Rotating a Matrix 90° Clockwise in LeetCode Challenge

🚀 LeetCode Day Problem Solving 🚀 Day-70 📌 Problem: Rotate a box 90° clockwise and simulate gravity on stones 🧱 👉 Grid contains: # → stone * → obstacle . → empty 🧠 Example Insight: 👉 After rotation, stones will fall downward 👉 But obstacles * stay fixed 💡 Key Insight (Important 🔥): ✔ Rotation + Gravity = 2-step thinking 👉 First handle gravity row-wise 👉 Then perform matrix rotation ⚡ Core Idea: Step 1️⃣ Apply Gravity 👉 For each row (left → right): ✔ Stones # fall to the rightmost possible position ✔ Stop when: obstacle * another stone # 👉 Treat each segment between * separately Step 2️⃣ Rotate Matrix 👉 Convert m x n → n x m ✔ New position: 👉 (i, j) → (j, m - 1 - i) 🔥 Why it works? ✔ Gravity in original box = right shift ✔ After rotation → becomes downward fall ⚡ Algorithm: 1️⃣ For each row → simulate gravity (right shift stones) 2️⃣ Create new matrix of size n x m 3️⃣ Rotate using index transformation 📊 Complexity Analysis: ⏱ Time Complexity: O(m × n) 📦 Space Complexity: O(m × n) 🧠 What I Learned: ✔ Matrix rotation technique ✔ Simulation problems (gravity logic) ✔ Handling obstacles in grid 🔥 Day 70 Completed! 2 concepts → 1 clean solution 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency #MilanSahoo 🚀

🚀 LeetCode Day Problem Solving 🚀 Day-68 📌 Problem: Rotate an n x n matrix by 90° clockwise ⚠️ Condition: 👉 Do it in-place (no extra matrix allowed) 🧠 Example: Input: 1 2 3 4 5 6 7 8 9 ✅ Output: 7 4 1 8 5 2 9 6 3 💡 Key Insight (Important 🔥): 👉 Rotation = Transpose + Reverse rows ⚡ Step 1: Transpose Matrix 👉 Convert rows → columns 1 2 3 1 4 7 4 5 6 → 2 5 8 7 8 9 3 6 9 ⚡ Step 2: Reverse Each Row 1 4 7 7 4 1 2 5 8 → 8 5 2 3 6 9 9 6 3 🔥 Algorithm: 1️⃣ Transpose: 👉 swap(matrix[i][j], matrix[j][i]) 2️⃣ Reverse each row ⚡ Why this works? ✔ Transpose flips diagonal ✔ Row reverse completes 90° rotation 📊 Complexity Analysis: ⏱ Time Complexity: O(n²) 📦 Space Complexity: O(1) (in-place) 🧠 What I Learned: ✔ Matrix manipulation tricks ✔ In-place transformations ✔ Transpose concept ✅ Day 68 Completed 🚀 Improving in Matrix Problems 💪 #Leetcode #coding #learning #CodingChallenge #ProblemSolving #DSA #Math #Simulation #Digits #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency #MilanSahoo 🚀

🚀 #LeetCode Day Problem Solving 🚀 Day-50 📌 Problem: Rotate a box 90° clockwise and simulate gravity on stones 🧱 👉 Grid contains: # → stone * → obstacle . → empty 🧠 Example Insight: 👉 After rotation, stones will fall downward 👉 But obstacles * stay fixed 💡 Key Insight (Important 🔥): ✔ Rotation + Gravity = 2-step thinking 👉 First handle gravity row-wise 👉 Then perform matrix rotation ⚡ Core Idea: Step 1️⃣ Apply Gravity 👉 For each row (left → right): ✔ Stones # fall to the rightmost possible position ✔ Stop when: obstacle * another stone # 👉 Treat each segment between * separately Step 2️⃣ Rotate Matrix 👉 Convert m x n → n x m ✔ New position: 👉 (i, j) → (j, m - 1 - i) 🔥 Why it works? ✔ Gravity in original box = right shift ✔ After rotation → becomes downward fall ⚡ Algorithm: 1️⃣ For each row → simulate gravity (right shift stones) 2️⃣ Create new matrix of size n x m 3️⃣ Rotate using index transformation 📊 Complexity Analysis: ⏱ Time Complexity: O(m × n) 📦 Space Complexity: O(m × n) 🧠 What I Learned: ✔ Matrix rotation technique ✔ Simulation problems (gravity logic) ✔ Handling obstacles in grid 🔥 Day 50 Completed! 2 concepts → 1 clean solution 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency

