Minimum Distance Between Three Equal Elements II LeetCode Solution

Day: 94/365 📌 LeetCode POTD: Minimum Distance Between Three Equal Elements I Easy Key takeaways/Learnings from this problem: 1. This problem is a nice reminder that tracking indices smartly is often more important than the values themselves. 2. Keeping the last few occurrences of each element helps you quickly check valid triplets without brute force. 3. It shows how a simple hashmap or array tracking can turn an O(n³) idea into something efficient. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

Day 42 of consistency 💻🔥 Solved Add Two Numbers using linked lists — a classic problem that really tests understanding of pointers, carry handling, and edge cases. Key takeaways: Handling carry efficiently is crucial Dummy nodes make list problems much cleaner Iterative approach keeps space optimal Not the fastest runtime yet, but improvement is a process — optimizing step by step. Consistency > Perfection. #Day42 #LeetCode #DSA #Java #CodingJourney #ProblemSolving #100DaysOfCode

🚀 Day 28/100 – #100DaysOfCode Today’s problem: Reverse Linked List (LeetCode 206) 💡 What I learned: How to reverse a singly linked list using an iterative approach Managing pointers efficiently (prev, curr, next) Understanding how links change direction step by step 🧠 Key Idea: Instead of creating a new list, we reverse the links in-place: Store next node Reverse current pointer Move forward 📊 Complexity: Time: O(n) Space: O(1) Staying consistent, one problem at a time... #Day28 #100DaysOfCode #LeetCode #DSA #LinkedList #Java #CodingJourney #ProblemSolving #TechSkills

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

🚀 Day 65/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2535. Difference Between Element Sum and Digit Sum of an Array Used a single loop + digit extraction approach to calculate both element sum and digit sum efficiently. ⏱️ Time Complexity: O(n × d) 📦 Space Complexity: O(1) Strengthening understanding of number manipulation and digit-based operations. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

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 92/100 ✅ 𝐏𝐫𝐨𝐛𝐥𝐞𝐦: 𝐓𝐡𝐢𝐫𝐝 𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐍𝐮𝐦𝐛𝐞𝐫 Today’s problem was about finding the third distinct maximum number in an array. If it doesn’t exist, we return the maximum number instead. 💡 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠𝐬: Maintain three variables to track top 3 distinct values Handle 𝐝𝐮𝐩𝐥𝐢𝐜𝐚𝐭𝐞𝐬carefully Use 𝐋𝐨𝐧𝐠.𝐌𝐈𝐍_𝐕𝐀𝐋𝐔𝐄 to avoid edge case issues Single pass solution → 𝐎(𝐧) 𝐭𝐢𝐦𝐞 𝐜𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Instead of sorting, we efficiently track max1, max2, and max3 while iterating through the array. This improves performance and avoids unnecessary computations. #Day92 #100DaysOfCode #LeetCode #Java #DSA #CodingJourney #ProblemSolving #TechGrowth

🚀 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 17 of #100DaysOfCode Solved Plus One (Easy) today. A simple but important problem that reinforces handling edge cases carefully. Key Idea: Traverse from the last digit, handle carry properly. Approach: Start from the end of the array If digit < 9 → increment and return If digit == 9 → make it 0 and carry forward If all digits are 9 → create new array with leading 1 Learning: Even the easiest problems test your attention to edge cases. Consistency > Complexity. #LeetCode #DSA #CodingJourney #Java #ProblemSolving #Consistency

🚀 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