Generate n x n Matrix in Spiral Order with LeetCode #59

🚀LeetCode Daily Challenge 🧩 Problem: Number of Ways to Paint N × 3 Grid You are given an integer n, representing the number of rows in a 3-column grid. You need to paint the grid using 3 different colors such that: ✔ No two adjacent cells in the same row have the same color ✔ No two adjacent cells in the same column have the same color 🎯 Goal: Return the total number of valid ways to paint the grid, modulo 10⁹ + 7. 🧠 Key Intuition Each row can be painted in only 12 valid color patterns that satisfy the row constraint. The problem becomes a row-by-row transition problem: A pattern in the current row is valid only if no column color matches the pattern in the previous row. This forms a classic Dynamic Programming with State Transitions problem: State → (row, previousPattern) Transition → Try all compatible patterns for the next row Memoization avoids recomputing overlapping subproblems, making the solution efficient. 🛠 Approach Predefine all 12 valid color patterns for a single row. Use DP + recursion: dp[row][prevPattern] stores the number of ways to paint from row onward. For each row: Try all patterns different from the previous one. Ensure no column color clashes with the previous row. Base case: When all rows are painted → return 1. Sum all valid configurations starting from row 0. 📈 Complexity Analysis Time Complexity: O(n × 12 × 12) ≈ O(n) Space Complexity: O(n × 12) 🔗 Problem https://lnkd.in/dk9hb-m7 💻 Solution https://lnkd.in/dA6cvig8 #LeetCode #Cpp #DSA #DynamicProgramming #Memoization #Recursion #Grids #Combinatorics #ProblemSolving #CodingChallenge #LeetCodeDailyChallenge #Algorithms

Headline: Keep It Simple: The Power of Sorting! 📉🚀 LeetCode #1464 – Maximum Product of Two Elements #DSAwithEdSlash | edSlash | LeetCode | Day 8/100 Week 2 is here! 🚀 For today’s problem, I focused on writing clean, readable code using C++ STL (Standard Template Library). The goal was to find the maximum product of two elements, which logically means finding the two largest numbers. 💡 Technical Approach: Instead of managing complex conditional logic, I used a Sorting Approach: Sort: Utilized std::sort() to arrange the array in ascending order. Access: Immediately accessed the last two elements (nums[n-1] and nums[n-2]) since they are guaranteed to be the maximums. Compute: Applied the required formula (max-1) * (second_max-1) and returned the result. 📊 Complexity Analysis: Time: O(N log N) — Dominated by the sorting algorithm, but extremely efficient to write. Space: O(1) — Solved in-place without using extra arrays. 🧠 Key Takeaways: Readability: Sometimes the best code is the one that is easiest to read and debug. STL Power: Mastering built-in functions like sort() is a superpower in C++. 👇 Discussion: Do you prefer the raw speed of a linear scan O(N) or the readability of sorting O(N log N) for Easy problems? Let me know! #LeetCode #DataStructures #Algorithms #Cpp #ProblemSolving #100DaysOfCode #CodingLife #edSlash #CareerGrowth

🚀LeetCode Daily Challenge 🧩 Problem: Count Negative Numbers in a Sorted Matrix You are given an m × n matrix where: Each row is sorted in non-increasing order Each column is also sorted in non-increasing order 🎯 Goal: Count the total number of negative numbers in the matrix efficiently. 🧠 Key Intuition A brute-force approach would scan every element → O(m·n), which is not optimal. Because the matrix is sorted both row-wise and column-wise, we can do much better: Start from the top-right corner If the current element is negative: ✔ All elements below it in the same column are also negative If the current element is non-negative: ✔ Move down to the next row This allows us to eliminate entire sections of the matrix at once. 🛠 Approach (Two-Pointer Technique) Initialize: row = 0 (top row) col = n - 1 (last column) While row < m and col >= 0: If grid[row][col] < 0: Add (m - row) to the count Move left (col--) Else: Move down (row++) Return the total count This strategy ensures we never revisit any cell. 📈 Complexity Analysis Time Complexity: O(m + n) Space Complexity: O(1) 🔗 Problem https://lnkd.in/dY9jsTkg 💻 Solution https://lnkd.in/daQrMf4A #LeetCode #Cpp #DSA #TwoPointers #Matrices #Arrays #ProblemSolving #CodingChallenge #LeetCodeDailyChallenge #Algorithms

