Mastering Array Rotation with LeetCode 189 Solution

🚀 01/04/26 — Navigating the Cycle: Mastering Circular Array Loops Today marks the start of a new month and a massive continuation of my journey—Day 45 of my coding streak! I tackled the Circular Array Loop (LeetCode 457), a complex problem that requires tracking directions and cycle lengths within a wrapping array. 🔄 Circular Array Loop (LeetCode 457) The challenge is to find if there is a cycle in the array where the movement is either entirely forward or entirely backward, and the cycle length is greater than one. The Logic: Traversal & Validation Next Index Calculation: I implemented a helper function calNxt to handle the circular movement. For positive jumps, it uses a simple modulo. For negative jumps, it adds the array length to the modulo result to ensure the index stays within positive bounds. State Tracking: For every starting index, I use a HashSet to keep track of visited nodes in the current path. Directional Consistency: I capture the direction at the start of each path (isPos). If the movement ever switches direction (e.g., a positive jump leads to a negative value), the loop is invalidated. Cycle Length Check: A valid cycle cannot have a length of 1. I ensure this by checking curr != next before returning true. Complexity Metrics: Time: O(n^2) as we potentially start a traversal from every index. Space: O(n) to store the HashSet for visited indices. 📊 Implementation Highlights FeatureLogic AppliedCircular Jump(curr + (seq % len) + len) % lenPath TrackingHashSet<Integer> for current iterationCondition 1Must stay in original directionCondition 2Cycle length must be > 1 📈 Consistency Report Reaching a 45-day streak feels incredible! Today's problem combined the Two-Pointer-like movement of jumping through indices with the HashSet logic I mastered earlier this month. Handling the "wrapping" indices correctly was a great refresher on modulo arithmetic. Huge thanks to Anuj Kumar (a.k.a CTO Bhaiya on YouTube) for the foundational roadmap. Moving into these more advanced circular logic problems shows how much my core array skills have grown. My tested implementation for the circular loop detection is attached below! 📄👇 #DSA #Java #LeetCode #CircularArray #HashSet #CycleDetection #CodingJourney #Consistency #45DayStreak #LearningInPublic #CTOBhaiya

🚀 Day 30/60 — LeetCode Discipline Problem Solved: 3Sum Closest Difficulty: Medium Today’s problem pushed beyond exact answers and focused on finding the closest possible sum to a given target — a subtle yet powerful variation of the classic 3Sum problem. By sorting the array and using a two-pointer approach, the solution efficiently explores combinations while continuously updating the closest result. This problem highlights how small modifications in logic can significantly change the nature of a problem — from exact matching to optimal approximation. 💡 Focus Areas: • Strengthened two-pointer technique • Practiced working with sorted arrays • Improved handling of optimization conditions • Learned to track and update closest values dynamically • Reinforced problem-solving under constraints ⚡ Performance Highlight: Achieved efficient runtime with optimized traversal. Not every problem asks for perfection — sometimes, the goal is to get as close as possible. #LeetCode #60DaysOfCode #100DaysOfCode #DSA #TwoPointers #Algorithms #ProblemSolving #CodingJourney #SoftwareEngineering #Java #Developers #Consistency #TechGrowth #LearnToCode

Solved Majority Element II using an optimized Boyer–Moore Voting Algorithm approach. The solution identifies elements that appear more than n/3 times without using extra space. Instead of using a HashMap, the algorithm maintains two potential candidates and their counts during traversal, then validates them in a second pass. Time Complexity: O(n) Space Complexity: O(1) #Java #DSA #ProblemSolving #Coding #LeetCode #Developers

Many developers use extra space (like HashMaps) to solve the majority element problem. But there’s a more optimal approach Moore’s Voting Algorithm. - O(n) time, O(1) space - Uses a candidate + count mechanism - Cancels out different elements - Requires a final verification step Key insight: At any point, the majority element will survive the cancellation process. This is a classic interview problem that tests your understanding of optimization. Have you used Moore’s Algorithm before? #DSA #Java #CodingInterview #Algorithms #SoftwareDevelopment #Programming #Developers

