LeetCode Binary Reduction Steps

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gx6PQVMB 💡 My thought process: The brute force method checks all possible rotations of the binary string by moving the first character to the end repeatedly. For each rotated string, it counts the number of mismatches needed to create an alternating pattern that starts with 1. The opposite alternating pattern, which starts with 0, would require different changes. The method takes the minimum of the two counts for each rotation and returns the lowest value overall. The optimized method does not generate all rotations. Instead, it doubles the string and uses a sliding window of size n to represent every possible rotation. It keeps track of the mismatch count for forming an alternating pattern that starts with 1. As the window slides, it updates this count by subtracting the contribution of the character that leaves the window and adding the incoming character. At each step, it compares both alternating patterns to find the minimum number of flips needed. 👉 My Solution: https://lnkd.in/gP2FWVYP If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/g5nYsHBR 💡 My thought process: To form a decimal number n using the fewest deci-binary numbers, the key factor is the highest digit in n. Each deci-binary number can contribute a maximum of 1 to any digit position. So, if a digit in n is 7, we need at least 7 deci-binary numbers to reach that sum. Thus, the minimum number of deci-binary numbers needed is simply the highest digit in the provided string. The approach is simple: start with a variable result set to 0. Go through each character in the string, convert it to an integer using n[i] - '0', update result with the greater value between result and digit, and finally return result. 👉 My Solution: https://lnkd.in/ggv3b48t If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gDzrB9cW 💡 My thought process: This solution creates the binary string step by step, starting from "0". For each level from 2 to n, it forms a new string by taking the previous string, inserting "1" in the middle, and then adding the reversed and bit-flipped version of the previous string (0 changes to 1 and 1 changes to 0). After each construction, it checks if the string has reached or exceeded the required k position. If it has, it immediately returns the k-th character. This approach prevents any extra construction once the answer is found. 👉 My Solution: https://lnkd.in/gFXgkTpm If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gJUJE-YA 💡 My thought process: The provided code uses a backtracking approach to generate all possible strings of length n that satisfy the happy string condition: no two adjacent characters can be identical. The algorithm builds these strings in lexicographical order by exploring 'a', 'b', and then 'c'. The main logic is in the solve function, which acts as a recursive explorer. It takes the current string and the number of characters left to add. At each step, it tries appending 'a', 'b', and 'c'. Before adding a character, it checks for safety by ensuring that the character is not the same as the last one in the string. This prevents violating the happy constraint. Once a character is added, the function calls itself with n-1 to fill the next position. After finishing that branch of the search, it removes the character to backtrack and try the next alphabetical option. In the getHappyString function, the process starts by launching three separate recursive chains: one with 'a', one with 'b', and one with 'c'. Since the loops and initial calls follow alphabetical order, the allStrings vector is filled with every valid combination in sorted order. Finally, the code checks if the requested index k exists within the total count of generated strings. If k is within bounds, it returns the string at index k-1; otherwise, it returns an empty string. 👉 My Solution: https://lnkd.in/gKDqj84r If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/g6uRPkE6 💡 My thought process: The first approach calculates the bitwise complement by examining each bit of the number. It starts by determining how many bits are needed to represent n using log2(n) + 1. For each bit position i, a mask (1 << i) is used to isolate that bit from n. If the extracted bit is 0, the result for that position is set by adding the same mask, effectively flipping the bit. This creates a new number where all bits within the effective bit length of n are inverted, resulting in the complement. The second approach uses the XOR operation to flip all relevant bits at once. It first calculates the number of bits in n and creates a mask filled with 1s in all those positions using (1 << digits) - 1. This mask represents the highest value that can be formed with the same number of bits. Applying n ^ mask flips every bit within that range because XOR with 1 changes the bit and XOR with 0 leaves it the same, giving the bitwise complement of n. 👉 My Solution: https://lnkd.in/gGCdUW3H If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 Day 27/60 — LeetCode Discipline Problem Solved: Length of Last Word Difficulty: Easy Today’s problem looked simple, yet it emphasized a subtle but important skill — handling edge cases cleanly. The goal was to find the length of the last word in a string, ignoring any trailing spaces. Instead of splitting the string or using extra space, the solution efficiently traverses from the end, skipping unnecessary characters and counting only what truly matters. It’s a reminder that strong coding is not always about complexity — sometimes, it’s about precision and clarity in small details. 💡 Focus Areas: • Practiced string traversal from end • Improved handling of trailing spaces • Reinforced clean and efficient logic • Avoided unnecessary extra space usage • Strengthened edge-case thinking ⚡ Performance Highlight: Achieved 0 ms runtime (100% performance). Small problems, when approached with discipline, sharpen the instincts that solve big ones. #LeetCode #60DaysOfCode #100DaysOfCode #DSA #Strings #Algorithms #ProblemSolving #CodingJourney #SoftwareEngineering #Java #Developers #Consistency #TechJourney #LearnToCode

