LeetCode: Longest Consecutive Sequence with Hash Set

🚀 50 Important Coding Questions – Question 36/50 🔹 Next Greater Element I | LeetCode A classic Monotonic Stack problem used frequently in coding interviews. 📌 Problem Statement You are given two arrays nums1 and nums2. For each element in nums1, find the next greater element in nums2. The next greater element of x is the first element to the right of x in nums2 that is greater than x. If it does not exist → return -1. Example: Input nums1 = [4,1,2] nums2 = [1,3,4,2] Output [-1,3,-1] 💡 Approach We use a Monotonic Decreasing Stack. Steps: 1️⃣ Traverse nums2 from right to left 2️⃣ Remove all elements from stack ≤ current element 3️⃣ The stack top becomes the next greater element 4️⃣ Store results in a map/array 5️⃣ Finally answer queries for nums1 This avoids checking every element repeatedly. ⏱ Time Complexity: O(n + m) 📦 Space Complexity: O(n) 📌 LeetCode Result: ✔ Accepted ⚡ Runtime: 0 ms 🧠 Concepts Strengthened ✔ Monotonic Stack ✔ Stack-based optimization ✔ Array traversal techniques ✔ Precomputation for faster queries 📍 Question 36 of 50 in my “50 Important Coding Questions” series. Every problem solved improves algorithmic thinking step by step 💯 👉 Question 37 coming next! #DSA #LeetCode #Stack #MonotonicStack #CodingInterview #ProblemSolving #CPlusPlus #TechJourney

🚀 Day 131 – LeetCode 150 Days Challenge #LeetCode #DSA #DynamicProgramming #Recursion #CodingJourney 🧩 Unique Paths II (With Obstacles) Today’s problem was Unique Paths II, where we need to count the number of ways to reach the bottom-right cell of a grid, but some cells contain obstacles. This is a classic extension of Unique Paths, but with an additional constraint: 👉 If a cell contains 1, it is blocked. 👉 If it contains 0, we can step on it. 💡 Approach – Recursion + Memoization (Top-Down DP) At every cell (i, j): We can move right → (i, j+1) Or move down ↓ (i+1, j) So the recurrence becomes: ways(i, j) = ways(i+1, j) + ways(i, j+1) But only if: We are inside the grid. The current cell is NOT an obstacle. Very Important: Order of Base Conditions This problem looks simple, but the order of checks is extremely important. 🔎 Why Order Matters? ❌ Mistake: Checking Destination First if(i==m-1 && j==n-1) return 1; But what if the destination cell itself is an obstacle! Then we would incorrectly return 1. That’s why in uniquePathsWithObstacles() we handle this first: if(obstacleGrid[m-1][n-1] == 1) return 0; 🧠 Important Learning: Condition Priority When writing recursive grid solutions, always think in this order: 1️⃣ Out of bounds check i >= m || j >= n 2️⃣ Obstacle check obstacleGrid[i][j] == 1 3️⃣ Destination check i == m-1 && j == n-1 Because: If you access obstacleGrid[i][j] before boundary check → ❌ Runtime error. If you check destination before obstacle → ❌ Wrong answer. If you forget to handle blocked destination → ❌ Wrong answer. Order is not just syntax. Order defines correctness. ⏱️ Time & Space Complexity Time: O(m × n) Space: O(m × n) for DP + recursion stack 🎯 Key Takeaway In DP problems: Writing recurrence is easy. Writing correct base cases in the correct order is the real skill. Today reinforced one important lesson: Small condition mistakes → Big logical errors. Day 131 completed ✅ On to Day 132 🔥

