Check Binary String for One Segment of Ones in Java

🚀 Day 47/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 448. Find All Numbers Disappeared in an Array Used an in-place marking technique by treating indices as a hash map and marking visited numbers as negative to identify missing elements. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) (excluding output list) Strengthening understanding of array manipulation and in-place hashing tricks. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 24/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2859. Sum of Values at Indices With K Set Bits Used bit manipulation to count the number of set bits in each index and summed the values whose index contains exactly k set bits. ⏱️ Time Complexity: O(n log n) 📦 Space Complexity: O(1) Strengthening understanding of bitwise operations and binary representation through consistent practice. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 46/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2315. Count Asterisks Used a boolean flag approach to track whether the current position is inside a pair of '|' and count only the valid '*' outside those sections. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) Strengthening understanding of state-based string traversal and conditional counting. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 48/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 1534. Count Good Triplets Used a brute-force triple nested loop to check all possible triplets and validate the given conditions using absolute differences. ⏱️ Time Complexity: O(n³) 📦 Space Complexity: O(1) Strengthening understanding of nested loop patterns and condition-based filtering. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

Day 88 - LeetCode Journey Solved LeetCode 24: Swap Nodes in Pairs in Java ✅ This problem is a great exercise in pointer manipulation in linked lists. Instead of changing values, I swapped the actual nodes by carefully adjusting the next pointers. Using a dummy node made the process much cleaner and helped handle edge cases smoothly. Step by step, I swapped pairs while moving forward in the list. Key takeaways: • Deep understanding of pointer manipulation • Importance of dummy node in linked list problems • Clean handling of edge cases • Iterative approach for swapping nodes ✅ All test cases passed ⚡ Efficient O(n) time and O(1) space Linked list problems like this really sharpen your fundamentals 🔥 #LeetCode #DSA #Java #LinkedList #Pointers #ProblemSolving #CodingJourney #InterviewPrep #Consistency #100DaysOfCode

🚀 Day 34/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 709. To Lower Case Used ASCII manipulation to convert uppercase letters to lowercase by adding 32, without using built-in functions. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) Strengthening understanding of character encoding and string manipulation basics. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

Day 14/100 – LeetCode Challenge 🚀 Problem: #169 Majority Element   Difficulty: Easy   Language: Java   Approach: Sorting + Middle Element   Time Complexity: O(n log n)   Space Complexity: O(1) 🔍 Key Insight: The majority element appears **more than ⌊n / 2⌋ times** in the array. If we **sort the array**, the majority element must occupy the **middle position** because it appears more than half of the time. Therefore, the element at index **n/2** will always be the majority element. 🧠 Solution Brief: First sorted the array using `Arrays.sort()`. Since the majority element appears more than half of the array length, it will always be positioned at the middle index. Finally returned `nums[nums.length / 2]` as the majority element. 📌 What I Learned: Understanding problem constraints can simplify the solution significantly. Sometimes a simple observation (like majority occupying the middle after sorting) can avoid more complex implementations. #LeetCode #Day14 #100DaysOfCode #Java #DSA #Arrays #ProblemSolving #CodingJourney

Day 59 — LeetCode Progress (Java) Problem: Count the Number of Consistent Strings Required: Given a string allowed and an array of strings words, return the number of consistent strings (strings that contain only characters from allowed). Idea: Use a HashSet for fast lookup to check whether each character in a word is allowed. Approach: Store all characters of allowed in a HashSet. Iterate through each word: Check every character in the word If any character is not in the set, the word is invalid Count all valid (consistent) words. Time Complexity: O(n × m) Space Complexity: O(1) #LeetCode #DSA #Java #HashSet #Strings #Algorithms #CodingJourney #100DaysOfCode

Day 12/100 – LeetCode Challenge Problem: Middle of the Linked List Today’s problem focused on finding the middle node of a singly linked list. Approach: Used the Two-Pointer Technique (Slow & Fast pointers). slow moves one step at a time fast moves two steps at a time When fast reaches the end of the list, slow will be at the middle node Complexity: Time: O(n) Space: O(1) Concepts Practiced: Linked List traversal Two-pointer technique Efficient single-pass solution #100DaysOfCode #LeetCode #DSA #Java #LinkedList #ProblemSolving #CodingJourney

🚀 Day 41/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 3427. Sum of Variable Length Subarrays Used a nested loop approach to calculate subarray sums for each index by dynamically determining the starting point using max(0, i - nums[i]). ⏱️ Time Complexity: O(n²) 📦 Space Complexity: O(1) Strengthening understanding of subarray problems and index-based range calculations. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency