Day: 45/365 📌 LeetCode POTD: Special Binary String Hard Key takeaways/Learnings from this problem: 1. This one teaches that recognizing a balanced structure (like valid parentheses) inside a binary string is the real trick. 2. Big takeaway: recursion + sorting substrings smartly can turn a complex greedy-looking problem into something surprisingly elegant. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷
LeetCode POTD: Special Binary String Challenge
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𝗬𝗼𝘂 𝗵𝗮𝘃𝗲 𝗯𝗲𝗲𝗻 𝘂𝘀𝗶𝗻𝗴 𝗕𝗶𝗻𝗮𝗿𝘆 𝗦𝗲𝗮𝗿𝗰𝗵 𝘄𝗿𝗼𝗻𝗴. It is not just for sorted arrays. The real definition — "Eliminate HALF the search space with each decision." That one shift in thinking unlocks 20+ LeetCode problems. Swipe to see the template, live code, and 7 problems you can now solve with one pattern. 👇 Save this. 🔖 #DSA #BinarySearch #Java #LeetCode #CodingInterview #DebugWithPurpose
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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 21 - Binary Search Key Idea: Implemented an iterative binary search to locate the target in a sorted array. Reduced the search interval by half in every iteration to achieve logarithmic runtime. Time Complexity: O(log n) #Day21 #LeetCode #Java #BinarySearch #DSA #Algorithms
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🚀 Day 31/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 69. Sqrt(x) Used the Binary Search approach to find the integer square root without using built-in power functions. The search space is reduced by checking mid * mid against x. ⏱️ Time Complexity: O(log n) 📦 Space Complexity: O(1) Strengthening problem-solving skills with binary search patterns on answer space. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency
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Day 35/100 – LeetCode Challenge 🚀 Problem: 3Sum Approach: Sorted the array Fixed one element and used two pointers for the remaining two Skipped duplicates to ensure unique triplets Used early stopping when the current number became positive Time Complexity: O(n²) Space Complexity: O(1) Key takeaway: Many 3-element sum problems reduce to sorting + two-pointer pattern. #LeetCode #100DaysOfCode #DSA #Java #TwoPointers #ProblemSolving #InterviewPrep
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🚀 Quick Sort in Java Quick Sort uses Divide & Conquer — Pick a pivot ➝ Partition ➝ Recursively sort. ⚡ Avg Time: O(n log n) 🔥 Fast & efficient for large datasets static void quickSort(int arr[], int low, int high) { if (low < high) { int pi = partition(arr, low, high); quickSort(arr, low, pi - 1); quickSort(arr, pi + 1, high); } } Clean. Powerful. Classic. 💻✨ #Java #QuickSort #Coding #DSA
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LeetCode Problem || Check if Binary String Has at Most One Segment of Ones(1784)🚀 we need to check: The string should have only one continuous block of '1's. After a '0' appears, '1' should never appear again. ✨ Insight: Instead of manually counting segments using loops, we can simply check if "01" exists in the string. 📌 Time Complexity: O(n) Practicing problems like these helps improve pattern recognition and problem-solving efficiency. #LeetCode #DSA #Java #CodingPractice #ProblemSolving
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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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Day 36/100 – LeetCode Challenge 🚀 Problem: 3Sum Closest Approach: Sorted the array Fixed one element and applied the two-pointer technique Tracked the closest sum by comparing absolute differences Returned immediately if an exact match was found Time Complexity: O(n²) Space Complexity: O(1) Key takeaway: Many optimization problems are variations of classic patterns. Understanding 3Sum deeply makes solving 3Sum Closest straightforward. #LeetCode #100DaysOfCode #DSA #Java #TwoPointers #ProblemSolving #InterviewPrep
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Leetcode Problem || Complement of Base 10 Integer(1009) 🚀 The Bitwise Complement problem using Java. 🔹 Idea: The complement of a number means flipping all bits in its binary representation (0 → 1 and 1 → 0). Instead of manually converting the number to binary, I used a bit manipulation trick: 1️⃣ Find the highest set bit using Integer.highestOneBit(n) 2️⃣ Create a mask with all bits set to 1 up to that position 3️⃣ Use XOR (^) with the mask to flip the bits #LeetCode #Java #BitManipulation #CodingPractice #ProblemSolving #SoftwareDevelopment
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