"Day 19: LeetCode Challenge - Max Frequency After Operations"

🚀 LeetCode POTD — 1611. Minimum One Bit Operations to Make Integers Zero 🎯 Difficulty: Hard | Topics: Bit Manipulation, Recursion, Mathematical Patterns 🔗 Solution Link: https://lnkd.in/g4RHfa8h Today’s #LeetCode Problem of the Day was one of those that really make you stop and think. The problem looked simple on the surface — transform a number into 0 using a defined set of bit operations — but understanding how the bits interact recursively was the real challenge. At first, I tried approaching it directly by simulating the operations… but quickly realized that wasn’t scalable. I took a hint, tried to reason it out again, but still couldn’t fully connect the dots. Then I watched Mazhar Imam Khan’s explanation — and everything finally clicked! 🎯 The key insight was understanding that for a number like 1100, you can derive its result by simply calculating: 👉 f(1100) = f(1000) - f(100) That single conceptual shift completely changed how I looked at the problem. 💡 Core Idea: Build an array (holder) representing the number of operations required for each bit position. Traverse from the most significant bit downward. If the current bit is set, adjust the result using a sign-flip mechanism to handle transitions between set bits. 🧠 Key Learnings: Many bit-manipulation problems are built on mathematical symmetry, not just logical operations. Understanding the pattern behind Gray Code transformations helps reveal the structure. Sometimes, one line of conceptual clarity from a great teacher saves hours of debugging. 🕒 Complexity: Time: O(size of binary transformation ) Space: O(size of binary transformation + 1) 💬 Takeaway: Don’t hesitate to seek clarity when stuck — a well-explained insight can bridge the gap between confusion and understanding. #LeetCode #ProblemOfTheDay #BitManipulation #HardProblem #DSA #C++ #CodingJourney #Consistency #ProblemSolving #LearningInPublic #100DaysOfCode #SoftwareEngineering #TechCommunity

🥷 Day 164 of My 365 LeetCode Challenge is done! 💥 Kicked off strong, solved my problem, and the coding vibe is awesome! 🧠 💡 LeetCode 1625 — Lexicographically Smallest String After Applying Operations Today’s challenge was an interesting blend of string manipulation, BFS traversal, and modular arithmetic — the kind of problem that truly sharpens both your algorithmic thinking and pattern recognition skills. 🧠✨ We’re given a numeric string s and two integers a and b. Two operations can be performed repeatedly in any order: 1️⃣ Add a to all digits at odd indices (with digits wrapping around modulo 10). 2️⃣ Rotate the string to the right by b positions. The goal? 🔍 Find the lexicographically smallest string possible after performing any number of these operations. Sounds simple, right? But here’s the catch — since operations can be performed infinitely, the problem hides a state-space exploration challenge. The same string can appear multiple times via different paths, so blindly iterating could easily lead to infinite loops or redundant computations. ⚙️ Approach & Thought Process: To systematically explore all possible transformations, I used Breadth-First Search (BFS) — perfect for enumerating all reachable states in minimal steps. 🔸 Step 1: Start from the original string and push it into a queue. 🔸 Step 2: For each string, perform both allowed operations: - Add a to digits at odd indices (handling digit wraparound using (digit + a) % 10). - Rotate the string by b positions. 🔸 Step 3: Use a set to track all previously visited strings to prevent cycles. 🔸 Step 4: Continuously compare and update the smallest lexicographic string seen so far. Once the queue is exhausted, the smallest recorded string is our answer ✅ 🔥 Key Learning: This problem elegantly demonstrates how graph traversal (BFS) can be applied beyond traditional graphs — even on state transformations of strings. Recognizing this pattern is a huge step toward mastering advanced algorithmic thinking. ✨ Every problem like this one is a reminder that great coding isn’t about memorizing — it’s about seeing structure where others see chaos. #LeetCode #CPlusPlus #CodingChallenge #ProblemSolving #Algorithms #BFS #SoftwareEngineering #Programming #LearningJourney #CodeNewbie #DataStructures

🚀 How I Solved LeetCode’s Word Search Problem Without DFS or Recursion When I came across Word Search (LeetCode 79), I noticed that almost every solution used DFS and backtracking. But I wanted to solve it my own way — using index tracking and adjacency logic instead of recursion. My intuition was simple: If a word exists in the grid, every next letter must lie right next to the previous one — either up, down, left, or right. That means their coordinate difference should satisfy: abs(x1 - x2) + abs(y1 - y2) == 1. This small observation became the foundation of my entire approach. 💡 My Thought Process 1️⃣ I mapped every letter in the grid to its coordinates. 2️⃣ I started from all positions of the first letter in the word. 3️⃣ For every next letter, I checked which positions are adjacent and not already used. 4️⃣ I extended paths step by step — no recursion, only iteration. 5️⃣ If any path reached the last letter → the word exists. ⚙️ My Implementation I used a BFS-style iterative search, where each state holds: the current cell (x, y) the current index in the word and a bitmask representing already-used cells. This replaced recursion and heavy visited[][][] arrays with a compact integer mask. Each move just extended to adjacent cells if the next character matched. It handled tricky cases like "ABCB" perfectly → returned false, which even some DFS implementations fail on. ✅ Why It Worked ✔️ No recursion = no stack overflow. ✔️ Bitmask = minimal memory use. ✔️ Adjacency rule = exactly matches grid movement. ✔️ Iterative BFS = simple, efficient, and logical. 🎯 What I Learned You don’t always need to follow the standard algorithms. Sometimes, just trusting your intuition and building around your idea leads to something even better. This problem taught me that creativity and logic go hand in hand. And when you believe in your approach — it truly works. 💪 #Coding #LeetCode #ProblemSolving #Cplusplus #Innovation #DSA #Learning #DeveloperJourney

