Minimize Grid Elements with Integer Operations

Day 39 #SDE Problems on mathematical optimization and dynamic programming on strings. Solved: • Count Good Numbers • Word Break Key Learning: • “Count Good Numbers” uses fast exponentiation (binary exponentiation) to efficiently compute large powers under modulo. • “Word Break” is a classic DP problem, where we check if a string can be segmented using a dictionary — reinforcing recursion + memoization patterns. #LeetCode #DSA #DynamicProgramming #Recursion #Algorithms #Java #SoftwareEngineering

The High Cost of the Wrong Initial Choice I have seen complex computation logic for over a million records take hours to run simply because of inefficient, row-by-row processing that is offered by database stored procedures and traditional java/python code. It is slow, impossible to scale, and eventually kills your product's edge in the market. By moving to vectorized logic with tools like NumPy or Polars, you can turn those hours of computation into milliseconds. This is not just about using newer tools, it is about leveraging modern execution like #SIMD (Single Instruction Multiple Data) and #multithreading to gain a massive competitive advantage. If your backend is computation heavy and you're still stuck in legacy loops, you are leaving performance, market edge and money on the table. References: https://lnkd.in/g8j3A5Bn https://lnkd.in/gaPnFUQm https://lnkd.in/gWt9WHnp #DataEngineering #SystemDesign #NumPy #PerformanceOptimization #Scalability #PythonProgramming #TechStrategy #Vectorization #TechToolChoice

🔥 DSA Challenge – Day 131/360 🚀 📌 Topic: Backtracking / Recursion 🧩 Problem: N-Queens Problem Statement: Place N queens on an N×N chessboard such that no two queens attack each other. 🔍 Example: Input: n = 4 Output: [[".Q..","...Q","Q...","..Q."], ["..Q.","Q...","...Q",".Q.."]] 💡 Approach: Optimized Backtracking (Hashing) 1️⃣ Step 1 – Try placing a queen column by column 2️⃣ Step 2 – Use arrays to check if row & diagonals are safe in O(1) 3️⃣ Step 3 – Place queen → recurse → backtrack if needed ⏱ Complexity: Time: O(N!) Space: O(N) + recursion stack 📚 Key Learning: Using hashing for rows & diagonals avoids repeated checks and makes backtracking much faster ⚡ #DSA #Java #Coding #InterviewPrep #ProblemSolving #TechJourney #131DaysOfCode #LeetCode

🔥 DSA Challenge – Day 120/360 🚀 📌 Topic: Math + Recursion 🧩 Problem: Count Good Numbers Problem Statement: Count total valid numbers of length n where even indices have 5 choices and odd indices have 4 choices. 🔍 Example: Input: n = 4 Output: 400 💡 Approach: Optimized (Binary Exponentiation) 1️⃣ Step 1 – Count even positions → use (n + 1) / 2 2️⃣ Step 2 – Count odd positions → use n / 2 3️⃣ Step 3 – Compute power using fast exponentiation and multiply ✔ Use formula: 5^(ceil(n/2)) × 4^(floor(n/2)) ✔ Apply modulo to handle large numbers ✔ Use recursion to reduce time complexity ⏱ Complexity: Time: O(log n) Space: O(log n) 📚 Key Learning: Binary Exponentiation helps reduce power calculation from O(n) to O(log n), which is very useful in large constraints problems. #DSA #Java #Coding #InterviewPrep #ProblemSolving #TechJourney #360DaysOfCode #LeetCode

🚀 Day 86 – DSA Journey | Path Sum in Binary Tree Continuing my daily DSA practice, today I worked on a problem that strengthened my understanding of recursion and root-to-leaf traversal. 📌 Problem Practiced: Path Sum (LeetCode 112) 🔍 Problem Idea: Determine if there exists a root-to-leaf path in a binary tree such that the sum of node values equals a given target. 💡 Key Insight: Instead of tracking the full path, we can reduce the problem by subtracting the current node’s value from the target as we traverse down the tree. 📌 Approach Used: • If the node is null → return false • Check if it is a leaf node – If yes, compare node value with remaining target • Subtract current node value from target • Recursively check left and right subtrees • If any path matches → return true 📌 Concepts Strengthened: • Tree traversal • Recursion • Root-to-leaf path logic • Problem reduction technique ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(h) (recursion stack) 🔥 Today’s takeaway: Breaking down a problem step by step (reducing the target at each node) makes recursion much more intuitive. On to Day 87! 🚀 #Day86 #DSAJourney #LeetCode #BinaryTree #Recursion #Java #ProblemSolving #Coding #LearningInPublic #Consistency

