Minimum Distance Between Three Equal Elements in Array with Java

🚀 Code 3 – #50LeetCodeChallenge Problem: 4Sum Given an array of integers and a target value, find all unique quadruplets that sum up to the target. Each element must be used only once, and the solution set should not contain duplicate combinations. 💡 Approach: Sort the array and use nested loops along with a two-pointer technique to find combinations efficiently. Skip duplicate elements to ensure only unique quadruplets are included. 📚 Key Takeaway: Combining sorting with the two-pointer approach helps reduce complexity and is highly effective for solving multi-sum problems like 4Sum. #LeetCode #Java #Coding #ProblemSolving #Arrays #TwoPointers #Programming

🚀 Code 5– #50LeetCodeChallenge 🧩 Problem: Remove Element Given an array and a value val, remove all occurrences of val in-place and return the count of remaining elements. The order of elements can be changed. 💡 Approach: Use a two-pointer technique—one pointer iterates through the array, while the other keeps track of the position to place elements that are not equal to val. 📚 Key Takeaway: In-place modification with two pointers helps achieve O(n) time complexity and O(1) space, making it efficient for array filtering problems. #LeetCode #Java #Coding #ProblemSolving #Arrays #TwoPointers #Programming

🚀 Day 15 of 180 — 3Sum Closest ✅ LeetCode 16 — 3Sum Closest First sort the array. Then fix one element and use two pointers for the remaining two — one at left, one at right. For every triplet I calculate the sum and check how far it is from the target: distance = |target - sum| If this distance is less than my current minimum difference, I update my answer. Moving pointers is simple — if sum is greater than target → move right pointer left if sum is less than target → move left pointer right if sum equals target → that's the closest it can get, return immediately The key thing I made sure — keep tracking the minimum difference throughout and update result whenever a closer sum is found. Day 15 done. 165 to go. 🔥 #180DaysDSA #Day15 #LeetCode #Java #DSA #TwoPointers #Sorting #ThreeSum #DSAJourney #CodingJourney #Programming #DataStructures #Algorithms #ProblemSolving #BuildInPublic #CodeNewbie #LearnToCode #100DaysOfCode #SoftwareDevelopment #Developer #StudentDeveloper #TechCommunity #LinkedInTech #CompetitiveProgramming

Find the minimum distance between a given start index and any occurrence of a target in the array. Leetcode problem link: https://lnkd.in/gZFkirHw 🔍 Key Takeaways: Used a two-pointer approach to scan from both ends. Kept updating the minimum absolute distance from start. Included the condition left != right to avoid checking the same element twice. ✅ Why this works: Instead of scanning only from one side, this approach checks both ends in each iteration, making the logic structured and efficient. 📘 What I like about this solution: Simple and readable Avoids redundant checks Great example of combining two pointers with distance calculation #Java #LeetCode #DSA #Programming #SoftwareEngineering #Developers #loveToCode

🚀 Day 568 of #750DaysOfCode 🚀 🔍 Problem Solved: Two Furthest Houses With Different Colors Today’s problem looked simple at first, but it had a nice twist that tested observation skills more than brute force thinking. 💡 Key Insight: To maximize the distance between two houses with different colors, we don’t need to check all pairs. The answer will always involve either: the first house, or the last house Why? Because the maximum distance comes from the edges of the array. ⚡ Approach: Compare every house with the first and last house If colors are different → calculate distance Keep track of the maximum distance 🧠 Optimization: Instead of an O(n²) brute-force approach, we can solve this in O(n) time with constant space. 📈 Complexity: Time: O(n) Space: O(1) ✨ Takeaway: Sometimes the best solution isn’t about trying everything — it’s about spotting the right pattern. #LeetCode #Java #DSA #CodingJourney #ProblemSolving #100DaysOfCode #Programming #Tech #LearningEveryday

🚀 Code 4 – #50LeetCodeChallenge 🧩 Problem: Remove Duplicates from Sorted Array Given a sorted array, remove duplicates in-place so that each unique element appears only once. Return the count of unique elements while maintaining the original order. 💡 Approach: Use the two-pointer technique—one pointer tracks the position of unique elements, while the other scans through the array. When a new unique element is found, place it at the correct position. 📚 Key Takeaway: Two-pointer approach is highly efficient for in-place array modifications, reducing space complexity to O(1) and time complexity to O(n). #LeetCode #Java #Coding #ProblemSolving #Arrays #TwoPointers #Programming

🚀 Day 23 of #50DaysOfCode Solved Daily Temperatures (LeetCode 739) 🌡️ Today’s focus was on mastering the Monotonic Stack concept — one of the most powerful patterns in DSA. Learned how to efficiently find the next greater element by storing indices and resolving them smartly instead of brute force. 💡 Key Learnings: • Stack helps reduce time complexity from O(n²) → O(n) • Always think in terms of “pending answers” • Monotonic stacks are 🔥 for interview questions ✅ Status: Accepted ✔️ ⏱️ Optimized approach implemented Every day getting better at problem-solving and consistency 💪 #DSA #LeetCode #Java #CodingJourney #Consistency #100DaysOfCode #Programming

🚀 Day 8 of #LeetCode Challenge 🔍 Problem: Climbing Stairs 💡 Approach: Used an iterative dynamic programming approach (Fibonacci pattern). Each step = sum of previous two steps Used variables instead of array to optimize space Updated values using a loop 📌 Learning: This problem is a classic example of Dynamic Programming, where we reuse previously computed results instead of recalculating. 🔥 #Day8 #LeetCode #Java #DSA #DynamicProgramming#Consistency

🚀 Day 6 of Solved LeetCode Problem 66 – Plus One 💡 📌 Problem Summary: Given an array of digits representing a large number, increment the number by 1 and return the updated array. 🔍 Key Learning: Traverse from the last digit (right to left) Handle carry (especially when digit = 9) If all digits are 9 → create a new array like [1,0,0,...] Time Complexity: O(n) () 💻 Approach: Instead of converting to an integer (which may overflow), we simulate manual addition digit by digit. #Day6 #LeetCode #Java #CodingJourney #ProblemSolving #Programming #DSA

🚀 Day 13 of 180 — 3Sum ✅ Today I did the 3Sum problem. How I did it? 1. Using the two pointer technique. 2. Ran a for loop from i = 0 to n - 2, where n = array length. Checked two things: 3. If current element > 0 → not possible to get a valid answer, so break. 4. Avoid same ith number by comparing it with the previous value. If previous value is same as current value, move to next ith element. Initialized pointers: left = i + 1 right = n - 1 5. Found sum: If it is 0 → added triplet to list 6. Checked for duplicates: If left points to same second number → move left If right points to same third number → move right 7. If sum < 0 → left++ to increase sum 8. If sum > 0 → right-- to decrease sum 9. At last, returned the list. Day 13 done. 167 to go. 🔥 #180DaysDSA #Day12 #LeetCode #Java #DSA #TwoPointers #Strings #DSAJourney #CodingJourney #Programming #DataStructures #Algorithms #ProblemSolving #BuildInPublic #CodeNewbie #LearnToCode #100DaysOfCode #SoftwareDevelopment #Developer #StudentDeveloper #TechCommunity #LinkedInTech #CompetitiveProgramming