Aakriti Jha’s Post

🚀 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

🚀 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 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

🚀 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 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 

🚀 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

DAY 3 OF 70-DSA PARTNER CHALLENGE Starting from the ground up and that means Loops in Java first. Before you touch arrays, trees, or dynamic programming, you need to speak the language fluently. Most people skip the fundamentals and wonder why they struggle later. We're not doing that. Drop a 1 in the comments if you're following along with this challenge! #codingjourney #jobready #coding

When coding.... it's always important to give meaningful names to your variables. This improves human readability. So the time I create an array of booleans, I'm going to call it aBunchOfBool. Uhm.... well maybe not. #programming #coding #humor #silly #codingJoke #codingHumor #java #javaProgramming #ProgrammingTips

Recursion vs Iteration — a common question in programming. Both approaches can solve the same problem, but they come with different trade-offs. In this short video, I covered: - Recursion (clean and elegant, but uses stack memory) - Iteration (faster and memory efficient) - Fibonacci example comparison - When to use each approach Understanding this difference helps you write efficient, optimized, and scalable code. Explore structured DSA in Java roadmap + practice: www.quipoin.com #DSA #Java #Programming #Coding #SoftwareEngineering #Recursion #InterviewPreparation