Day 105: Circular Arrays & Shortest Paths 🔄 Problem 2515: Shortest Distance to Target String in a Circular Array Today’s challenge involved finding the minimum steps to reach a target string in a circular array, allowing movement in both directions. The Strategy: • Bidirectional Search: Since the array is circular, the distance can be calculated in two ways: moving forward or moving backward. • Modular Arithmetic: I used (dist + n) % n to handle the wrap-around logic seamlessly, ensuring the index stays within bounds regardless of the direction. • Optimization: By iterating once through the array and comparing the distances for every occurrence of the target, I maintained an O(N) time complexity. Sometimes the most elegant way to handle a "circular" problem is simply embracing the symmetry of the path. 🚀 #LeetCode #Java #Algorithms #ProblemSolving #DailyCode

🔥 Day 83 of my LeetCode Journey 🔥 📘 Problem: 367. Valid Perfect Square 🎯 Difficulty: Easy 🔹 Problem Statement: Given a positive integer num, return true if num is a perfect square, otherwise return false. A perfect square is an integer that is the square of another integer. You must not use any built-in library function like sqrt. 🔹 Approach Used: Apply Binary Search on range 0 to num Find mid and compute mid * mid If equal to num → return true If mid * mid is greater → search left half If smaller → search right half Repeat until condition satisfies or range ends 🔹 Key Concepts: Binary Search Handling large numbers (use long to avoid overflow) Mathematical reasoning Efficient searching 🔹 Learning: This problem shows how binary search can be applied beyond arrays to mathematical problems. It also emphasizes handling overflow carefully when dealing with large 📌 Think beyond arrays — binary search works on answer space too 🚀 #LeetCode #Day83 #Java #BinarySearch #Math #DSA #ProblemSolving

Solved Binary Search problem today and strengthened my understanding of one of the most fundamental algorithms in computer science. Binary Search works on sorted arrays and reduces the search space by half in every step. This makes it highly efficient with a time complexity of O(log n), which is much faster than linear search for large datasets. Key takeaways from today: • Importance of sorted data for applying Binary Search • How to correctly calculate mid to avoid overflow • Understanding left and right pointer movement • Writing clean and optimized code in Java Practicing problems like this really builds strong problem-solving skills and improves logical thinking. Consistency is the key. One problem at a time. #DataStructures #Algorithms #BinarySearch #Java #CodingPractice #LeetCode #ProblemSolving #TechJourney

🚀 LeetCode Challenge 5/50 🔍 Problem: Missing Number Today’s problem was a great example of how mathematical optimization can simplify logic and improve efficiency. 💡 Approach: Instead of using extra space or sorting, I applied the Mathematical Formula method: Calculated the expected sum of numbers from 0 to n Found the actual sum of the given array The difference between them gives the missing number ⚡ This approach avoids unnecessary iterations and extra memory usage. 📊 Complexity Analysis: Time Complexity: O(n) Space Complexity: O(1) 📚 Key Learning: Sometimes the most optimal solution is not complex—it’s about recognizing patterns and applying the right formula. This problem reinforced the importance of mathematical thinking in algorithms. Consistency continues… 💪 #LeetCode #Algorithms #DataStructures #ProblemSolving #CodingJourney #Java #100DaysOfCode #StudentDeveloper #Learning

🔥 DSA Challenge – Day 119/360 🚀 📌 Topic: Recursion / Math 🧩 Problem: Pow(x, n) Problem Statement: Find x raised to the power n (xⁿ) efficiently, even for large values of n. 🔍 Example: Input: x = 2, n = 5 Output: 32 💡 Approach: Optimized (Binary Exponentiation) 1️⃣ Step 1 – If n is negative, convert → x = 1/x and n = -n 2️⃣ Step 2 – Recursively calculate half = power(x, n/2) 3️⃣ Step 3 – • If n is even → return half × half • If n is odd → return half × half × x ⏱ Complexity: Time: O(log n) Space: O(log n) 📚 Key Learning: Breaking a problem into halves can drastically improve performance 🚀 #DSA #Java #Coding #InterviewPrep #ProblemSolving #TechJourney #119DaysOfCode #LeetCode