Converting integers to Roman numerals in Java

🔢 Today I worked on a classic matrix problem in Java — Diagonal Sum. The goal was simple: Calculate the sum of both the primary and secondary diagonals of a square matrix. Initially, I used a nested loop approach with O(n²) complexity. But then I optimized it to a clean O(n) solution by directly accessing diagonal indexes. While optimizing, I made an interesting mistake: ❌ I wrote if (i != matrix[i][matrix.length - 1 - i]) Here I accidentally compared an index with an element value. ✔ Correct approach: if (i != matrix.length - 1 - i) This ensures the center element in odd-sized matrices isn’t counted twice. 🧠 Key Learning: Indexes and values may look similar, but mixing them can break logic silently. Optimization is not just about speed — it’s about accuracy. #Java #leetcode #Coding #DSA #ProblemSolving #LearningInPublic #SoftwareEngineering

🧠 Daily LeetCode Grind — Java Edition Today’s challenge: ✅ Container With Most Water (#11 - Medium) 📌 Goal: Given an integer array height[], find two lines that, together with the x-axis, form a container that holds the most water. 📌 Approach: 🔹 Use the Two-Pointer Technique to optimize the search for the maximum area. 🔹 Start with two pointers at both ends of the array and calculate the area at each step. 🔹 Move the pointer pointing to the smaller height inward to maximize potential area. 🔹 Continue until both pointers meet. 🧩 Test Case: Input: height = [1,8,6,2,5,4,8,3,7] Output: 49 💡 Key Takeaways: 🔹 Learned to balance area optimization using width × height. 🔹 Reinforced the power of the two-pointer approach for reducing time complexity. 🔹 Strengthened problem-solving speed through mathematical reasoning. 💻 Language: Java 🧠 Complexity: O(n) — single pass using two pointers. #LeetCode #Java #CodingPractice #ProblemSolving #TwoPointer #Algorithm #DeveloperLife #CybernautEdTech #AcceptedSolution

✨ Ever wondered what each part of `public static void main(String[] args)` means in Java? ✨ Check out this colorful visual breakdown! 🎨👇 🟡 **public** – Access specifier: Makes the method accessible from anywhere. 🟠 **static** – Keyword: Allows JVM to call main() without creating an object. 🟢 **void** – Return type: Specifies that main() doesn’t return any value. 🔵 **main** – Identifier: Predefined name recognized by the JVM as the entry point. 🟣 **String[] args** – Parameter: Accepts command-line arguments for the Java program. 📝 The code inside `{}` is the function body where your program execution starts! The `main()` method is the heart of every Java application. Understanding it is your first step to writing powerful Java programs! 🚀 #Java #Programming #Learning #CodeNewbie #MainMethod #JavaBasics Anand Kumar Buddarapu sir Saketh Kallepu sir

Day 91 of #100DaysOfCode Solved Base 7 in Java 🔢 Approach The challenge was to convert a given integer (num) to its base 7 string representation. Conversion Method The core of the solution lies in the standard algorithm for base conversion: repeated division and remainder collection. Handle Zero and Negatives: If the input num is 0, the result is immediately "0". I determined if the number is negative and stored this in a boolean flag, then proceeded with the absolute value of the number (num = Math.abs(num)) for the conversion logic. Conversion Loop: I used a while loop that continues as long as num > 0. In each iteration: The remainder when num is divided by 7 (num % 7) gives the next digit in base 7. This digit is appended to a StringBuilder. num is then updated by integer division by 7 (num /= 7). Final Result: Since the remainders are collected in reverse order, I called sb.reverse(). If the original number was negative, I prepended a hyphen (-) to the reversed string. Finally, I returned the result as a string. This simple and efficient implementation had a very fast runtime, beating 77.39% of submissions. #Java #100DaysOfCode #LeetCode #CodingChallenge #Algorithms #BaseConversion #ProblemSolving

🚀Day 94/100 #100DaysOfLeetCode 🧩Problem: Reverse Words in a String III✅ 💻Language: Java 🔍Approach: 🔹Split the given string by spaces to separate each word. 🔹For every word, reverse it using a StringBuilder. 🔹Append each reversed word back to the result string, adding spaces between them. 🔹Finally, return the complete reversed word string. 🧠Key Takeaways: 🔹Practiced efficient string manipulation using StringBuilder. 🔹Learned how splitting and reversing operations can be optimized for clarity. 🔹Strengthened understanding of text processing in Java. ⚡Performance: ⏱️Runtime: 4 ms (Beats 86.86%) 💾Memory: 45.49 MB (Beats 47.42%) #100DaysOfLeetCode #Java #CodingJourney #LeetCode #CodingChallenge #ProblemSolving