Leetcode Practice - 18. 4Sum Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that: ✅ 0 <= a, b, c, d < n ✅ a, b, c, and d are distinct. ✅ nums[a] + nums[b] + nums[c] + nums[d] == target You may return the answer in any order.   Example 1: Input: nums = [1,0,-1,0,-2,2], target = 0 Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]] Example 2: Input: nums = [2,2,2,2,2], target = 8 Output: [[2,2,2,2]]   Constraints: 1 <= nums.length <= 200 -109 <= nums[i] <= 109 -109 <= target <= 109 #LeetCode #Java #CodingPractice #ProblemSolving #DSA #Array #DeveloperJourney #TechLearning

Solved the Copy List with Random Pointer problem using an in-place approach. Instead of using extra space like HashMap, the solution inserts copied nodes in between original nodes, sets random pointers, and then separates both lists. This keeps the solution efficient in both time and space. Time Complexity: O(n) Space Complexity: O(1) #Java #DSA #LinkedList #LeetCode #Coding

🚀 Code 6 – #50LeetCodeChallenge 🧩 Problem: Search Insert Position Given a sorted array of distinct integers and a target value, return its index if found. If not, return the position where it should be inserted to maintain sorted order. 💡 Approach: Use Binary Search to efficiently locate the target or its correct insertion position in O(log n) time. 📚 Key Takeaway: Binary search is the go-to technique for problems involving sorted arrays, especially when optimal time complexity is required. #LeetCode #Java #Coding #ProblemSolving #BinarySearch #Arrays #Programming

✅ Solved: Palindrome Number — LeetCode #9 Accepted with all 11,511 / 11,511 test cases passed! 🎯 The problem asks: Given an integer x, return true if x is a palindrome, and false otherwise. My approach: Immediately return false for negative numbers (can't be palindromes) Convert the integer to a String using String.valueOf() Reverse the string by iterating from end to start Compare original and reversed — if equal, it's a palindrome ✔️ Runtime: 16 ms | Memory: 46.48 MB A clean beginner-friendly solution. Next, I'd like to optimize it using a mathematical digit-reversal approach — no String conversion needed, better memory efficiency! 🚀 If you're grinding LeetCode, don't skip the easy problems. They build the intuition you need for the hard ones. 💪 #LeetCode #Java #DSA #DataStructures #Algorithms #CodingChallenge #Programming #SoftwareDevelopment #ProblemSolving #PalindromeNumber

🚀 Day 26/100 Days of #CodeChallenge Today’s problem: Isomorphic Strings (Leetcode 205) This problem helped me understand how to map characters between two strings while maintaining consistency and order. 💡 Key Concept: Two strings are isomorphic if characters in one string can be replaced to get the other string — with: ✔️ One-to-one mapping ✔️ No two characters mapping to the same character ✔️ Order preserved 🧠 What I Learned: How to use mapping (arrays/hashmaps) efficiently Importance of bidirectional checking Handling edge cases like unequal lengths ⚡ Approach: Compare lengths first Track character mappings using arrays Ensure consistency in both directions ⏱️ Complexity Analysis: Time Complexity: O(n) → We traverse the strings once Space Complexity: O(1) → Fixed-size arrays (256 characters) ✅ Successfully solved and understood the logic! Every day is a step closer to mastering problem-solving 💪 #Day26 #100DaysOfCode #Java #DSA #LeetCode #CodingJourney #ProblemSolving #TechLearning

🚀 Solved Linked List Cycle Detection using Fast & Slow Pointers Used Algorithm are: - Slow pointer moves 1 step - Fast pointer moves 2 steps - If they meet → cycle exists To find cycle start: - Move slow to head - Move both one step → meeting point = cycle start ⚡ Time: O(n) | Space: O(1) #DSA #LinkedList #Java #LeetCode #Coding