LeetCode Challenge: String to Integer Problem Solution

🚀 Day 29/60 — LeetCode Discipline Problem Solved: Longest Common Prefix Difficulty: Easy Today’s problem focused on identifying the longest common starting pattern across multiple strings — a classic exercise in string comparison and iteration. The approach involved checking characters position by position across all strings until a mismatch is found. This reinforces how simple iterative logic can elegantly solve problems involving multiple inputs. It’s a reminder that sometimes, clarity in approach matters more than complexity in code. 💡 Focus Areas: • Strengthened string traversal techniques • Practiced comparing multiple inputs efficiently • Improved understanding of edge cases (empty strings, mismatch handling) • Reinforced early stopping conditions for optimization • Focused on clean and readable logic ⚡ Performance Highlight: Achieved efficient runtime with minimal overhead. Finding common ground across multiple inputs is not just a coding skill — it’s a pattern recognition mindset. #LeetCode #60DaysOfCode #100DaysOfCode #DSA #Strings #Algorithms #ProblemSolving #CodingJourney #SoftwareEngineering #Java #Developers #Consistency #TechGrowth #LearnToCode

🚀 Day 30/60 — LeetCode Discipline Problem Solved: 3Sum Closest Difficulty: Medium Today’s problem pushed beyond exact answers and focused on finding the closest possible sum to a given target — a subtle yet powerful variation of the classic 3Sum problem. By sorting the array and using a two-pointer approach, the solution efficiently explores combinations while continuously updating the closest result. This problem highlights how small modifications in logic can significantly change the nature of a problem — from exact matching to optimal approximation. 💡 Focus Areas: • Strengthened two-pointer technique • Practiced working with sorted arrays • Improved handling of optimization conditions • Learned to track and update closest values dynamically • Reinforced problem-solving under constraints ⚡ Performance Highlight: Achieved efficient runtime with optimized traversal. Not every problem asks for perfection — sometimes, the goal is to get as close as possible. #LeetCode #60DaysOfCode #100DaysOfCode #DSA #TwoPointers #Algorithms #ProblemSolving #CodingJourney #SoftwareEngineering #Java #Developers #Consistency #TechGrowth #LearnToCode

🧠 LeetCode POTD — The Biggest Hint Was Hidden in One Line 1855. Maximum Distance Between a Pair of Values Today's problem looked tricky at first considering the time complexity. We need to find indices (i, j) such that: 👉 i <= j 👉 nums1[i] <= nums2[j] And maximize: 👉 j - i ━━━━━━━━━━━━━━━━━━━ My first instinct? Try brute force combinations. Check many pairs. Maybe binary search. But I was missing the most important line in the question: Both arrays are non-increasing (sorted). That one detail changes everything. ━━━━━━━━━━━━━━━━━━━ 💡 Once I noticed that, the solution became a clean two-pointer approach. Start with: 👉 i = 0 in nums1 👉 j = 0 in nums2 Then: If nums1[i] <= nums2[j] 👉 This pair is valid 👉 Update answer with j - i 👉 Move j forward to try for a bigger distance Else: 👉 Current i is too large 👉 Move i forward Because arrays are sorted, once a pair fails, moving i is the only useful move. ━━━━━━━━━━━━━━━━━━━ 📌 What I liked about this problem: The challenge wasn't the coding. It was reading carefully enough to spot the property that unlocks the solution. ✅ Sometimes the answer is already in the problem statement. ✅ You just have to notice it. Curious — did anyone else miss the sorted hint at first? 👀 #LeetCode #TwoPointers #ProblemSolving #SoftwareEngineering #DSA #Java #c++ #SDE

🚀 Daily Coding Progress Solved a problem on LeetCode today: 🔹 Problem: Maximum Value of a String in an Array (Easy) 💻 Example 1: Input: ["alic3","bob","3","4","00000"] "alic3" → contains letters → value = 5 (length) "bob" → only letters → value = 3 "3" → only digits → value = 3 "4" → only digits → value = 4 "00000" → only digits → value = 0 ✅ Output: 5 🧠 Approach: Loop through each string in the array. Check if the string is numeric using Character.isDigit(). If numeric → value = integer value of the string. Else → value = length of the string. Track the maximum value while iterating. ⚡ Why this approach? • Time Complexity: O(n * m) – n strings of max length m ≤ 9. • Space Complexity: O(1) – only tracking max value. • Simple, efficient, and handles leading zeros perfectly. 💡 Key Learning: Checking if a string is numeric with Character.isDigit() avoids regex overhead. Handling mixed alphanumeric strings elegantly by using string length. Iterating through an array while maintaining a running maximum is a fundamental technique. Consistency pays off 🔁 Step by step, improving problem-solving skills and coding fluency. #LeetCode #DSA #CodingJourney #100DaysOfCode #SoftwareEngineering #Java #ProblemSolving

🚀 Solved “Search Insert Position” using Binary Search on LeetCode! Today I worked on an interesting problem where the goal is to find the index of a target element in a sorted array. If the target is not present, we return the index where it should be inserted to maintain sorted order. 🔍 Approach: Instead of using a linear search (O(n)), I implemented an efficient Binary Search (O(log n)) approach. 💡 Key Learning: One important detail is calculating the middle index safely: mid = low + (high - low) / 2 This avoids potential integer overflow compared to (low + high) / 2 and ensures correct behavior even for large inputs. ⚙️ Logic: If target == arr[mid] → return mid If target > arr[mid] → search right half Else → search left half If not found → return ‘low’ as the correct insert position 📈 Result: Achieved 100% performance on LeetCode 🚀 This problem reinforced my understanding of: ✔ Binary Search fundamentals ✔ Edge case handling ✔ Writing optimized and safe code Looking forward to solving more problems and improving problem-solving skills! #LeetCode #BinarySearch #Java #DataStructures #Coding #ProblemSolving