I’ve been giving the 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐖𝐞𝐞𝐤𝐥𝐲 𝐂𝐨𝐧𝐭𝐞𝐬𝐭 regularly for the last 2–3 weeks, but couldn’t share my thoughts earlier due to time constraints. Today I thought I should finally post my approaches and also learn from how other developers solved the same problems. This week I secured 𝐑𝐚𝐧𝐤 763 Here’s a quick look at how I approached the problems: 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 1 – 𝐌𝐢𝐧𝐢𝐦𝐮𝐦 𝐂𝐚𝐩𝐚𝐜𝐢𝐭𝐲 𝐁𝐨𝐱 I iterated through the array and kept track of the smallest capacity that could fit the item. Whenever I found a better candidate, I updated the minimum capacity and stored its index. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 2 – 𝐒𝐦𝐚𝐥𝐥𝐞𝐬𝐭 𝐁𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐈𝐧𝐝𝐞𝐱 I maintained the running left sum and precomputed the right-side products using a reverse pass. Then I checked each index to see where the left sum equals the right product and returned the smallest one. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 3 – 𝐌𝐢𝐧𝐢𝐦𝐮𝐦 𝐎𝐩𝐞𝐫𝐚𝐭𝐢𝐨𝐧𝐬 𝐭𝐨 𝐒𝐨𝐫𝐭 𝐒𝐭𝐫𝐢𝐧𝐠 I compared the string with its sorted version and located the mismatched section. Based on how the mismatch appears, I determined whether the string could be fixed in 1, 2, or 3 operations. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 4 – 𝐌𝐢𝐧𝐢𝐦𝐮𝐦 𝐏𝐚𝐫𝐭𝐢𝐭𝐢𝐨𝐧 𝐂𝐨𝐬𝐭 Using prefix sums, I quickly counted sensitive elements in any segment. Then I recursively checked whether splitting a segment into halves reduces the cost compared to keeping it as one. Contests like these always help sharpen problem-solving skills. Also curious to see how others approached these problems — always something new to learn. #LeetCode #DSA #CompetitiveProgramming #ProblemSolving #Coding

Ever stared at your own code a month later and wondered, "What was I thinking?!" 😵💫 We've all been there! One of the quickest wins for cleaner, more maintainable code is embracing **meaningful and descriptive variable names**. ✨ Instead of using vague names like `x`, `data`, or `temp`, think about what the variable *represents* in your specific context. This makes your code self-documenting, easier to debug, and a joy for your future self (and your teammates!) to read. Take a look at the difference: ```python # Before (unclear purpose) def calc(x, y): return x * y # After (crystal clear!) def calculate_rectangle_area(length, width): return length * width ``` See how much easier it is to understand the "After" example's intent without any extra comments? That's the power of good naming! What's YOUR favorite clean code practice that changed your development life? Share your wisdom below! 👇 #CleanCode #ProgrammingTips #SoftwareDevelopment #DevTips #CodingBestPractices #CodeQuality #DeveloperLife #TechEd

🚀 Day 6 of 100 Days LeetCode Challenge Problem: Check if Binary String Has at Most One Segment of Ones Today’s problem is all about pattern observation in strings—simple, but easy to overthink. 💡 Key Insight: The string should contain only one continuous block of '1's. 👉 That means: Once a 0 appears after a 1, There should be no more '1's later 🔍 Simplest Trick: Just check if the pattern "01" appears more than once OR even better → check if "10" appears followed by another "1" 💡 Cleaner approach: Traverse the string Count transitions from 1 → 0 If you ever see 1 again after that → ❌ Invalid 🔥 What I Learned Today: Many problems are just pattern validation Clean logic beats complex conditions Always try to reduce the problem to a simple rule 📈 Challenge Progress: Day 6/100 ✅ Consistency building strong! LeetCode, Strings, Pattern Recognition, Greedy, DSA Practice, Coding Challenge, Problem Solving, Algorithm Thinking, Programming #100DaysOfCode #LeetCode #DSA #CodingChallenge #Strings #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep

Day 6 of My LeetCode Consistency Journey Today I solved Longest Consecutive Sequence (LeetCode 128). At first, my instinct was to sort the array and then check for consecutive elements. But sorting leads to O(n log n) time complexity, while the problem expects an O(n) solution. After thinking through it, I realized the key idea is to use hashing. 💡 Main insight: A number should only start a sequence if (num - 1) does not exist. This ensures we only start counting from the beginning of a sequence and avoid unnecessary checks. From there, the sequence can be expanded using num + 1 until the chain breaks. Key learnings from today: • Using a hash set for O(1) lookup can drastically improve performance. • Identifying the start of a sequence avoids redundant computation. • Small implementation details matter — like not modifying the loop variable during iteration. This problem was a great reminder that sometimes the real challenge is recognizing the pattern, not writing the code. Day by day, the goal remains the same: consistency and better problem-solving thinking. #LeetCode #DSA #ProblemSolving #CodingJourney #Consistency