🚀 Day 67 of 100 Days LeetCode Challenge Problem: Rotating the Box Day 67 is a really interesting Matrix Simulation + Gravity Mechanics problem 🔥 💡 Key Insight: The problem has TWO major operations: 1️⃣ Simulate gravity on stones # 2️⃣ Rotate the matrix 90° clockwise The tricky part: 👉 Obstacles * block falling stones 🔍 Core Observation: Instead of rotating first: ✅ Simulate stone movement row by row THEN ✅ Rotate the final box This simplifies the logic significantly 🚀 🔍 Step 1: Simulate Gravity For every row: Process from right → left Maintain an empty position pointer Rules: # → falls to nearest empty spot * → acts as barrier/reset point Example: # . # * . # After gravity: . # # * . # 🔍 Step 2: Rotate the Matrix 90° clockwise rotation formula: rotated[j][m-1-i] = box[i][j] 💡 Why This Works: Gravity only affects rows before rotation Obstacles naturally partition movement regions 🔥 Time Complexity: O(m × n) Efficient simulation + transformation ✅ 🔥 What I Learned Today: Simulation problems become easier when broken into stages Matrix rotation patterns are extremely reusable Obstacle-based movement often needs pointer/reset logic 📈 Challenge Progress: Day 67/100 ✅ Closing in on Day 70 🚀 🔍 Keywords: LeetCode, Matrix Simulation, Grid Problems, Arrays, Two Pointers, Gravity Simulation, Matrix Rotation, DSA Practice, Problem Solving #100DaysOfCode #LeetCode #DSA #CodingChallenge #Matrix #Simulation #GridProblems #Arrays #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep

Day 103 on LeetCode Reveal Cards In Increasing Order 🃏📈✅ Another beautiful problem where simulation + queue intuition worked together perfectly 💯 🔹 Reveal Cards In Increasing Order The goal was to arrange the deck such that: • Reveal top card • Move next top card to bottom • Continue the process And the revealed sequence becomes sorted in increasing order. 🔸 Approach Used in My Solution: • First sort the deck • Use a queue to track indices of positions • Place smallest available card at current front index • Simulate the reveal process by: – Removing front index – Moving next index to back 💡 Instead of simulating cards directly, the smart part was simulating the positions. 🔹 Core Concepts Used • Queue simulation • Sorting • Index mapping Key idea: • Queue stores available positions • Sorted values are assigned according to reveal order 🔹 What Made This Problem Interesting Initially the problem feels confusing because: • Forward simulation becomes messy • Card movement creates unusual ordering But once the queue of indices idea clicks, the whole problem becomes clean and elegant. ⚡ Complexity: • Time Complexity: O(n log n) due to sorting • Space Complexity: O(n) 💡 Key Takeaways: • Sometimes simulating positions is easier than simulating values • Queue problems improve process visualization skills • Breaking a confusing problem into small operations changes everything These implementation and simulation problems are slowly improving: • Thinking ability • Visualization • Handling conditions • Writing complete logic independently Maybe not perfect yet. Maybe not always optimal yet. But definitely progressing step by step. 🔥 Small daily improvements eventually become big changes. #LeetCode #DSA #Algorithms #DataStructures #Queue #Simulation #Sorting #ProblemSolving #Coding #Programming #Cpp #STL #SoftwareEngineering #ComputerScience #CodingPractice #DeveloperLife #TechJourney #CodingDaily #Consistency #100DaysOfCode #BuildInPublic #AlgorithmPractice #CodingSkills #Developers #TechCommunity #SoftwareDeveloper #EngineeringJourney