📌 Day 35/150 – Remove Nth Node From End of List (LeetCode #19) Today’s problem was a classic linked list challenge that tested my understanding of two-pointer techniques — a powerful pattern for solving traversal-based problems efficiently. ⚙️ The task? Given the head of a linked list, remove the n-th node from the end and return the updated list. At first glance, it seems simple — just find the node from the end and delete it. But the twist lies in doing it in a single pass for optimal performance. 🚀 🔹 Brute Force Idea Traverse the list once to count total nodes, then again to remove the (length–n)th node. ✅ Easy to implement ❌ Requires two passes → not optimal 🔹 Optimal Approach – Two Pointer Technique To achieve this in one pass, we use two pointers: fast and slow. 👉 Move the fast pointer n steps ahead. 👉 Then move both fast and slow together until fast reaches the end. 👉 The slow pointer will be just before the target node — remove it cleanly. This elegant trick ensures we traverse the list only once! 🧠 Example: Input: head = [1, 2, 3, 4, 5], n = 2 Output: [1, 2, 3, 5] Here, the 2nd node from the end (4) is removed using the one-pass logic. ⏱️ Time Complexity: O(L) 📦 Space Complexity: O(1) 💡 Learning: Linked list problems often look tricky because of pointer manipulation, but patterns like slow & fast pointers simplify them beautifully. Recognizing and reusing such patterns across problems is a major step in mastering DSA. 💪 #150DaysOfCode #LeetCode #ProblemSolving #LinkedList #TwoPointer #CodingChallenge #SoftwareEngineering #DSA #Cplusplus #Developers #TechCommunity #LearningJourney 🚀

🚀 Day 5 – LeetCode Problem Solving Journey 💻 Today’s problem was “Remove Element”, an easy-level array problem that focuses on in-place modification — a key concept for memory-efficient programming. Problem Statement: Given an array nums and a value val, remove all occurrences of val in-place and return the number of elements that are not equal to val. The relative order of elements may be changed. Example 1: Input: nums = [3,2,2,3], val = 3 Output: 2, nums = [2,2,_,_] Example 2: Input: nums = [0,1,2,2,3,0,4,2], val = 2 Output: 5, nums = [0,1,4,0,3,_,_,_] 🧠 Concept Used: Two-pointer technique In-place replacement of valid elements Time Complexity: O(n) Space Complexity: O(1) Every problem builds consistency, clarity, and confidence — one step closer to mastering DSA. 💪 #LeetCode #Day5 #ProblemSolving #CodingJourney #100DaysOfCode #Python #DSA #Programming #DevelopersCommunity #SoftwareEngineering #TechLearning #CareerGrowth #ipec #ipec30

Day 26 of #50DaysLeetCodeChallenge 🧩 Problem: 561. Array Partition Today’s problem was a refreshing one — an Easy-level LeetCode challenge that reinforces how sorting and pairing can optimize outcomes. It’s one of those problems that looks simple but beautifully demonstrates the power of greedy algorithms. 🧠 Thought Process: 🔹 The task is to group integers into pairs such that the sum of the smaller element in each pair is maximized. 🔹 Sorting the array ensures each pair is as balanced as possible. 🔹 Once sorted, we simply take every alternate element (starting from index 0), since that will always represent the smaller element in each optimal pair. 🔹 This method eliminates unnecessary complexity and ensures maximum total sum. ⚙️ My Approach: 🔸 Sort the array using Arrays.sort(nums). 🔸 Initialize sum = 0. 🔸 Loop through the array with a step of 2, adding nums[i] to the sum each time. 🔸 Return the final sum — the maximum possible result. 📈 Performance: ✅ Accepted ⚡ Runtime: 17 ms (Beats 13.97%) 💾 Memory: Efficient and clean implementation 💡 Reflection: Sometimes, simplicity is the key to optimization. Even without complex data structures, understanding the problem’s core logic can lead to elegant solutions. Each day’s challenge sharpens both reasoning and coding clarity. Grateful to Trainer Shishir chaurasiya and PrepInsta for guiding me through consistent, concept-driven learning in DSA 🙏 #LeetCode #50DaysLeetCodeChallenge #ProblemSolving #Java #AlgorithmDesign #CodingChallenge #PrepInsta #DeveloperJourney #LogicBuilding #DSA #CleanCode #LearningEveryday

