Balancing a Binary Search Tree with Inorder Traversal

🚀 Day 100 of #100DaysOfLeetCode Problem: 127. Word Ladder Difficulty: Hard Key Insight: This is a shortest-path problem where each valid word transformation is an unweighted edge — BFS guarantees the minimum sequence length. Approach: Used BFS starting from beginWord. At each step, generated all possible words by changing one character (a–z). Used a HashSet for O(1) lookup and removed visited words to avoid revisits. Complexity: Time: O(N × L × 26) Space: O(N) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

#Day58 of my second #100DaysOfCode Today’s problem was about carefully handling overlapping intervals. DSA • Solved Merge Intervals (LeetCode 56) • Implemented a brute force approach by checking and merging overlapping intervals → O(2n) + O(n log n) time • Implemented the optimal approach by sorting intervals and merging them in a single traversal → O(n log n) + O(n) time • Key idea: once intervals are sorted by start time, overlapping ranges can be merged efficiently in one pass These interval problems really test attention to detail with edge cases. #DSA #Algorithms #LeetCode #Java #100DaysOfCode #WomenWhoCode #BuildInPublic #LearningInPublic

Day 60/100 – LeetCode Challenge ✅ Problem: #67 Add Binary Difficulty: Easy Language: Java Approach: Reverse Iteration with Carry Time Complexity: O(max(n, m)) Space Complexity: O(max(n, m)) Key Insight: Binary addition follows same rules as decimal: Sum bits + carry Result bit = sum % 2 New carry = sum / 2 Solution Brief: Iterated from rightmost bits of both strings. Tracked carry and built result from right to left using StringBuilder. Reversed final string for correct order. #LeetCode #Day60 #100DaysOfCode #Binary #Java #Algorithm #CodingChallenge #ProblemSolving #AddBinary #EasyProblem #StringManipulation #BitManipulation #DSA

Day 71/100 – LeetCode Challenge ✅ Problem: #1356 Sort Integers by The Number of 1 Bits Difficulty: Easy Language: Java Approach: Custom Comparator with Bit Counting Time Complexity: O(n log n) Space Complexity: O(n) Key Insight: Sort integers based on number of set bits (1s in binary representation). If bit counts are equal, sort by natural integer order. Use Integer.bitCount() for efficient population count. Solution Brief: Converted primitive array to Integer[] for custom sorting. Applied Arrays.sort() with comparator: Compare bitCount(a) vs bitCount(b) If equal, compare integers directly using a.compareTo(b) Converted back to primitive array. #LeetCode #Day71 #100DaysOfCode #Sorting #Java #Algorithm #CodingChallenge #ProblemSolving #SortByBits #EasyProblem #BitManipulation #Comparator #DSA

🚀 Day 36 of #100DaysOfCode 🌱 Topic: Strings / Pattern Checking ✅ Problem Solved: LeetCode 1784 – Check if Binary String Has at Most One Segment of Ones 🛠 Approach: Traversed the binary string and checked the transition between characters. Iterated through the string from left to right. Looked for the pattern "01" → followed by "1" again, which indicates a new segment of ones. If a 0 → 1 transition occurs after already seeing ones, it means more than one segment exists. #100DaysOfCode #Day36 #DSA #Strings #LeetCode #Java #ProblemSolving #CodingJourney #Consistency

Day 7 of Daily DSA 🚀 Solved LeetCode 167: Two Sum II – Input Array Is Sorted Approach: Used the two-pointer technique leveraging the sorted nature of the array. Moved pointers inward based on the comparison with the target to find the answer in one pass. Complexity: • Time: O(n) • Space: O(1) (constant extra space) LeetCode Stats: • Runtime: 2 ms (Beats 96.10%) • Memory: 48.54 MB (Beats 38.87%) This problem highlights how understanding input constraints can turn a brute-force solution into an optimal one. #DSA #LeetCode #Java #TwoPointers #ProblemSolving #Consistency

Leetcode Problem || Number of Steps to Reduce a Number in Binary Representation to One(1404)🚀 Today I solved a binary manipulation problem where we must: If number is even → divide by 2 If odd → add 1 Count steps until number becomes 1 💡 Learned: How to simulate binary addition using carry Avoid integer overflow Optimize string-based numeric problems #Java #DSA #BitManipulation #LeetCode #ProblemSolving

🎯 Day 100 of #100DaysOfCode 🔥 What a way to wrap it up! Solved LeetCode #3666 – Minimum Operations to Equalize Binary String ✅ A problem that blends math, parity logic, and careful case analysis—not your usual binary flip question. Key takeaways: -> Count-based optimization over brute force -> Handling even/odd operation patterns smartly -> Early exits = cleaner & faster logic -> Thinking in terms of operations feasibility rather than simulation 🧠 Language: Java -> Runtime: 0 ms (Beats 83.87%) -> Memory: 47.90 MB 100 days. Countless problems. One habit built: consistency 💪 Onward to harder problems and deeper concepts 🚀 #LeetCode #Java #DSA #BinaryStrings #ProblemSolving #Consistency #100DaysChallenge

Day 5 – Longest Consecutive Sequence. Today's problem focused on finding the longest consecutive sequence in an unsorted array, which helps strengthen understanding of sorting, edge cases, and sequence tracking. Approach Step 1 – Sort the Array Step 2 – Traverse the Array While iterating: 1️⃣ If elements are equal → skip duplicates 2️⃣ If nums[i] == nums[i-1] + 1 → increase sequence count 3️⃣ Otherwise → reset the count. Step 3 – Track Maximum Length. Github Code :- https://lnkd.in/gF8DqckJ #LeetCode #DSA #StriverA2Z #CodingJourney #Java #Arrays #ProblemSolving #CodingPractice #SoftwareEngineering #TechInterviewPrep #100DaysOfCode #DeveloperJourney #ComputerScience

🚀 Day 98 of #100DaysOfCode Solved LeetCode #1356 – Sort Integers by The Number of 1 Bits ✅ A clean bit manipulation + sorting problem that rewards thinking beyond plain comparisons. Key Takeaways: -> Using Integer.bitCount() to count set bits efficiently -> Encoding sort priority directly into values for simplicity -> Leveraging built-in sorting for clean and fast solutions -> Small tricks can lead to elegant code ✨ Language: Java -> Runtime: 6 ms (Beats 93.70%) ⚡ -> Memory: 47.42 MB Almost there. Staying consistent till the end 💻🔥 #LeetCode #Java #BitManipulation #Sorting #ProblemSolving #DSA #100DaysOfCode