🚀 #LeetCode Day Problem Solving 🚀 Day-48 📌 Problem: Rotate an n x n matrix by 90° clockwise ⚠️ Condition: 👉 Do it in-place (no extra matrix allowed) 🧠 Example: Input: 1 2 3 4 5 6 7 8 9 ✅ Output: 7 4 1 8 5 2 9 6 3 💡 Key Insight (Important 🔥): 👉 Rotation = Transpose + Reverse rows ⚡ Step 1: Transpose Matrix 👉 Convert rows → columns 1 2 3 1 4 7 4 5 6 → 2 5 8 7 8 9 3 6 9 ⚡ Step 2: Reverse Each Row 1 4 7 7 4 1 2 5 8 → 8 5 2 3 6 9 9 6 3 🔥 Algorithm: 1️⃣ Transpose: 👉 swap(matrix[i][j], matrix[j][i]) 2️⃣ Reverse each row ⚡ Why this works? ✔ Transpose flips diagonal ✔ Row reverse completes 90° rotation 📊 Complexity Analysis: ⏱ Time Complexity: O(n²) 📦 Space Complexity: O(1) (in-place) 🧠 What I Learned: ✔ Matrix manipulation tricks ✔ In-place transformations ✔ Transpose concept ✅ Day 48 Completed 🚀 Improving in Matrix Problems 💪 #Leetcode #coding #learning #CodingChallenge #ProblemSolving #DSA #Math #Simulation #Digits #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency

Day 94 on LeetCode — Daily Temperatures 🌡️📈✅ Another solid day working on stack + pattern recognition problems 💯 🔹 Daily Temperatures A classic problem that introduces the concept of monotonic stack. 🔸 Approach Used in My Solution: • Traverse the array of temperatures • Use a stack to store indices of unresolved days • For each current temperature:   – Compare with stack top   – If current is higher → resolve previous days   – Calculate distance and update result • Push current index to stack 💡 This ensures we efficiently find the next greater temperature for each day. 🔹 Core Concept Used • Monotonic decreasing stack • Next greater element pattern Key idea: • Store indices instead of values • Resolve elements only when a greater value appears • Avoid unnecessary nested loops ⚡ Complexity: • Time Complexity: O(n) • Space Complexity: O(n) 💡 Key Takeaways: • Monotonic stack is powerful for “next greater element” problems • Using indices helps calculate distances easily • Converts brute force O(n²) → optimized O(n) solution 🔥 Moving from brute force thinking → to optimized pattern-based solutions. #LeetCode #DSA #Algorithms #DataStructures #Stack #MonotonicStack #Arrays #ProblemSolving #Coding #Programming #Cpp #STL #SoftwareEngineering #ComputerScience #CodingPractice #DeveloperLife #TechJourney #CodingDaily #Consistency #100DaysOfCode #BuildInPublic #AlgorithmPractice #CodingSkills #Developers #TechCommunity #SoftwareDeveloper #EngineeringJourney