🚀Day 95/100 #100DaysOfLeetCode 🔍Problem: Length of Last Word✅ 💻Language: Java 🧠Approach: 1️⃣ First, trim the string to remove trailing and leading spaces. 2️⃣ Initialize a counter to 0. 3️⃣ Traverse the string from the end. 4️⃣ Count characters until the first space is encountered. 5️⃣ Return the count as the length of the last word. 💡Key Takeaways: 🔹Trimming the input ensures no trailing spaces interfere. 🔹Backward traversal avoids splitting the entire string. 🔹Simple yet efficient one-pass solution. ⚡Performance: ⏱️Runtime: 0 ms (Beats 100.00%) 💾Memory: 41.58 MB (Beats 91.52%) #100DaysOfLeetCode #Java #CodingJourney #ProblemSolving #LeetCode #CodingChallenge

🧠 Java with DSA – Day 38: Linked Logic Today’s focus: building the backbone—singly linked lists in Java. - Created custom Node class with data and next—because every journey starts with a pointer. - Linked nodes manually to form a chain—visualizing memory as a trail of connections. - Printed intermediate nodes to trace flow—because debugging is storytelling. - Realized that linked lists aren’t just data—they’re dynamic, flexible, and foundational. DSA is about more than structures—it’s about understanding how data moves, grows, and adapts. 🚀 All code + notes on GitHub – https://lnkd.in/gGWnjGxk #LinkedListBasics #NodeByNode #CodeWithClarity #ShouryaCodes #100DaysOfCode #DSAFoundation #LinkedInDevJourney #OneQuesADay

🧠 Daily LeetCode Grind — Java Edition Today’s challenge: ✅ Palindrome Number (#9 - Easy) 📌 Goal: Check whether an integer reads the same backward as forward without converting it to a string. 📌 Approach: 🔹 Handle negatives and numbers ending in 0 (except 0 itself). 🔹 Reverse only half of the digits using modulo and division. 🔹 Compare the original and reversed halves for equality. 🧩 Test Cases: Input: 121 → Output: true Input: -121 → Output: false Input: 10 → Output: false 💡 Key Takeaways: 🔹 Strengthened arithmetic-based problem-solving (no string ops). 🔹 Learned efficient O(1) space reversal logic. 🔹 Improved understanding of numeric pattern recognition. 💻 Language: Java 🧠 Complexity: O(log₁₀ n) — reverse digits only once. #LeetCode #Java #CodingPractice #ProblemSolving #DSA #PalindromeNumber #DeveloperLife #AcceptedSolution #CybernautEdTech

🚀 Day 98 of #160DaysOfGFG 📊 Problem: Find Median in a Stream 💻 Language: Java 🔗 Practice here: GeeksforGeeks - Find Median in a Stream 🧩 Problem Description: Given an integer stream, we need to find the median of the numbers at every insertion step. In other words — after inserting each element, what’s the current median of all elements seen so far? 💡 Key Idea: Use two heaps (priority queues) to efficiently maintain the lower and upper halves of the data: A max-heap (left side) to store the smaller half of elements. A min-heap (right side) to store the larger half. By keeping the heaps balanced (their sizes differ by at most 1), the median can be found in O(1) time after each insertion, and the overall stream processing runs in O(N log N).

(Largest Perimeter Triangle — Java 🔺) Hey everyone 👋 Today I explored how to find the largest perimeter triangle from a given array of side lengths using Java. While solving this, I understood how a simple mathematical condition can help in optimizing the entire logic 🔥 Here’s the breakdown 👇 🧠 Concept: To form a valid triangle from 3 sides, the sum of any two sides must be greater than the third one. → a + b > c → a + c > b → b + c > a 💡 Approach 1: Sort the array in descending order and check triplets from the start. If any 3 consecutive sides satisfy the triangle condition, their sum gives the largest perimeter. ⚡ Approach 2 (Optimized): Instead of reversing the array, just sort it in ascending order and iterate from the end — gives the same result but with cleaner logic and less work. 🧩 Time Complexity: O(N log N) 💾 Space Complexity: O(1) This problem helped me understand how sorting order and traversal direction can make an algorithm more elegant and efficient 🚀 📁 Code is available on my GitHub repo: https://lnkd.in/ekjD9s22 #Java #DSA #ProblemSolving #Arrays #CodingJourney #Developers #LearningByDoing #FullStackDeveloper #MCA #GitHub