"Adapting LCS to Minimum ASCII Sum Problem"

Day 9 of #30DaysOfCode 📐 From Brute Force to Ultra-Optimized: Maximum Rectangle Area Problem Just cracked an interesting geometric problem that taught me the power of micro-optimizations in competitive programming! The Challenge: Find the largest axis-aligned rectangle from a set of points where no other points lie inside or on the borders. The Journey: Started with O(n⁴) brute force approach Optimized to O(n³) using diagonal-pair strategy Applied competitive programming micro-optimizations Key Optimizations That Made the Difference: - unordered_set with reserve() → 3-5x faster lookups vs set - Bit packing coordinates → (x,y) into single 64-bit integer for faster hashing - Const references → Reduced array access overhead - Early break/continue → Eliminated wasted iterations - Direct comparison → Avoided max() function overhead The Results: ⚡ 6-8x performance improvement 💾 Same O(n) space complexity 🎯 Runs in ~100-200 operations for n ≤ 10 Key Takeaway: For small constraints, the right data structure + micro-optimizations matter more than asymptotic complexity. Sometimes it's not about changing the algorithm, but making every operation count! #CompetitiveProgramming #CPP #AlgorithmOptimization #DSA #ProblemSolving #SoftwareEngineering #CodingInterview #PerformanceOptimization Educative #educative

🚀 #Day57 of #100DaysOfLeetCodeHard 📘 Problem: [LeetCode 2547 – Minimum Cost to Split an Array] My Submission:https://lnkd.in/gB2aQ9TS This one was a really interesting DP + preprocessing problem that tested both efficiency and clarity of thought. 💡 Approach: The key idea was to precompute the cost of every possible subarray [i...j], where cost represents the number of elements that appear more than once (the trimmed part). Once precomputed, the problem transforms into a simple Dynamic Programming formulation: dp[i] = min( k + cost[i][j] + dp[j+1] ) for all j ≥ i This way, we can efficiently find the optimal points to split the array, minimizing the total cost. Given the constraints (n ≤ 1000), an O(n²) preprocessing and DP approach works perfectly. 🧮 Complexity: Time: O(n²) Space: O(n²) A neat problem that beautifully combines frequency tracking, preprocessing, and classic DP — a great example of how strong intuition can simplify what initially looks intimidating. #100DaysOfLeetCodeHard #Day57 #LeetCode #DynamicProgramming #ProblemSolving #DSA #CompetitiveProgramming #Precomputation

💡 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐃𝐚𝐢𝐥𝐲 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞 -𝟒𝟕𝟒. 𝐎𝐧𝐞𝐬 𝐚𝐧𝐝 𝐙𝐞𝐫𝐨𝐞𝐬 We’re given a list of binary strings and two integers m and n. The task is to find the largest subset of strings such that the total number of 0s ≤ m and 1s ≤ n. 🧩 My Approach: I solved it using Top-Down DP (Memoization) : • For each string, I had two choices - include it (if within limits) or skip it. • Used a 3D memo[i][m][n] to store results for subproblems (index i, remaining 0s, remaining 1s). • Compared both choices and took the maximum. • This way, we efficiently explore all possibilities while avoiding redundant recalculations. Learning: Dynamic programming shines when you break problems into overlapping subproblems and optimize with memoization. #LeetCode #DSA #DynamicProgramming #ProblemSolving #Consistency

🚀 Day 132 of #160DaysOfGFG – Word Break 📌 Problem: Given a string s and a dictionary of valid words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. 🧠 Solution Approach: ✅ Use Dynamic Programming ✅ Define dp[i] = true if the substring s[0...i] can be segmented using dictionary words. ✅ For each i, check every j < i → if dp[j] is true and s[j...i] is in the dictionary, set dp[i] = true. 🕒 Time Complexity: O(n²) 💾 Space Complexity: O(n) 💡 Learning: A perfect example of how subproblems and prefix checks combine elegantly in DP. Once you understand this, you’ll find similar logic in Palindrome Partitioning and Sentence Formation problems. 🔖 #Day132 #160DaysOfGFG #DynamicProgramming #WordBreak #ProblemSolving #GeeksForGeeks #DSA #CodingChallenge #InterviewPrep #DailyDSA

✨ Day 85 of #100DaysOfCode ✨ Today I worked on the Word Search problem using C++ and Backtracking. In this problem, we’re given a 2D grid of characters and need to check if a given word can be formed by sequentially adjacent letters (up, down, left, or right). 🔹 My approach: Start from every cell that matches the first character of the word. Use recursion to explore all four directions — top, right, bottom, and left. Temporarily mark a cell as visited (using ‘!’) to prevent revisiting it during the same search path. If we reach the end of the word, we return true. After exploring, we backtrack by restoring the cell’s original value. This solution is a great example of depth-first search (DFS) combined with backtracking — a powerful pattern for solving grid-based problems. 💡 Key Concepts Used: Recursion Backtracking 2D Array Traversal Boundary Condition Handling #100DaysOfCode #DSA #CPlusPlus #Backtracking #Recursion #ProblemSolving #CodeNewbie #CodingJourney

