LeetCode 88: Merge Sorted Array in-place

🚀 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 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

🚀 Day 546 of #750DaysOfCode🚀 🔥 Solved: Check if Strings Can be Made Equal With Operations I (LeetCode Easy) 💡 Problem Insight We are allowed to swap characters at indices where: 👉 j - i = 2 This means: Index 0 ↔ 2 (even positions) Index 1 ↔ 3 (odd positions) 🚫 But we cannot mix even and odd indices 🧠 Key Observation The string is divided into 2 independent groups: Even indices → (0, 2) Odd indices → (1, 3) 👉 We can rearrange within each group freely 👉 So both groups must match between s1 and s2 ⚡ Approach Extract characters: Even indices from both strings Odd indices from both strings Sort both groups Compare: Even parts must match Odd parts must match 📈 Complexity Time: O(1) Space: O(1) 💬 Key Takeaway Sometimes problems look like string manipulation, but the real trick is: 👉 Understanding constraints → grouping → independent transformations 🔁 Consistency check ✔️ Another day, another step forward 🚀 #LeetCode #DataStructures #Algorithms #Java #CodingChallenge #ProblemSolving #100DaysOfCode #Consistency

Day 78 - LeetCode Journey Solved LeetCode 142: Linked List Cycle II (Medium) today — a powerful extension of the cycle detection problem. Earlier, we learned how to detect if a cycle exists. Now the challenge is to find the exact node where the cycle begins. 💡 Core Idea (Floyd’s Algorithm Extended): 1) Use slow & fast pointers to detect a cycle 2) Once they meet, reset one pointer to the head 3) Move both pointers one step at a time 4) The point where they meet again = start of the cycle 🤯 Why it works? Because of the mathematical relationship between distances traveled inside the cycle — both pointers align perfectly at the cycle entry point. ⚡ Key Learning Points: • Advanced use of two-pointer technique • Understanding the mathematics behind cycle detection • Solving without modifying the list • Maintaining O(n) time and O(1) space This is not just coding — this is algorithmic thinking at a deeper level. Also, this pattern connects with: -> Find duplicate number (cycle in array) -> Happy Number problem -> Loop detection in graphs ✅ Stronger conceptual clarity ✅ Better problem-solving depth ✅ Confidence with tricky pointer problems From detecting a cycle to pinpointing its start — that’s real progress 🚀 #LeetCode #DSA #Java #LinkedList #TwoPointers #Algorithms #ProblemSolving #CodingJourney #Consistency #100DaysOfCode #InterviewPreparation #DeveloperGrowth #KeepCoding

Day 20 of my #30DayCodeChallenge: Efficiency through Pruning! Today's challenge was Word Search II-a complex puzzle that tests how you manage massive search spaces. The Logic: 1. Trie Integration: I stored the dictionary in a Trie to check prefixes in O(1) time. 2. DFS & Backtracking: Explored the grid cell by cell, but with a twist... 3. Intelligent Pruning: If a path doesn't match a Trie prefix, the search stops immediately. This turns an exponential problem into something much more manageable. Coding isn't just about finding the answer; it's about finding it before your timer runs out! #Java #DataStructures #Backtracking #Trie #Algorithms #CleanCode #150DaysOfCode

🚀 LeetCode Problem || Fancy Sequence (1622) Today I worked on an interesting design problem where we need to maintain a sequence that supports the following operations efficiently: • append(val) → Append a number to the sequence • addAll(inc) → Add a value to every element in the sequence • multAll(m) → Multiply every element by a value • getIndex(idx) → Retrieve the value at a specific index 💡 Optimization Idea Instead of modifying every element, we represent the sequence transformation using a mathematical form: value=a×x+bvalue = a \times x + bvalue=a×x+bWhere: a tracks the cumulative multiplication b tracks the cumulative addition x is the stored base value #LeetCode #Java #Algorithms #DataStructures #ProblemSolving #CodingJourney

🚀 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 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

📌 LeetCode Daily Challenge — Day 27 Problem: 3786. Cyclic Shifts of a Matrix Topic: Array, Matrix, Math, Simulation   📌 Quick Problem Sense: You're given an m × n matrix and an integer k. Every step: even-indexed rows shift left, odd-indexed rows shift right — repeated k times. Return true if the matrix after k steps is identical to the original. The catch? k can be huge — so simulating step by step is a dead end. 🚫   🧠 Approach (Simple Thinking): 🔹 A row of length n always returns to original after n shifts — but it might return sooner! 🔹 The minimum cyclic period of a row = the smallest divisor d of n such that the row is just a block of size d repeated 🔹 Check it: row[j] == row[j % d] for all j — if true, d is the period! 🔹 Direction (left vs right) doesn't affect the period — both complete their cycle in the same number of steps 🔹 The matrix restores to original iff k % period == 0 for every row 🔹 Get all divisors of n once in O(√n), check each row against them — no simulation needed! 🎯   ⏱️ Time Complexity: Divisors of n → O(√n) Per row period check → O(d(n) × n) All rows → O(m × d(n) × n) Blazing fast — no k-step simulation ever needed!   📦 Space Complexity: Just a divisors list → O(√n) Zero extra matrix copies or simulation buffers!   I wrote a full breakdown with dry run, real-life analogy, and step-by-step code walkthrough here 👇 https://lnkd.in/gbqN_Y7a If you solved it using the KMP failure function to find the string period in O(n), I'd love to see that approach, drop it in the comments 💬 See you in the next problem 👋 #LeetCode #DSA #CodingChallenge #Java #ProblemSolving