #Java #ProblemSolving #CodingJourney #LearningInPublic #Consistency

🔥 Day 140/360 – This Trick Generates All Subsets Instantly 🚀 Most people solve this using recursion… But today I learned how to generate all subsets using Bit Manipulation👇 📌 Topic: Bit Manipulation + Array 🧩 Problem: Subsets (Power Set) 📝 Problem Statement: Given an integer array, return all possible subsets (the power set). 🔍 Example: Input: [1, 2, 3] Output: [[], [1], [2], [1,2], [3], [1,3], [2,3], [1,2,3]] 💡 Approach: Bit Masking (Optimized) ✔ Step 1 – Calculate total subsets = 2ⁿ using (1 << n) ✔ Step 2 – Loop from 0 to 2ⁿ - 1 (each number represents a subset) ✔ Step 3 – Use bits to decide whether to include an element ⚡ Key Idea: Each number (mask) represents a subset in binary form. If a bit is ON → include that element. ⏱ Complexity: Time → O(n × 2ⁿ) Space → O(n × 2ⁿ) 📚 What I Learned: Bit manipulation can replace recursion in subset generation and makes the logic easier to visualize in binary. 🚀 Why This Matters: This concept is widely used in backtracking, DP, and interview problems. #DSA #Java #Coding #ProblemSolving #InterviewPrep #LeetCode #BitManipulation #TechJourney #140DaysOfCode

🔥 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 37 #SDE backtracking and subset generation patterns. Solved: • Power Set in Lexicographic Order • Combination Sum Key Learning: • Generating a power set strengthens understanding of recursion trees and how to systematically explore all subsets in a structured (lexicographic) manner. • “Combination Sum” reinforces backtracking with choices and pruning, where we explore combinations while respecting constraints like target sum. #LeetCode #DSA #Backtracking #Recursion #Algorithms #Java #SoftwareEngineering

#100DaysOfCode – Day 2 Today I worked on a DSA problem based on arrays: Check if an array is sorted and rotated 🔍 Approach: Instead of finding the exact rotation point, I focused on identifying a pattern: In a sorted and rotated array, the order should break at most once. So, I checked how many times an element is greater than the next element while traversing the array in a circular manner. ✔️ If the count of such breaks is 0 or 1 → valid ❌ If it’s more than 1 → not a sorted rotated array 🧠 Key Takeaway: This problem taught me how pattern observation can simplify logic and avoid unnecessary complexity. Sometimes the best solution is not the most obvious one! 📈 Staying consistent and improving step by step 💪 #100DaysOfCode #DSA #DataStructures #Algorithms #Java #CodingJourney #ProblemSolving #LeetCode #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