🔗 Day 63 of #100DaysOfCode 🔗 🔹 Problem: Valid Parentheses – LeetCode ✨ Approach: Implemented a stack-based validation to ensure every opening bracket has a matching closing one in correct order. Each character is checked systematically — pushing opens and popping closes — making it both clean and efficient! ⚡ 📊 Complexity Analysis: Time Complexity: O(n) — single traversal of the string Space Complexity: O(n) — stack usage for unmatched brackets ✅ Runtime: 2 ms (Beats 97.47%) ✅ Memory: 41.96 MB 🔑 Key Insight: Stacks are the unsung heroes of structured logic — from parentheses validation to syntax parsing, they keep the balance just right. 🧠 #LeetCode #100DaysOfCode #ProblemSolving #DSA #Stack #AlgorithmDesign #CodeJourney #ProgrammingChallenge #LogicBuilding #Efficiency #CodingDaily

Day 25 of 100 Days of DSA: Matrix Chain Multiplication & Merge K Sorted Linked Lists Today I solved two important problems that helped me understand Dynamic Programming and Heaps more clearly. 🔹 1️⃣ Matrix Chain Multiplication (MCM) We are given an array arr[] representing the dimensions of matrices such that matrix i has dimensions arr[i-1] × arr[i]. Since matrix multiplication is associative, the order of multiplication affects the total cost. Idea: 👉 Try every possible partition k between i and j 👉 Calculate the cost = cost(left) + cost(right) + (arr[i-1] × arr[k] × arr[j]) 👉 Take the minimum among all partitions I used a bottom-up DP approach (gap method) where dp[i][j] stores the minimum cost to multiply matrices from i to j. ⏱ Time Complexity: O(n³) 🔹 2️⃣ Merge K Sorted Linked Lists To merge multiple sorted linked lists efficiently, I used a Min Heap (Priority Queue). Steps: Push the head of each list into the heap. Repeatedly extract the smallest node and insert its next node into the heap. Continue until all nodes are processed. ⏱ Time Complexity: O(N log K) where N is the total number of nodes and K is the number of lists. 🔑 Takeaway: Dynamic Programming helps in breaking complex problems into smaller parts, while Heaps make optimization easier and faster. Thanks to Ravi Raushan and the InterviewEasy DSA Sheet (https://intervieweasy.in) for keeping this 100-day journey well-structured! 🙌 #100DaysOfDSA #DynamicProgramming #Heaps #MatrixChainMultiplication #ProblemSolving #LearningJourney #InterviewPreparation #DSA

🎯 LeetCode Daily – Combination Sum IV (Day 125) Today’s problem was a powerful exploration of Dynamic Programming with sequence-based combinations - Combination Sum IV. ✅ My Approach: 🔹 Problem – Combination Sum IV: • The goal was to find the number of possible ordered combinations that sum up to a given target using elements from an array. • Implemented Memoization (Top-Down DP) to recursively explore all ways to reach the target, caching results to prevent recomputation. • Then optimized with Tabulation (Bottom-Up DP), where each state dp[i] represents the number of combinations that sum to i. • For every value from 0 to target, iterated through all numbers and accumulated valid combinations. 🧠 Key Insight: Unlike the traditional Subset Sum or Coin Change problems, this problem treats order as significant, making it a classic case of permutation-based DP. The key is understanding how smaller sub-targets build up to form the total - a perfect demonstration of the power of recursive relationships. 📊 Time Complexity: O(N × Target) 📦 Space Complexity: O(Target) #LeetCode #DSA #DynamicProgramming #Combinatorics #ProblemSolving #DailyCoding #LearningInPublic #Day125

🚀 Day 130 of #160DaysOfGFG – Matrix Chain Multiplication 📌 Problem: Given dimensions of matrices, the goal is to find the minimum number of scalar multiplications needed to multiply them all. The order of multiplication matters — and that’s what makes this problem tricky! 📝 Solution Approach: ✅ Use Dynamic Programming with a bottom-up approach. ✅ Define dp[i][j] as the minimum cost to multiply matrices from i to j. ✅ Recurrence: dp[i][j] = min(dp[i][k] + dp[k+1][j] + arr[i-1]*arr[k]*arr[j]) for all k between i and j-1. ✅ Iterate by chain length and build the solution progressively. ⏱ Time Complexity: O(n³) 💾 Space Complexity: O(n²) 💡 Learning: Matrix Chain Multiplication is the foundation of many DP problems like Boolean Parenthesization, Optimal BST, and Palindrome Partitioning. It’s a masterclass in breaking a problem into optimal subproblems. 🔖 #Day130 #160DaysOfGFG #DynamicProgramming #MatrixChainMultiplication #DP #ProblemSolving #GeeksForGeeks #DSA #CodingChallenge #InterviewPrep #DailyDSA