𝗗𝗮𝘆 𝟮𝟰 𝗼𝗳 𝘁𝗵𝗲 #𝟱𝟬𝗗𝗮𝘆𝘀𝗢𝗳𝗗𝗦𝗔𝗖𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲:𝗩𝗮𝗹𝗶𝗱 𝗔𝗻𝗮𝗴𝗿𝗮𝗺 𝘿𝙖𝙮 𝟮𝟰:𝗩𝗮𝗹𝗶𝗱 𝗔𝗻𝗮𝗴𝗿𝗮𝗺 Returns true if strings s and t have the exact same characters with the exact same frequencies, otherwise return false. 𝗞𝗲𝘆 𝗜𝗻𝘀𝗶𝗴𝗵𝘁𝘀: 𝟭.𝗧𝗵𝗲 "𝗡𝗲𝘁 𝗭𝗲𝗿𝗼" 𝗟𝗼𝗴𝗶𝗰: By incrementing for s and decrementing for t, you are essentially looking for a perfect balance. An anagram is only confirmed if every "addition" is perfectly cancelled out by a "subtraction." 𝟮.𝗘𝗮𝗿𝗹𝘆 𝗘𝘅𝗶𝘁 𝗢𝗽𝘁𝗶𝗺𝗶𝘇𝗮𝘁𝗶𝗼𝗻: The if (arr[t[j]-'a'] < 0) check is a smart move. It allows the program to stop immediately if it detects that string t has a character that string s didn't provide enough of, saving unnecessary processing time. 𝟯.𝗔𝗦𝗖𝗜𝗜 𝗠𝗮𝗽𝗽𝗶𝗻𝗴: The s[i] - 'a' trick is the most efficient way to map characters to indices in C++, as it avoids the memory overhead of a Hash Map while maintaining O(1) access time. 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆 𝗧𝗶𝗺𝗲 O(n) 𝗦𝗽𝗮𝗰𝗲 O(1) #DSA #DataStructures #Algorithms #LeetCode  #CodingChallenge #SoftwareEngineering #TechCareer  #PowerSet

LeetCode #1 – Two Sum #DSAwithEdSlash | edSlash | LeetCode | Day 6/100 Revisited the most famous problem on LeetCode today. While there are many ways to solve "Two Sum," I decided to implement the Brute Force approach first to fully understand the underlying logic before optimizing. 💡 Technical Approach: I used a straightforward nested loop strategy to check every possible pair: Outer Loop: Picks the first number (nums[i]). Inner Loop: Scans all subsequent numbers (nums[j]) to see if they add up to the target. Result: If a match is found, return the indices immediately. 📊 Complexity Analysis: Time: $O(n^2)$ — Checks every pair. Great for small inputs! Space: $O(1)$ — Zero extra memory used. Unlike the Hash Map approach (which takes O(N) space), this solution is extremely memory efficient. 🧠 Key Takeaways: First Principles: Sometimes the most intuitive solution is the best place to start. Space-Time Tradeoff: This solution saves memory at the cost of speed. It’s a perfect example of the classic trade-off in computer science. 👇 Discussion: In an interview, I'd optimize this next using a Hash Map, but I love how clean the logic is here. Which do you prioritize in your first pass: Code readability or raw execution speed? #LeetCode #DataStructures #Algorithms #Cpp #ProblemSolving #100DaysOfCode #CodingLife #edSlash #CareerGrowth

Day 14 Problem Statement: You’re given multiple axis-aligned squares on a 2D plane. Squares may overlap. The task is to find the minimum y-coordinate of a horizontal line such that the total area covered by the squares above the line equals the total area covered below the line, counting overlapping regions only once. Approach: Since overlapping areas must not be double-counted, we compute the union area of all squares. Using a sweep line along the y-axis, each square generates an entry and exit event. Between consecutive y-events, the active squares remain the same, so we merge their x-intervals to get the total covered width. The area of each horizontal strip is simply width × height. After calculating the total union area, we scan these strips again to find the exact y-coordinate where the cumulative area reaches half of the total, giving the required answer with precision. #LeetCode #Algorithms #DataStructures #SweepLine #Geometry #ProblemSolving #Java#DailyCodingChallenge #SoftwareEngineering #DSA #CompetitiveProgramming #LearningJourney

Day 22 – Linked List Transformations and Efficient Sorting 🌀🔗 Today’s session focused on advanced linked list manipulations—from rotational transformations to sorting algorithms tailored for linear structures. 🌀 Rotate List Transformed a linked list by rotating nodes right k places. Used two-pointer traversal to find length and compute effective rotation, then rewired pointers in O(n) time. A clever exercise in modular arithmetic and pointer management that avoided unnecessary node copying while preserving structure. 🔗 Sort List Implemented merge sort for linked lists—achieving O(n log n) time with O(log n) recursion stack space. Employed fast-slow pointer technique to split lists, recursively sorted halves, and merged them in sorted order. Demonstrated how divide-and-conquer adapts elegantly to linear structures without random access. Both problems reinforced how pointer manipulation and algorithmic adaptation turn complex list operations into systematic, efficient processes. From rotation to sorting, the journey through linked structures continues—one node, one connection at a time. 💻 #LeetCode #Coding #Algorithms #DataStructures #LinkedList #TwoPointers #MergeSort #ProblemSolving #SoftwareEngineering #Programming #100DaysOfCode