🚀 50 Important Coding Questions – Question 34/50 🔹 Valid Parentheses | LeetCode A fundamental Stack-based problem that checks whether parentheses in a string are properly balanced. 📌 Problem Statement Given a string containing just the characters ( ) { } [ ], determine if the input string is valid. A string is valid if: ✔ Open brackets are closed by the same type ✔ Open brackets are closed in the correct order ✔ Every closing bracket has a matching opening bracket Example: Input: s = "()[]{}" Output: true 💡 Approach 1️⃣ Traverse the string character by character 2️⃣ Use a stack to store opening brackets 3️⃣ When a closing bracket appears, check the top of the stack 4️⃣ If they form a valid pair → pop from stack 5️⃣ Otherwise → invalid string ✔ Stack helps track unmatched opening brackets. ⏱ Time Complexity: O(n) 📦 Space Complexity: O(n) 📌 LeetCode Result: ✔ Accepted ⚡ Efficient stack-based implementation 🧠 Concepts Strengthened ✔ Stack data structure ✔ String traversal ✔ Matching pair logic ✔ Edge case handling 📍 Question 34 of 50 in my “50 Important Coding Questions” series. Every problem solved builds stronger algorithmic thinking 💯 👉 Question 35 coming next! #DSA #LeetCode #Stack #CodingInterview #ProblemSolving #CPlusPlus #TechJourney

🚀 Day 12/50 – LeetCode Challenge 🧩 Problem: Generate Parentheses Today’s problem focused on generating all possible combinations of valid parentheses for a given number n. This problem is a great example of backtracking and recursion. 📌 Problem Summary: Given n pairs of parentheses, generate all combinations of well-formed parentheses. For example: If n = 3, the valid combinations are: ((())), (()()), (())(), ()(()), ()()() 🔍 Approach Used (Backtracking) ✔ Start building the string step by step ✔ Add an opening bracket ( if the count is less than n ✔ Add a closing bracket ) only if it keeps the sequence valid ✔ Continue exploring possibilities recursively until the string length reaches 2n Backtracking ensures that we only build valid parentheses combinations. ⏱ Time Complexity: Approximately O(4ⁿ / √n) 📦 Space Complexity: O(n) 💡 Key Learning ✔ Understanding backtracking technique ✔ Building valid combinations with constraints ✔ Managing recursion and decision trees Problems like this improve logical thinking and help understand how recursive exploration works. Consistency in practice leads to stronger problem-solving skills 🚀 🔗 Problem Link: https://lnkd.in/gR-8BihF #50DaysOfLeetCode #LeetCode #DSA #Backtracking #Recursion #ProblemSolving #CodingJourney #FutureAIEngineer #Consistency

🥷 Day 301 of My 365 LeetCode Challenge is done! 💥 Kicked off strong, solved my problem, and the coding vibe is awesome! 🧠 📌 LeetCode 1758 – Minimum Changes To Make Alternating Binary String Sometimes the simplest problems offer the best opportunity to strengthen your fundamentals. Today’s challenge focused on string manipulation and pattern recognition — two concepts that frequently appear in coding interviews. 🧠✨ 🔍 Problem Overview We’re given a binary string consisting only of 0s and 1s. The goal is to determine the minimum number of changes required to make the string alternating. An alternating string means: ✔ No two adjacent characters are the same ✔ Valid patterns would look like "010101..." or "101010..." So essentially, the string must follow one of two possible patterns. ⚙️ Key Idea Behind the Solution Instead of trying complex transformations, we can simply compare the string against the two valid alternating patterns: 1️⃣ Pattern starting with 0 → 010101... 2️⃣ Pattern starting with 1 → 101010... By counting how many positions differ from each pattern, we get the number of changes required for each scenario. The final answer is simply the minimum of the two counts. 💡 Why This Problem Matters Even though it’s categorized as an easy problem, it reinforces several important skills: • Pattern-based thinking • Efficient string traversal • Writing concise and optimized logic • Recognizing that sometimes brute-force comparison can still be optimal 🧠 Key Takeaway Not every coding challenge requires complex algorithms. Many problems reward clear observation and structured thinking. Mastering these fundamentals builds a strong foundation for tackling more advanced algorithmic challenges. On to the next challenge! 💪🔥 #LeetCode #DSA #Strings #Algorithms #ProblemSolving 🧠 #CodingJourney #InterviewPrep #CPlusPlus 🚀

