Minimum Distance Between Three Equal Elements LeetCode Solution

Day: 95/365 📌 LeetCode POTD: Minimum Distance Between Three Equal Elements II Medium Key takeaways/Learnings from this problem: 1. This one highlights how tracking indices smartly is more important than checking all triplets brute-force. 2. Using previous occurrences helps you narrow down valid triples quickly instead of re-scanning the array. 3. Big takeaway: problems with “equal elements” often reduce to index management + pattern observation. 4. Also a good reminder that optimizing from O(n³) thinking to near O(n) comes down to storing the right info while iterating. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

🚀 Day 21 of #128DaysOfCode 🔍 Key Learnings: Efficient searching in O(log n) Using low & high to narrow the range Finding correct position even if target is absent 🧠 Approach: Compare mid with target → move left/right → return index or insert position Consistency is key 🔥 #DSA #Java #CodingJourney #PlacementPreparation

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 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

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 75 – Linked List Cycle Detection Solved a problem to detect whether a linked list contains a cycle using an efficient two-pointer approach. Key Learnings: Applied Floyd’s Cycle Detection Algorithm Understood how different pointer speeds help detect loops Achieved optimal solution with O(1) space and O(n) time complexity #DSA #Java #LinkedList #TwoPointers #ProblemSolving #CodingPractice

🚀 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 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 58/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 278. First Bad Version Used binary search on answer space to efficiently find the first bad version while minimizing API calls. ⏱️ Time Complexity: O(log n) 📦 Space Complexity: O(1) Strengthening understanding of binary search optimization and decision-based problems. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 57/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 242. Valid Anagram Used a HashMap frequency counting approach (Unicode-safe using code points) to compare character frequencies of both strings. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) Strengthening understanding of string frequency analysis and Unicode handling. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency