Optimize LeetCode Problem with Efficient Formula

🌟 Day 29/30 – Add Two Numbers Approach: Instead of handling multiple loops for different list lengths, I used one single loop. At each step, I: Added values from both lists (if present) Included the carry Created a new node with sum % 10 Updated carry using sum / 10 Missing nodes are treated as 0, which simplifies the logic a lot. Complexity: Time: O(max(n, m)) Space: O(max(n, m)) Key Takeaway: A dummy node + single loop can replace multiple edge-case loops and make the solution much cleaner. #DSA #LeetCode #Java #LinkedList #ProblemSolving #Day29 #30DayChallenge #Coding

🔹 Day 85 – #100DaysOfLeetCode Problem: 3010. Divide an Array Into Subarrays With Minimum Cost I Difficulty: Easy Key Insight: The cost of a subarray depends only on its first element. Since the first subarray always starts at index 0, the problem reduces to selecting the two smallest possible starting elements from the remaining array. Approach: Fix the first subarray cost as nums[0] Find the smallest and second smallest values in nums[1…n-1] Add them to get the minimum total cost Time Complexity: O(n) Space Complexity: O(1) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

Day 13/100 – LeetCode Challenge 🚀 Problem: Binary Search Approach: Maintained left and right pointers Compared middle element with target Reduced the search space by half each step Time Complexity: O(log n) Space Complexity: O(1) Key takeaway: Binary search is the foundation for solving problems with logarithmic efficiency. #LeetCode #100DaysOfCode #DSA #Java #ProblemSolving #InterviewPrep #100DaysOfLeetCode

🧠 Day 76 of #100DaysOfLeetCode 📌 Problem: Combination Sum II 📊 Difficulty: Medium 💡 Key Insight: Sorting the array helps handle duplicates. While backtracking, duplicates are skipped only in the “not pick” path to ensure unique combinations. 🛠️ Approach: Sort the candidates array Use backtracking to explore combinations Each number can be used once → move to index + 1 Skip duplicate elements to avoid repeated combinations ⏱️ Complexity: Time: O(2^n) Space: O(n) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

🌡️ Daily Temperatures – Stack Based Problem (LeetCode 739) Solved Daily Temperatures using Monotonic Stack approach. Learned how to efficiently find the next greater element in future days and calculate waiting days in O(n) time. 📌 Key Concepts: Monotonic Stack Next Greater Element Pattern Time Optimization (O(n) vs O(n²)) #DSA #LeetCode #MonotonicStack #Java #ProblemSolving #CodingJourney

Day 5/100 – LeetCode Challenge 🚀 Problem: Intersection of Two Arrays Optimized approach: Used HashSet to store elements of the first array Checked presence while iterating the second array Stored unique intersection values Time Complexity: O(n + m) Space Complexity: O(n) Key takeaway: Hashing removes the need for sorting and nested loops, leading to cleaner and faster solutions. #LeetCode #100DaysOfCode #DSA #Java #ProblemSolving #InterviewPrep #100DaysOfLeetCode

🚀 Day 2 | Problem 2 of my 15-Day 50+ String DSA Challenge Problem: Count Vowels in a String (Java) Today I worked on a basic but important string problem and focused on getting the logic and flow right instead of just writing code. Key learnings: • Iterating through each character of a string • Correct use of loop boundaries to avoid runtime errors • Understanding array indexing (0 to length-1) • Applying conditional checks cleanly for vowel detection Strengthening fundamentals step by step 💪 #DSA #Java #StringProblems #LearningInPublic #CodingJourney

🚀 Day 2 of #100DaysOfCode Solved Contains Duplicate on LeetCode ✅ 🧠 Key takeaway: Using a HashSet helps quickly check if any element appears more than once by leveraging constant-time lookups. ⚙️ Approach: 🔹Traverse the array 🔹Add elements to a set 🔹If an element already exists → duplicate found ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) #100DaysOfCode #LeetCode #DSA #Java #HashSet #ProblemSolving #LearningInPublic

🚀 Day 147 | Frequency Counting Across Multiple Strings Today’s problem was a clean exercise in comparing character frequencies across multiple inputs. 🧩 Problem Solved: 1002. Find Common Characters 🔍 Approach: • Initialized a frequency array using the first word. • For each subsequent word, counted character frequencies and kept the minimum count for each character. • Constructed the result by adding each character as many times as its final minimum frequency. ✨ Key Insight: Using minimum frequency across all strings ensures only truly common characters are included. 📚 Topics: Hashing · Strings · Frequency Counting 💻 Platform: LeetCode #LeetCode #DSA #ProblemSolving #DailyCoding #Strings #Hashing #Java #Consistency

LeetCode Practice - 1047. Remove All Adjacent Duplicates In String ✅ Key idea Instead of using a real stack: 🔷We use a character array 🔷And an integer pointer (top) 🔷This behaves exactly like a stack. 🧩 What we maintain 🔷char[] stack → stores final characters 🔷top → points to the last inserted character 🔷top = -1 → stack is empty #LeetCode #Java #StringHandling #CodingPractice #ProblemSolving #DSA #DeveloperJourney #TechLearning