Most developers believe that nested loops are unavoidable. However, they are actually a design mistake. When your code includes: - A loop inside another loop - Repeated scanning - O(n²) complexity You are not solving the problem efficiently; you are simply adhering to a habit. In my research, I explored: - Why nested loops occur - The underlying root causes - How indexing can reduce complexity from O(n²) to O(n) The most significant realization? Performance is determined before writing code, based on how data is structured. Nested loops are not merely a coding issue; they are a problem of thinking. I have shared the full research as a document and would appreciate your thoughts. How frequently do you encounter nested loops in production code #SoftwareEngineering #Performance #CleanCode #Java #Backend #SystemDesign #Developers #TechInsights #JavaDevelopment #NestedLoops #CodeOptimization #SoftwareEngineering #ProgrammingTips #JavaTips #PerformanceTuning #EfficientCoding #TechInsights #DeveloperCommunity #CodingBestPractices #SoftwareDevelopment #JavaProgramming #TechOptimization #DevLife

The Ash framework in Elixir is an excellent example of declarative programming in action. I’ve been exploring it by modeling Yggdrasil, the World Tree from Norse mythology 🌳. Part 1 of the series is now up, where I dive into creating a domain and the first resource in Ash, along with basic CRUD operations. Check it out here: https://lnkd.in/e3MdRic5 #AshFramework #ElixirLang #DeclarativeProgramming

Day 80 - LeetCode Journey 🚀 Solved LeetCode 876: Middle of the Linked List (Easy) — a classic problem that introduces one of the most powerful patterns in linked lists. At first, finding the middle seems straightforward. But doing it efficiently in a single pass is where the real insight comes in. 💡 Core Idea (Slow & Fast Pointer Technique): Initialize two pointers: slow and fast Move slow by 1 step and fast by 2 steps When fast reaches the end, slow will be at the middle If there are two middle nodes, this approach naturally returns the second one. 🤯 Why it works? Because the fast pointer covers double the distance of the slow pointer, so when it finishes traversing the list, the slow pointer is exactly halfway. ⚡ Key Learning Points: • Efficient single-pass traversal • Mastering the slow-fast pointer pattern • Handling even and odd length lists • Achieving O(n) time and O(1) space • Building intuition for pointer-based problems This is a foundational pattern used in many advanced problems. Also, this pattern connects with: Linked List Cycle detection Palindrome Linked List Reorder List Finding intersection of two linked lists ✅ Stronger grasp of two-pointer technique ✅ Better problem-solving efficiency ✅ Cleaner and optimized linked list logic Simple problem, powerful pattern — this is how concepts build up 🚀 #LeetCode #DSA #Java #LinkedList #TwoPointers #Algorithms #ProblemSolving #CodingJourney #Consistency #100DaysOfCode #InterviewPreparation #DeveloperGrowth #KeepCoding

🚀 Day 2 of My Coding Challenge Improving my problem-solving skills step by step! Today I solved an important array problem from LeetCode. 🔹 Platforms: LeetCode & GeeksforGeeks 🔹 Problem: LeetCode #189 – Rotate Array 🔹 Problem Statement: Rotate the array to the right by k steps. 🔹 Approach: I used the Reversal Algorithm for an optimized solution: 1️⃣ Reverse the entire array 2️⃣ Reverse first k elements 3️⃣ Reverse remaining elements 🔹 Example: Input: [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] 🔹 What I learned: ✔ Optimized approach from brute force to O(n) ✔ Importance of array manipulation techniques ✔ Handling edge cases like k > n 💻 Code: import java.util.Arrays; public class RotateArrayLeetCode189 { ``` public static void reverse(int[] arr, int start, int end) { while (start < end) { int temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; start++; end--; } } public static void rotate(int[] arr, int k) { int n = arr.length; k = k % n; if (k < 0) { k = k + n; } // Step 1: Reverse entire array reverse(arr, 0, n - 1); // Step 2: Reverse first k elements reverse(arr, 0, k - 1); // Step 3: Reverse remaining elements reverse(arr, k, n - 1); } public static void main(String[] args) { int[] arr = {1, 2, 3, 4, 5, 6, 7}; int k = 3; rotate(arr, k); System.out.println(Arrays.toString(arr)); } ``` } 🔗 GitHub: https://lnkd.in/g-wNSrPq #Java #DSA #LeetCode #CodingChallenge #50DaysOfCode #LearningJourney #PlacementPreparation

“What I learned after solving 300+ LeetCode problems 👇” At the beginning, I made a mistake. I kept solving random problems every day thinking: “More problems = more skill” But I was wrong. I realized that solving problems randomly is mostly a waste of time if you don’t understand the pattern behind them. Everything changed when I started focusing on patterns instead of problems. Here are some important patterns I learned: • Sliding Window • Two Pointers • Prefix Sum • Binary Search • Recursion & Backtracking • Linked List Patterns • Stack & Monotonic Stack • Hashing / Frequency Count Once you understand these patterns: You don’t need to solve 1000 problems. You start recognizing solutions instantly. Now when I see a problem, I don’t panic. I ask: 👉 “Which pattern does this belong to?” That one question changed everything for me. Still learning, but now with direction 🚀 #DSA #LeetCode #Java #BackendDevelopment #CodingJourney