LeetCode Problem: Fancy Sequence Optimization

🚀 Day 63 of #100DaysOfLeetCode ✅ Problem Solved: Unique Binary Search Trees (LeetCode 96) Today’s problem was a great example of how Dynamic Programming and mathematical patterns (Catalan Numbers) come together. 🔍 Key Insight: For every node chosen as root, the number of unique BSTs is: 👉 Left Subtrees × Right Subtrees This leads to the recurrence: dp[n] = Σ (dp[left] × dp[right]) 💡 What I learned: Breaking problems into smaller subproblems makes complex structures easier Recognizing patterns like Catalan Numbers is a game changer DP is not just about arrays, it's about thinking smart ⚡ Result: ✔️ Runtime: 0 ms (Beats 100%) ✔️ Clean and optimized solution Consistency is slowly turning into confidence 💪 #LeetCode #DataStructures #DynamicProgramming #CodingJourney #ProblemSolving #Java #100DaysOfCode

🚀 LeetCode – Subsets | Backtracking Approach Given an integer array nums of unique elements, return all possible subsets (the power set). The solution set must not contain duplicate subsets. Return the solution in any order. 💡 Approach Used – Backtracking I used a recursive backtracking strategy where:  1. Start with an empty subset  2. Add the current subset to the result  3. Choose an element and explore further combinations  4. Backtrack by removing the element to try other possibilities This ensures that we explore all possible combinations systematically. Since each element has two choices (include or exclude), the total subsets become 2ⁿ. #LeetCode #DSA #Backtracking #Java #CodingPractice #Algorithms #ProblemSolving #SoftwareEngineering #TechLearning

🚀 Day 3/100 – LeetCode Challenge 🔍 Problem: Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be inserted to maintain the sorted order. 💡 Key Concepts Used: • Binary Search • Time Complexity Optimization – O(log n) • Efficient searching in sorted arrays 📚 What I Learned: • How binary search reduces search time significantly compared to linear search • Handling edge cases when the target element is not present • Determining the correct insertion index while maintaining sorted order hashtag #100DaysOfCode #LeetCode #Java #DataStructures #Algorithms #BinarySearch #CodingJourney #ProblemSolving

🔥 Day 59/100 – LeetCode Challenge 📌 Problem Solved: Remove Duplicates from Sorted Array II (Medium) Today’s problem was a great exercise in in-place array manipulation and two-pointer technique. 💡 Key Idea: Since the array is already sorted, duplicates are adjacent. Instead of removing all duplicates, we allow each element to appear at most twice. 👉 The trick is to compare the current element with the element at index k-2. If they are the same → skip ❌ If different → keep it ✅ ⚙️ Approach: Initialize pointer k = 2 Traverse from index 2 Copy valid elements forward Maintain order without extra space 🧠 What I learned: How to efficiently handle constraints like “at most twice” Importance of thinking in terms of index relationships (k-2) Writing optimal O(n) solutions with O(1) space 📊 Performance: ⚡ Runtime: 0 ms (100%) 💾 Memory: 48.46 MB 💻 Tech Used: Java Consistency is key 🔑 — 59 days done, 41 more to go! #100DaysOfCode #LeetCode #Java #DataStructures #CodingChallenge #ProblemSolving #TechJourney

🚀 Day 25 of my #100DaysOfCode Journey Today, I solved the LeetCode problem Valid Perfect Square. Problem Insight: Given a positive integer, the task is to determine whether it is a perfect square without using built-in functions like sqrt(). Approach: Used Binary Search to efficiently find whether a number has an integer square root. Initialized search range from 0 to num Calculated mid and checked mid * mid If equal → return true If square is smaller → move right (low = mid + 1) If square is larger → move left (high = mid - 1) Used long for multiplication to avoid overflow issues. Time Complexity: O(log n) — efficient binary search approach Takeaway: Binary Search is not just for arrays — it can be applied to mathematical problems to optimize performance and avoid brute force. #DSA #Java #LeetCode #CodingJourney #100DaysOfCode #BinarySearch

🚀 Day 44 of #100DaysOfCode Solved 1011. Capacity To Ship Packages Within D Days on LeetCode 📦 🧠 Key Insight: We need to find the minimum ship capacity such that all packages can be shipped within D days. This is a classic case of Binary Search on Answer. ⚙️ Approach: 1️⃣ The minimum capacity = max(weight) (since one package must fit) 2️⃣ The maximum capacity = sum of all weights 3️⃣ Apply Binary Search between these bounds 4️⃣ For each capacity mid, simulate shipping: 🔹Keep adding weights until capacity exceeds 🔹Move to the next day when exceeded 5️⃣ If we can ship within D days → try smaller capacity 6️⃣ Else → increase capacity This helps us find the minimum valid capacity efficiently. ⏱️ Time Complexity: O(n log(sum)) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #BinarySearch #Algorithms #Java #InterviewPrep #CodingJourney

Day 36/50 🚀 Solved “Valid Parentheses” today — a classic stack problem, but a great reminder that simple concepts can be powerful when applied correctly. 💡 Key takeaway: Using a stack makes it easy to track opening brackets and validate matching pairs efficiently. Every closing bracket should correspond to the most recent unmatched opening one — LIFO in action. ⚙️ What I focused on: Clean conditional checks Avoiding unnecessary complexity Writing readable, structured code 📈 Result: Accepted ✅ Optimized runtime & solid performance On to Day 37 🔥 #50DaysOfCode #DataStructures #Algorithms #Java #LeetCode #CodingJourney

🚀 Day 45 of #100DaysOfCode 🧩 Problem Solved: Longest Common Prefix Today’s challenge was to find the common prefix among a list of strings. I used an efficient approach by leveraging sorting. 💡 Approach: Sort the array of strings Compare the first and last strings Build the prefix character by character 🧠 Why this works: After sorting, the first and last strings are the most different. Their common prefix will be the answer for all strings. ⚡ Key Learning: Smart thinking > brute force. Using sorting can reduce unnecessary comparisons and simplify logic. 🏷️ Tags: #100DaysOfCode #Day45 #LeetCode #Java #DSA #CodingJourney #ProblemSolving #Algorithms #String #Array 📈 Staying consistent and improving step by step! 🚀

🚀 Day 46 of #100DaysOfCode Solved 719. Find K-th Smallest Pair Distance on LeetCode 📏 🧠 Key Insight: We need to find the k-th smallest absolute difference among all possible pairs in the array. Brute force (generating all pairs) would be O(n²) → not efficient. Instead, we use Binary Search on Answer + Two Pointers. ⚙️ Approach: 1️⃣ Sort the array 2️⃣ Define search space: 🔹left = 0 (minimum distance) 🔹right = max(nums) - min(nums) 3️⃣ Apply Binary Search: 🔹For a given mid, count how many pairs have distance ≤ mid 4️⃣ Counting pairs efficiently: 🔹Use two pointers 🔹For each i, move j while nums[j] - nums[i] ≤ mid 🔹Add (j - i - 1) to count 5️⃣ If count ≥ k → try smaller distance 6️⃣ Else → increase distance 🎯 Final answer = smallest distance satisfying the condition ⏱️ Time Complexity: O(n log n + n log D) 🔹Sorting + Binary Search with linear check 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #BinarySearch #TwoPointers #Algorithms #Java #InterviewPrep #CodingJourney