LeetCode Challenge: Sub-arrays with Average Greater than Threshold

💡 Day 48 of LeetCode Problem Solved! 🔧 🌟128. Longest Consecutive Sequence🌟 🔗 Solution Code: https://lnkd.in/gyN5ZJBr Task: Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence. You must write an algorithm that runs in O(n) time.   Example 1: Input: nums = [100,4,200,1,3,2] Output: 4 Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4. Example 2: Input: nums = [0,3,7,2,5,8,4,6,0,1] Output: 9 Example 3: Input: nums = [1,0,1,2] Output: 3 #LeetCode #Java #DSA #ProblemSolving #Consistency #100DaysOfChallenge #CodingJourney #KeepGrowing

Solved a problem where we need to check if two strings can be made equal using a special operation. The rule is: you can swap characters only if the distance between their positions is even. So basically, characters at even indices can only swap among themselves, and same for odd indices. Idea: Instead of actually swapping, I just counted characters separately for even and odd positions in both strings. If both match, then it’s possible — otherwise not. Simple concept, but interesting twist! 😊 #LeetCode #DSA #Java #ProblemSolving #CodingJourney

Day 31/50 🚀 — Reverse Vowels of a String (LeetCode 345) Today’s problem was a great reminder that sometimes the simplest approaches are the most efficient. 🔹 Used the two-pointer technique 🔹 Focused on in-place swapping 🔹 Optimized for both time (O(n)) and space (O(1)) Key takeaway: Instead of overthinking, break the problem into smaller checks—identify vowels, move pointers smartly, and swap only when needed. Clean and efficient 💡 Happy to see this solution performing well: ⚡ Runtime: 2 ms (faster than 99%+) 📦 Space: Decent optimization #Day31 #LeetCode #DSA #Java #CodingJourney #50DaysOfCode #ProblemSolving #SoftwareEngineering

🚀 Day 67 of #LeetCode Challenge ✅ Problem Solved: Check If Two String Arrays are Equivalent 💡 What I learned today: • Learned how to compare two string arrays without joining them • Understood how to traverse multiple strings using pointers • Improved handling of indices across arrays and strings • Realized the importance of edge cases to avoid runtime errors 🧠 Approach: • Used four pointers to track positions in both arrays and strings • Compared characters one by one • Moved to the next string when current string ends • Ensured both arrays are fully traversed at the end 📊 Key Takeaway: Efficient solutions avoid extra space — comparing character by character is better than building new strings 🔥 Consistency + small improvements every day = big progress #Day67 #LeetCode #CodingJourney #DSA #Java #ProblemSolving #Consistency

💡 Day 42 of LeetCode Problem Solved! 🔧 🌟1876. Substrings of Size Three with Distinct Characters🌟 Task : • A string is good if there are no repeated characters. • Given a string s, return the number of good substrings of length three in s. • Note that if there are multiple occurrences of the same substring, every occurrence should be counted. • A substring is a contiguous sequence of characters in a string.   Example 1: Input: s = "xyzzaz" Output: 1 Explanation: There are 4 substrings of size 3: "xyz", "yzz", "zza", and "zaz". The only good substring of length 3 is "xyz". Example 2: Input: s = "aababcabc" Output: 4 Explanation: There are 7 substrings of size 3: "aab", "aba", "bab", "abc", "bca", "cab", and "abc". The good substrings are "abc", "bca", "cab", and "abc". #LeetCode #Java #DSA #ProblemSolving #Consistency #100DaysOfChallenge #CodingJourney #KeepGrowing

Solved LeetCode 2839 – Check if Strings Can be Made Equal With Operations I today. In simple terms, we are given two strings and allowed to swap characters only at positions that are 2 indices apart (like index 0 with 2, 1 with 3). So instead of trying all swaps, the idea is to separate characters at even and odd positions and check if both strings have the same characters in those positions. If the frequency of characters matches for even and odd indices separately, then the strings can be made equal. A simple yet interesting problem that improves thinking around patterns and constraints. #LeetCode #DSA #Java #ProblemSolving #CodingJourney

LeetCode — Problem 189 | Day 3 💡 Problem: Rotate Array Given an array, rotate it to the right by k steps. 🧠 My Approach: - Used reverse technique for in-place rotation - First reversed the entire array - Then reversed first k elements - Finally reversed remaining elements - Handled k using k = k % n This problem gave a good understanding of: ✔️ Array manipulation ✔️ In-place optimization (O(1) space) ✔️ Reverse logic #LeetCode #DSA #Java #CodingJourney

🚀 Day 27/100 Days of Code Challenge Today’s problem: Find Peak Element (Leetcode 162) 🔍 What I learned: How to find a peak element efficiently without scanning the entire array Using Binary Search to reduce time complexity from O(n) to O(log n) Understanding how the “slope” of elements helps decide the search direction 🧠 Key Idea: Instead of checking every element, compare the middle element with its neighbor: If nums[mid] < nums[mid + 1] → move right Else → move left ✅ Example: Input: [1, 2, 3, 1] Output: 2 (index of peak element 3) Consistency is key 🔑 — improving problem-solving skills one day at a time! 💪 #Day27 #100DaysOfCode #LeetCode #BinarySearch #DSA #Java #CodingJourney