Solving LeetCode Problem 387: First Unique Character in a String

🚀 Code 7– #50LeetCodeChallenge 🧩 Problem: Plus One Given an array representing a large integer, increment the number by one and return the updated array of digits. 💡 Approach: Traverse the array from the end and handle the carry: • If the digit is less than 9, simply increment and return • If the digit is 9, set it to 0 and continue • If all digits are 9, create a new array with an extra digit at the front. 📚 Key Takeaway: Handling carry efficiently is key in digit-based problems, especially when working with large numbers represented as arrays. #LeetCode #Java #Coding #ProblemSolving #Arrays #Math #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 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

🚀 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 8 – #50LeetCodeChallenge 🧩 Problem: Merge Sorted Array Given two sorted arrays, merge them into a single sorted array in-place within nums1, without using extra space. 💡 Approach: Use the two-pointer technique from the end: • Start comparing elements from the back of both arrays • Place the larger element at the end of nums1 • Continue until all elements are merged 📚 Key Takeaway: Working from the end avoids overwriting existing elements and ensures an efficient O(m + n) solution with O(1) extra space. #LeetCode #Java #Coding #ProblemSolving #Arrays #TwoPointers #Programming 

🚀 Code 2 – #50LeetCodeChallenge Problem: "Longest Common Prefix" Given an array of strings, find the longest common prefix shared among all strings. If no common prefix exists, return an empty string "". 💡 Approach: Start with the first string as the prefix and compare it with the remaining strings. Gradually reduce the prefix until it matches the beginning of each string. 📚 Key Takeaway: String comparison and iterative reduction help efficiently solve prefix-based problems with optimal performance. #LeetCode #Java #Coding #ProblemSolving #Strings #Programming

🔥 Day 36 of #LeetCode Journey ✅ Problem: Minimum Distance Between Three Equal Elements I Today’s problem was about finding the minimum distance between three equal elements in an array. 💡 Key Idea: Instead of checking all combinations, we track indices efficiently: Store the last two occurrences of each number When a third occurrence appears, calculate the distance Keep updating the minimum distance 🧠 What I Learned: Optimizing from brute force to O(n) approach Using HashMap effectively for tracking indices Thinking in terms of patterns instead of combinations ⚡ Time Complexity: O(n) ⚡ Space Complexity: O(n) 🔍 Example: Input: [1,2,1,1,3,1] Output: 3 Small optimizations make a big difference 💡 Staying consistent and improving every day! #Java #DSA #LeetCode #CodingJourney #100DaysOfCode #Programming #ProblemSolving

🚀 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

🚀 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