🚀 Day 11 of my 120 Days LeetCode Challenge! 🔹 Problem: Container With Most Water Difficulty: Medium Language Used: C++ 🧩 Problem Description: You are given an array height[] where each element represents the height of a vertical line on the x-axis. The goal is to select two lines that, together with the x-axis, form a container that can hold the maximum amount of water. The container’s area is determined by the shorter line’s height and the distance between the two lines. We need to find the pair of lines that form the largest possible area. 💡 Example: Input: height = [1,8,6,2,5,4,8,3,7] Output: 49 Explanation: The maximum water area that can be contained is 49 units. 🧠 Approach Used – Two Pointer Technique: Instead of checking all possible pairs (which would take O(n²) time), I used a more efficient two-pointer approach: Start with two pointers — one at the beginning (left = 0) and one at the end (right = n-1). Calculate the area between the two lines. Keep track of the maximum area found so far. Move the pointer that points to the shorter line, since moving the taller line won’t increase the area. Continue until both pointers meet. This way, we get the answer in O(n) time with O(1) extra space. ✨ Key Takeaways: Smart pointer movement can save huge computation time. Always look for patterns that let you avoid brute-force methods. The two-pointer approach is powerful for array and interval problems. #LeetCode #Day11 #120DaysChallenge #DSA #CodingJourney #ProblemSolving #CPlusPlus #LearnToCode #Tech #Programmers

Tackling a Quick LeetCode Challenge: Final Value of Variable After Performing Operations (2011) It's amazing how a simple iteration can solve a coding problem! This one is a perfect example of keeping it simple: tracking a variable's state change based on a list of defined operations. The Problem: We start with x=0 and are given an array of string operations (like ++X, X++, --X, X--). We need to calculate the final value of x. My C++ Approach: I noticed that regardless of whether the increment/decrement is pre (++X, −−X) or post (X++, X−−), the effect is the same for the final value. Therefore, I only need to check what the operation is, not where the X is. I iterate through the operations array. Inside the loop, I check if the string contains two plus signs ("++"). An even simpler check, as shown in the code, is checking if the operation string equals "++X" or "X++" or even just checking for the presence of a '+'. If it's an increment operation, I use x++. If it's a decrement operation, I use x--. A neat little problem to practice basic array iteration and conditionals! Full code is accepted and running in the screenshot. Have you solved this one? What was your approach? #DataStructures #Algorithms #CodingInterviewPrep #SoftwareDevelopment #DailyCoding #C++

🚀 LeetCode Daily Challenge — 3726. Remove Zeros in Decimal Representation (Easy) Ever thought how a simple number like 1020030 can turn into 123? 🤔 That’s the power of string manipulation in programming — short, clean, and efficient! 💡 🧩 Problem Summary: You’re given a positive integer n. Your task: Remove all zeros from its decimal representation. Example: Input: n = 1020030  Output: 123 ✅ Hint: Convert the number to a string → remove all '0' → convert back to integer. That’s it — one-liner solution for clean data transformation. 🔥 💻 I’ve explained this problem with step-by-step approaches and code on my website 👉 algopush.com Check it out for brute force to optimized explanations! #LeetCode #LeetCodeChallenge #Coding #ProblemSolving #DSA #Programming #100DaysOfCode #CodeSeCareer #TechLearning #Developers #algopush #LeetCodeDaily #StringManipulation #LearnToCode #erdelhiboy #leetcodeeasy

🚀 Day 8 of #120DaysOfCode — String to Integer (atoi) 🧩 Today’s challenge was LeetCode Problem 8: “String to Integer (atoi)”, a classic string parsing problem that teaches how to handle input conversion safely and efficiently — something we often take for granted in programming. 🧠 Problem Summary: We need to implement the myAtoi(string s) function that converts a string into a 32-bit signed integer. The function must: Ignore leading whitespaces Handle optional + or - signs Extract the number until a non-digit is found Clamp (limit) the value within the 32-bit integer range Return the final integer result 📘 Example: Input → " -042" Output → -42 ⚙️ Approach I Used: I solved this problem using a step-by-step parsing approach: Trim spaces: Move through the string to skip leading whitespaces. Check the sign: If the next character is '-', mark the sign as negative, otherwise positive. Convert characters to digits: Loop through the characters while they are digits and build the result (result = result * 10 + digit). Handle overflow: Before updating the result, check if adding another digit would exceed the 32-bit integer range ([-2³¹, 2³¹ - 1]). Return the final value: Apply the sign and return the integer. This approach ensures correctness, prevents overflow, and follows a clean linear-time logic with O(n) complexity and O(1) space. 💡 Key Learning: Even a simple-looking task like string-to-integer conversion involves careful consideration of edge cases, boundaries, and input validation — all critical in real-world software systems. 💬 Next Goal: Continue improving my problem-solving skills with more logic-building and string manipulation challenges! #100DaysOfCode #120DaysOfCode #LeetCode #CProgramming #ProblemSolving #CodingChallenge #DeveloperJourney #LearningEveryday