🧮 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐃𝐚𝐢𝐥𝐲 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞 #𝟔𝟏 – 𝐑𝐨𝐭𝐚𝐭𝐞 𝐋𝐢𝐬𝐭 ⏺ 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 (𝐟𝐫𝐨𝐦 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞): Rotate a linked list to the right by k positions. 📖 𝐄𝐱𝐚𝐦𝐩𝐥𝐞: Input: [1,2,3,4,5], k = 2 Output: [4,5,1,2,3] 💡𝐄𝐚𝐬𝐲 𝐰𝐚𝐲 𝐭𝐨 𝐮𝐧𝐝𝐞𝐫𝐬𝐭𝐚𝐧𝐝: Think of the list as a circle 🔁. Rotate it, then break the circle at the correct spot. ✨ 𝐊𝐞𝐲 𝐈𝐝𝐞𝐚: 1. Find the list length 2. Use k % length 3. Connect tail → head (circular list) 4. Break at the new tail 📝 𝐂# 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:  Full code with comments is shown in the attached image 👇 📌 Tomorrow: 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 #𝟔𝟐 – 𝐔𝐧𝐢𝐪𝐮𝐞 𝐏𝐚𝐭𝐡𝐬 (C#). Stay tuned! #LeetCode #CSharp #CodingChallenge #LinkedList #ProblemSolving #Algorithms

🚀 LeetCode Daily Challenge 🧩 Problem: Magic Squares in Grid You are given an m × n grid of integers. A 3 × 3 magic square is defined as a grid where: All numbers are distinct Every number is in the range [1, 9] The sum of each row, each column, and both diagonals is the same 🎯 Goal: Count how many 3 × 3 magic squares exist inside the given grid. 🧠 Key Intuition A brute-force check of every 3 × 3 subgrid is acceptable because: Each subgrid is very small (fixed size) The number of possible positions is limited For each 3 × 3 window, we must verify: All values are between 1 and 9 All 9 values are distinct All row sums, column sums, and diagonal sums are equal If any of these conditions fail, the subgrid cannot be a magic square. 🛠 Approach Iterate over all possible 3 × 3 subgrids For each subgrid: Use a map to ensure distinct values Compute: Row sums Column sums Both diagonal sums Store row and column sums in sets to verify equality If: All sums are equal Diagonals match the same sum Values are valid and distinct → Count it as a magic square Return the total count 📈 Complexity Analysis Time Complexity: O(m × n) Space Complexity: O(1) 🔗 Problem https://lnkd.in/d39x_RRj 💻 Solution https://lnkd.in/dCr9MFA3 #LeetCode #Cpp #DSA #Matrices #BruteForce #Implementation #ProblemSolving #CodingChallenge #LeetCodeDailyChallenge #Algorithms

LeetCode #977 – Squares of a Sorted Array #DSAwithEdSlash | edSlash | LeetCode | Day 12/100 Day 12/100! 🚀 Today's challenge was to square a sorted array (including negative numbers). Input: [-4, -1, 0, 3, 10] If you just square them, you get [16, 1, 0, 9, 100]. The order is completely broken because large negatives become large positives! 💡 Technical Approach (Brute Force): While there are complex pointer-based solutions, I decided to stick to the most intuitive logic first: Iterate: Walk through the array and square every number. Sort: Use the Standard Template Library sort() function to fix the order. Sometimes, the best code is the code you can write quickly and understand easily. 📊 Complexity Analysis: Time: O(Nlog N) — The sorting step dominates the runtime. Space: O(1) — Done in-place (ignoring the internal stack used by sort). 🧠 Key Takeaways: Get it working first: Premature optimization is the root of all evil. I prioritized a working solution over a perfect one. STL Power: C++ sort() is incredibly optimized and handles the heavy lifting for us. Next Steps: I know an O(N) Two-Pointer solution exists—that is my next challenge to implement! 👇 Discussion: In a whiteboard interview, do you write the Brute Force solution first to show you understand the basics, or do you jump straight to the optimal solution? #LeetCode #DataStructures #Algorithms #Cpp #ProblemSolving #100DaysOfCode #CodingLife #edSlash #CareerGrowth