𝙔𝙤𝙪 𝙙𝙤𝙣’𝙩 𝙣𝙚𝙚𝙙 𝙩𝙤 𝙨𝙤𝙡𝙫𝙚 100𝙨 𝙤𝙛 𝙘𝙤𝙙𝙞𝙣𝙜 𝙦𝙪𝙚𝙨𝙩𝙞𝙤𝙣𝙨. 𝙔𝙤𝙪 𝙟𝙪𝙨𝙩 𝙣𝙚𝙚𝙙 𝙩𝙤 𝙪𝙣𝙙𝙚𝙧𝙨𝙩𝙖𝙣𝙙 𝙥𝙖𝙩𝙩𝙚𝙧𝙣𝙨. Most people do this ❌ → Random LeetCode questions Top candidates do this ✅ → Learn patterns → apply everywhere I came across 20 coding patterns and honestly… it simplifies everything Instead of memorizing problems focus on these 👇 ➥ Sliding Window → for subarrays & substrings ➥ Two Pointers → for sorted arrays ➥ Fast & Slow Pointers → cycle detection ➥ Merge Intervals → overlapping ranges ➥ Top K Elements → heap-based problems ➥ Binary Search → optimized searching …and many more patterns like these in doc Don’t ask → 𝗛𝗼𝘄 𝘁𝗼 𝘀𝗼𝗹𝘃𝗲 𝘁𝗵𝗶𝘀 𝗾𝘂𝗲𝘀𝘁𝗶𝗼𝗻? Ask → 𝗪𝗵𝗶𝗰𝗵 𝗽𝗮𝘁𝘁𝗲𝗿𝗻 𝗶𝘀 𝘁𝗵𝗶𝘀? Once you identify the pattern, the solution becomes much easier. Stop solving randomly. Start learning patterns. That’s the real shortcut Doc Credit - Design Gurus ♻️ Repost if you found this useful 🤝 Follow Sattari Sateesh Kumar for more  👨💻 For 1:1 guidance → https://topmate.io/sateesh #python #pyspark #pysparklearning #dataengineering #sqllearning #dataengineeringinterview #azuredataengineer #bigdata #spark #datalearning #datacareer #azuredataengineering #dataengineeringjobs #linkedinlearning #dataengineeringlearning

🥷 Day 307 of My 365 LeetCode Challenge is done! 💥 Kicked off strong, solved my problem, and the coding vibe is awesome! 🧠 📌 LeetCode 1009 – Complement of Base 10 Integer Today’s problem was a nice exercise in bit manipulation. Even though the problem looks simple, it highlights how understanding binary representation can make solutions much more efficient. 🧠 🔍 Problem Overview We are given a non-negative integer n. The task is to find its bitwise complement — meaning we flip every bit in its binary representation. Example: • 5 in binary → 101 • Flip all bits → 010 • Result in decimal → 2 However, the complement should only consider the significant bits of the number (the bits actually used in its binary representation). 💡 Key Idea Behind the Solution Instead of flipping bits blindly, we first determine how many bits are used to represent the number. Steps: 1️⃣ Build a mask that has all bits set to 1 for the length of the number. 2️⃣ Apply the bitwise NOT operation on the number. 3️⃣ Use the mask with AND operation to keep only the valid bits. This ensures we only flip the meaningful bits and ignore leading zeros. ⚙️ Concepts Used ✔ Bit Manipulation ✔ Bit Masking ✔ Binary Representation ✔ Efficient Bitwise Operations 🧠 Key Takeaway Problems like this reinforce how powerful bitwise operations can be. Understanding how numbers are represented in binary can often lead to clean and highly efficient solutions. Consistent daily practice continues to improve logical thinking and problem-solving skills. #LeetCode #DSA 🧠 #BitManipulation #Algorithms #ProblemSolving #CodingJourney #InterviewPrep 🚀