🚀 LeetCode Day Problem Solving 🚀 Day-83 📌 Problem: Find the smallest common element present in both sorted arrays 🔥 If no common element exists → return -1 🧠 Example: Input: nums1 = [1,2,3] nums2 = [2,4] ✅ Output: 2 💡 Why? Common elements are: [[2]] Smallest common element = 2 ✅ 💡 Key Insight (Very Important 🔥): 👉 Both arrays are already SORTED ⚡ So instead of brute force checking every pair ❌ We can use: #️⃣ Two Pointer Technique 🚀 ⚡ Core Idea: Maintain pointers: ✔ i for nums1 ✔ j for nums2 Case 1️⃣: If: [nums1[i] = nums2[j]] ✅ Found smallest common element immediately because arrays are sorted 🔥 Case 2️⃣: If: [nums1[i] < nums2[j]] 👉 Move i forward because smaller value cannot match future larger values Case 3️⃣: If: [nums1[i] > nums2[j]] 👉 Move j forward 🧠 Why This Works? Since arrays are sorted: ✔ Once a smaller value is skipped, it can never become useful later So we eliminate unnecessary comparisons efficiently ⚡ ⚡ Algorithm: 1️⃣ Initialize: ✔ i = 0 ✔ j = 0 2️⃣ Traverse both arrays: ✔ If equal → return value ✔ Else move pointer with smaller value 3️⃣ If traversal ends → return -1 🔥 Example Walkthrough: nums1 = [1,2,3,6] nums2 = [2,3,4,5] Step 1: Compare: [1 \text{ and } 2] 1 < 2 👉 Move first pointer Step 2: Compare: [2 \text{ and } 2] ✅ Match found Answer = 2 📊 Complexity Analysis: ⏱ Time Complexity: [O(n+m)] where n and m are array sizes 📦 Space Complexity: O(1) 🧠 What I Learned: ✔ Two Pointer Technique ✔ Efficient traversal of sorted arrays ✔ Avoiding brute force comparisons 🔥 Day 83 Completed! Sorted arrays often unlock elegant linear-time solutions 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency #MilanSahoo 🚀

🚀 Day 63 of 100 Days LeetCode Challenge Problem: Rotated Digits Day 63 is a fun Digit Validation + Simulation problem 🔥 💡 Key Insight: A number is considered good if: ✅ Every digit remains valid after rotation ✅ The rotated number becomes DIFFERENT from the original 🔍 Digit Rotation Rules: Valid rotations: 0 → 0 1 → 1 8 → 8 2 ↔ 5 6 ↔ 9 Invalid digits: 3, 4, 7 🔍 Core Observation: A number is GOOD if: It contains at least one changing digit: 2, 5, 6, 9 AND It contains no invalid digits 🔍 Core Approach: 1️⃣ Iterate from: 1 → n 2️⃣ Check Each Digit For every digit: If invalid → reject number ❌ If digit changes after rotation → mark as different ✅ 3️⃣ Count Valid Good Numbers Only count numbers that: Stay valid Become different after rotation 💡 Why This Works: Digits 0,1,8 alone don’t change the number At least one transformed digit is required 🔥 Optimization Thought: This can also be solved using: ⚡ Digit DP for larger constraints But direct simulation works efficiently here ✅ 🔥 What I Learned Today: Digit classification simplifies many number problems Simulation becomes powerful with clean rules Understanding conditions carefully is more important than coding complexity 📈 Challenge Progress: Day 63/100 ✅ Still rotating through challenges consistently 🚀 LeetCode, Digit Manipulation, Simulation, Math, Number Theory, Arrays, DSA Practice, Problem Solving #100DaysOfCode #LeetCode #DSA #CodingChallenge #DigitManipulation #Math #Simulation #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep

💡 LeetCode Daily — Problem Solved: Rotate Function (Medium). Today’s problem looked mathematical at first glance… but the real trick was pattern recognition + optimization. At face value, rotating the array and recalculating the function every time feels natural. But that leads straight into an O(n²) trap — not scalable. So I stepped back and asked: 👉 What actually changes when we rotate the array? 🔍 Key Insight (Simple Visualization). Instead of recomputing everything: Each rotation shifts elements, but the total sum stays constant. Only the weight contribution of elements changes. That leads to a powerful relation: Next rotation value can be derived from the previous one using a formula — no need to rebuild from scratch. ⚙️ My Approach Compute initial value F(0) using a loop. Use accumulate() (STL) to get total sum efficiently. Derive next values using a rolling formula. Track maximum along the way. This reduces complexity to O(n) — clean and efficient. 📊 Consistency Snapshot Total active days on LeetCode: 75 Max streak: 68 days 🧠 What I Took Away This problem wasn’t about rotations. It was about: Breaking brute-force thinking. Finding relationships between consecutive states. Trusting math over simulation. Sometimes, optimization isn’t about writing better code — it’s about seeing the problem differently. Still learning. Still refining how I think. #LeetCode #DSA #ProblemSolving #CPP #STL #DailyCoding #Consistency