Day 46: Binary Addition from scratch! 🔢 Problem 67: Add Binary Java's BigInteger is a thing, but manual bit manipulation is just more satisfying. My approach: 1. Padded both strings to the same length using character arrays. 2. Simulated schoolbook addition from right to left. 3. Used % 2 for the result bit and / 2 for the carry. It’s a clean O(N) solution that avoids overflow and keeps the logic transparent. No shortcuts, just pure logic. 🚀 #LeetCode #Java #Binary #Algorithms #CodingChallenge #ProblemSolving
Binary Addition from Scratch in Java
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Day 6/100 – LeetCode Challenge 🚀 Problem: #189 Rotate Array Difficulty: Medium Language: Java Approach: Array Reversal Technique Time Complexity: O(n) Space Complexity: O(1) 🔍 Key Insight: Instead of shifting elements one by one, the array can be rotated efficiently using a three-step reversal strategy. Steps: 1️⃣ Reverse the entire array 2️⃣ Reverse the first k elements 3️⃣ Reverse the remaining elements This achieves the required rotation in-place with constant extra space. 🧠 Solution Brief: Calculated k % n to handle cases where k is greater than array length. Reversed the entire array first. Then reversed the first k elements and the remaining n-k elements. This sequence correctly rotates the array to the right. 📌 What I Learned: Understanding patterns like array reversal can simplify problems that initially seem complex. Optimizing from brute force shifting to an in-place O(n) solution improves efficiency. #LeetCode #Day6 #100DaysOfCode #Java #DSA #Arrays #RotateArray #ProblemSolving #CodingJourney
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Strings look simple until edge cases show up 👀 Day 8 / #100DaysOfCode 🚀 Solved: Valid Palindrome (LeetCode 125) 🔹 Approach: Used the two-pointer technique starting from both ends. Ignored non-alphanumeric characters and compared characters after converting the string to lowercase. Time Complexity: O(n) Space Complexity: O(1) ✔ Practiced string traversal with pointers ✔ Handled real-world edge cases (spaces, symbols, cases) ✔ Avoided extra space by working in-place 💡 Takeaway: Clean pointer logic often beats preprocessing with extra data structures. Building strong fundamentals in Java, one problem at a time. #LearnInPublic #100DaysOfCode #DSA #Java #Strings #ProblemSolving #SoftwareEngineer #CodingJourney #Consistency
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Day 31/100 - LEETCODE Challenge ✅ Problem : Find First and Last Position of Element in Sorted Array Solved the Search for First and Last Position of Element in Sorted Array problem using an optimized Binary Search approach in Java. Instead of using a linear scan, I implemented two separate binary searches to efficiently find the first and last occurrence of the target element in O(log n) time complexity. This solution improves performance for large datasets and achieved 0 ms runtime (100% beats) on LeetCode. Problems like this help strengthen understanding of binary search variations and edge case handling in sorted arrays. #100DaysOfCode #java #Coding #SoftwareDeveloper
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🚀 Day 76 of #100DaysOfCode Today I solved a string manipulation problem: Clear Digits 🧠 Problem: Given a string, whenever a digit appears, remove the previous character. Digits act like a backspace operation. 💡 Key Insight: Instead of modifying the string directly (since Strings are immutable in Java), I used a StringBuilder, which works like a stack: If character → append (push) If digit → delete last character (pop) 📌 What I Learned: StringBuilder is powerful for string modifications Always handle edge cases (like digit at the start) Many string problems are actually stack problems in disguise ⏱ Complexity: Time: O(n) Space: O(n) #Java #DSA #LeetCode #CodingJourney #100DaysOfCode #ProblemSolving
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Day 24 of #100DaysOfLeetCode 💻✅ Solved #35. Search Insert Position on LeetCode using Java. Approach: • Applied Binary Search to achieve O(log n) time complexity • Initialized two pointers: left and right • Calculated mid using left + (right - left) / 2 to avoid overflow • Compared mid element with target to adjust search space • If target not found, returned left as the correct insert position Performance: ✓ Runtime: 0 ms (Beats 100% submissions) ✓ Memory: 43.2 MB Key Learning: ✓ Strengthened understanding of Binary Search pattern ✓ Learned how to determine insert position when element is absent ✓ Improved confidence in handling sorted array problems efficiently Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #ProblemSolving #CodingJourney #100DaysOfCode
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\🚀 Day 15 of My LeetCode Journey 🧩 Problem Solved: 1980. Find Unique Binary String 💻 Language: Java Today I worked on a problem where we are given n unique binary strings, each of length n, and we need to find another binary string of length n that does not exist in the list. 🔍 Key Insight: Instead of checking all possible binary combinations, we can build a new string by flipping the i-th bit of the i-th string. This guarantees the new string will differ from every string in the array at least at one position. ⚡️ Concept Used: Diagonalization technique 📈 Time Complexity: O(n) 📦 Space Complexity: O(n) 💡 This problem reminded me that sometimes the smartest solutions come from simple observations rather than brute force approaches. Consistency is key — solving problems daily to strengthen my Java and problem-solving skills. #LeetCode #Java #CodingJourney #ProblemSolving #100DaysOfCode #DataStructures #LearningInPublic
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🚀 #100DaysOfCode | Day 35 📌 LeetCode : Minimum Depth of Binary Tree Today I solved the Minimum Depth of Binary Tree problem using Java. The goal was to find the shortest path from the root node to the nearest leaf node. I applied a recursive approach while carefully handling edge cases where one subtree is null. Instead of directly taking the minimum of both sides, I ensured the solution correctly skips null paths to avoid incorrect depth calculations. 📌 Key takeaways: 🔹 Understood the difference between minimum and maximum depth logic 🔹 Learned the importance of handling null child nodes 🔹 Strengthened recursion and tree traversal concepts 🔹 Improved problem-solving accuracy in edge cases This problem helped me think more clearly about tree structures and reinforced the importance of precise base conditions in recursion. #Java #LeetCode #DSA #ProblemSolving #CodingJourney #100DaysOfCode
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🚀 Day 18/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 1528. Shuffle String Used a direct indexing approach by placing each character at its correct position using the given indices array. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) A simple yet important problem to strengthen understanding of arrays and index mapping. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency
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🚀 Day 96 of #100DaysOfCode Solved LeetCode #1461 – Check If a String Contains All Binary Codes of Size K ✅ A solid mix of sliding window + bit manipulation, where efficiency really matters. Key Takeaways: -> Using rolling hash / bitmask to track binary substrings -> Avoiding substring overhead for better performance -> Understanding the limit: total required codes = 2^k -> Early pruning with length checks for optimization Language: Java -> Runtime: 6 ms (Beats 100%) ⚡ -> Memory: 47.73 MB Almost at the finish line — one binary string at a time 💻🔥 #LeetCode #Java #BitManipulation #SlidingWindow #Strings #ProblemSolving #100DaysOfCode
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LeetCode Problem || Find Unique Binary String (1980)🚀 Today I solved the problem "Find Unique Binary String" using Java. 🔹 Problem: We are given an array of n binary strings, each of length n. The goal is to return a binary string of length n that does not exist in the array. 🔹 Approach (Diagonal Flip Technique): The idea is simple but powerful: Traverse the array using index i. Look at the i-th character of the i-th string (nums[i][i]). Flip the bit (0 → 1, 1 → 0). Append it to a result string. 💡 Time Complexity: O(n) Practicing problems like this strengthens logical thinking and problem-solving skills. #LeetCode #Java #CodingPractice #ProblemSolving #DSA
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