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• Solving LeetCode Problem | Day 13 • Problem: Fizz Buzz | LeetCode 412 • Approach: Simple implementation problem focused on conditional logic. For each number from 1 to n: Check divisibility by both 3 and 5 "FizzBuzz" If only divisible by 3 => "Fizz" If only divisible by 5 => "Buzz" Otherwise convert number to string. Key focus was writing clean and readable logic rather than overcomplicating it. • Time Complexity: O(n) • Space Complexity: O(n) • What I learned: Even easy problems matter — they test clarity of thinking. Writing clean conditions without redundancy is important. Don’t ignore basics while chasing harder problems. #leetcode #dsa #programming #java #coding #problemSolving #100daysofcode

💡#LeetCode Daily Challenge – Smart Optimization! Today I worked on a problem where we need to find the minimum distance between three equal elements in an array. At first, it looks like a brute-force problem, but the real trick is simplifying the formula.After observing carefully, the distance formula actually reduces to just twice the difference between the first and last indices. So the middle element doesn’t even matter! That insight helped me avoid unnecessary computations.I grouped indices of each number and checked only consecutive triples to get the minimum distance efficiently. This problem reminded me how powerful pattern recognition can be in coding. #LeetCode #ProblemSolving #DataStructures #Algorithms #CodingInterview #Java #Programming #CodingJourney #TechLearning #Developers #SoftwareEngineering

🚀 Day 561 of #750DaysOfCode 🚀 📌 Problem: Minimum Distance to the Target Element Today’s problem was simple yet a great reminder of how powerful basic iteration can be when applied correctly. 🔍 The task was to find the minimum distance between a given start index and any index i such that nums[i] == target. 💡 Key Insight: Instead of overthinking, just iterate through the array and track the minimum value of |i - start| whenever the target is found. Clean, efficient, and effective. 🧠 What I Learned: Sometimes brute force with clarity is the best solution Always look for opportunities to minimize operations with simple logic Writing clean and readable code matters as much as solving the problem ⚡ Approach: Traverse the array Check for target Update minimum distance ⏱️ Complexity: Time: O(n) Space: O(1) 💻 Consistency is key. Small steps every day build strong problem-solving skills over time. #leetcode #dsa #programming #java #coding #developers #softwareengineering #100daysofcode #codingjourney #tech #learning #growth

🚀 Day 16 of #100DaysOfCode Mission 100 days problem solve আজকের Problem: Rotate Function (rotate element) find out highest sum 🧠 Topic: array কে Clockwise rotate, প্রতিটা rotation এর জন্য একটা value calculate করতে হবে🚀 #coding #programming #developers #webdevelopment #javascript #typescript #dsa #datastructures #algorithms #codinginterview #leetcode #problemSolving

🚀 Day 13 / 50 – Wild Coding Kickoff 🔥 Today’s problem looked like it wanted a complicated solution… but it ended up teaching me something much simpler. I read the question and instantly thought: “Okay, substring search… this might get tricky.” 👀 Loops, comparisons, edge cases… my mind started going there. But then I stopped and asked: 👉 “Is there a simpler way?” That’s when I used this: class Solution { public int strStr(String haystack, String needle) { int index = haystack.indexOf(needle); return index; } } 💡 What does indexOf() actually do? It searches for the first occurrence of needle inside haystack Returns: ✅ The starting index if found ❌ -1 if the substring is not present 🔥 Example haystack = "sadbutsad" needle = "sad" 👉 Output: 0 (because "sad" starts at index 0) 🧠 What I learned today Sometimes we try to prove we know everything… But real skill is knowing when to keep things simple. Using built-in methods effectively is also a skill. #Day13 #50DaysOfCode #WildCodingKickoff #LeetCode #Java #CodingJourney #Consistency #KeepItSimple #ProblemSolving

• Solving LeetCode Problem | Day 13 • Problem: First Unique Character in a String | LeetCode 387 • Approach: Instead of checking every character multiple times (O(n²)), I used a frequency array. Step 1: Count occurrences of each character Step 2: Traverse again to find the first character with frequency = 1 This keeps the logic simple and avoids unnecessary comparisons. • Time Complexity: O(n) • Space Complexity: O(1) • What I Learned: Trying to force everything into one loop can make the solution messy. Breaking the problem into two clean passes often leads to better clarity and efficiency. #leetcode #dsa #java #programming #coding #algorithms #problemSolving #100daysofcode

Day 63 of Practicing DSA (LeetCode && Competitive Programming) Today’s progress: LeetCode • Longest Common Prefix – Easy Focused on: • Sorting-based approach for string comparison • Comparing first and last strings to find common prefix • Understanding optimization in string problems • Writing clean and concise logic This problem improved my understanding of efficient string handling techniques. Competitive Programming • Continued practicing problem-solving patterns Improved my thinking on: • Breaking down string problems logically • Writing optimized and readable code Simple problem, but strong fundamentals matter 💯 Consistency is the real key 🔥 #DSA #Java #LeetCode #Codeforces #Strings #ProblemSolving #Consistency #LearningInPublic 🚀

Day 14 of #50DaysOfLeetCode Challenge Today’s coding challenge was Rotate List (LeetCode 61). The goal is to rotate a linked list to the right by k places. At first, it sounds like you need to move nodes one by one, but that’s inefficient. The key to a high-performance solution is to treat the list like a loop! The Efficiency Strategy: 1. Calculate Length: First, find the total length of the list (n). 2. Handle Large K: If k is larger than n, we only need to rotate k % n times. This avoids unnecessary cycles. 3. Make it a Circle: Connect the last node's "next" pointer to the original head to create a temporary circular list. 4. Find the New Break: Move (n - (k % n)) steps from the head to find the new tail of the rotated list. 5.Break the Loop: Save the node after the new tail as the new head, and then set the new tail's "next" to null. #LeetCode #BitManipulation #Java #Coding #Efficiency #TechLearning

🚀 Code 2 – #50LeetCodeChallenge Problem: "Longest Common Prefix" Given an array of strings, find the longest common prefix shared among all strings. If no common prefix exists, return an empty string "". 💡 Approach: Start with the first string as the prefix and compare it with the remaining strings. Gradually reduce the prefix until it matches the beginning of each string. 📚 Key Takeaway: String comparison and iterative reduction help efficiently solve prefix-based problems with optimal performance. #LeetCode #Java #Coding #ProblemSolving #Strings #Programming

If you can't pronounce your variable names, you can't discuss your code — rename them to sound like natural language. Let's walk through it in the slides below. #cleancode #code #programming #developer