🚀 LeetCode Day Problem Solving 🚀 Day-14 📌 Problem: The complement of an integer is obtained by flipping all bits in its binary representation. 👉 Replace every 0 with 1 and every 1 with 0. Given an integer n, return its binary complement in decimal form. 🧠 Examples: 🔹 Input: n = 5 ✅ Output: 2 📖 Explanation: Binary of 5 → 101 Complement → 010 Decimal value → 2 🔹 Input: n = 7 ✅ Output: 0 📖 Explanation: Binary of 7 → 111 Complement → 000 Decimal value → 0 🔹 Input: n = 10 ✅ Output: 5 📖 Explanation: Binary of 10 → 1010 Complement → 0101 Decimal value → 5 💡 Key Insight: Steps to solve: ✔ Convert the number to binary ✔ Flip each bit (0 → 1, 1 → 0) ✔ Convert the flipped binary back to decimal In Python, we can also use a bitmask trick to flip only the significant bits. 📊 Complexity Analysis: ⏱ Time Complexity: O(log n) 📦 Space Complexity: O(1) Efficient for 0 ≤ n < 10⁹. 🧠 What I Learned: ✔ Binary manipulation problems often rely on bitwise operations ✔ Masks help flip only the required bits ✔ Understanding binary representation is essential for bit manipulation problems ✅ Day 14 Completed 14 days of consistent DSA practice! Every day getting better at problem solving and logical thinking 🚀🔥 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency 🚀

🚀 Day 60 of LeetCode Problem Solving Journey — 100 Days LeetCode Challenge Today, I solved LeetCode #485 — Max Consecutive Ones using C++, under the guidance of Trainer NEKAL SINGH SALARIA Singh at REGex Software Services. 🔍 Problem Summary: Given a binary array nums, return the maximum number of consecutive 1's in the array. 🧠 Approach Used (Linear Traversal): I used a simple counting technique: Traverse the array once If element is 1 → increment counter Update maximum length using max() If element is 0 → reset counter This approach is efficient, clean, and uses only one pass. 📚 Key Learnings of the Day ✔ Linear traversal is powerful for sequence problems ✔ Maintaining a running count avoids extra storage ✔ Reset logic is key when tracking consecutive patterns ✔ Simple logic often leads to optimal solutions ⏱ Complexity • Time Complexity: O(n) • Space Complexity: O(1) 💡 Optimization Insight: This is already the optimal solution because it uses a single pass and constant space. The main improvement focus would be writing cleaner or more readable code rather than reducing complexity. Consistency is the key — see you on Day 61 🔥 #Day60 #100DaysLeetCodeChallenge #LeetCode #RegexSoftwareServices #NekalSingh #ProblemSolving #DSA #CPlusPlus #CodingChallenge #ProgrammingJourney #Arrays #KeepGrowing

The secret isn't solving 500 problems randomly. It's solving the right problems with the right progression. . . 150 days of consistency can build the kind of thinking interviews actually test. 1. Arrays & Strings Most coding rounds start here. Practice problems on prefix sum, two pointers, sliding window, and subarrays. 2. Hashing HashMaps and HashSets help reduce time complexity drastically Be comfortable solving frequency and lookup based problems. 3. Recursion & Backtracking Important for problems like subsets, permutations, combinations, and N-Queens. 4. Linked List Reverse linked list, cycle detection, merging lists, and pointer manipulation. 5. Stack & Queue Learn monotonic stack, next greater element, and basic queue operations. 6. Trees & Binary Trees Tree traversals (inorder, preorder, postorder), height, diameter, LCA, and BFS. 7. Binary Search Not just on arrays. Also practice binary search on answer problems. 8. Dynamic Programming Start with memoization, then tabulation. Classic problems: knapsack, LIS, DP on grids. 9. Graphs BFS, DFS, shortest path, topological sort, union find. 10. Time & Space Complexity Always analyze the complexity of your solution before finishing. Connect Swadesh Kumar for more such content Comment 'MAANG' & I'll DM you the PDF directly 150 focused probelms> 300+ random question practice. If you're preparing seriously for MAANG, this roadmap can change your trajectory.