Avoid Inefficient Computation Logic with Vectorized Tools

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 10 of #50DaysOfLeetCode Challenge Just solved the "Leaf-Similar Trees" problem! This challenge was a perfect way to practice Depth First Search (DFS). The goal is to determine if two different binary trees have the same "leaf value sequence" when read from left to right. Key Takeaways: Tree Traversal: Used DFS to navigate down to the leaf nodes while maintaining the specific left-to-right order. Leaf Identification: A simple but effective check—a node is a leaf only if both its left and right children are null. Comparison Logic: Collecting the leaf sequences into lists and comparing them highlights how structural differences in trees don't always mean their "outputs" are different. #DataStructures #Algorithms #CodingJourney #Java #BinaryTree #DFS #LeetCode #TechLearning

Wow! That's maths. Impressive question. Had fun solving it. The problem asks for the minimum operations to make all elements in a 2D grid equal by adding or subtracting a given integer x. At first glance, you might think about finding the average of all the numbers. However, the mathematical trick to minimize absolute differences is actually to target the median. My Approach: Flatten & Sort: I converted the 2D grid into a 1D array and sorted it to easily find the median values. The Window Check: To be absolutely safe, my code checks a small window of elements right around the median (((m * n) / 2) - 2 to + 2). Modulo Validation: If the absolute difference between an element and the target isn't perfectly divisible by x (val % x != 0), it's mathematically impossible, so we break and return -1. Happy to get this one Accepted! #LeetCode #Java #Algorithms #ProblemSolving #DataStructures #CompetitiveProgramming

#CodeEveryday — My DSA Journey | Day 14 🧩 Problem Solved: Search a 2D Matrix (LeetCode #74) 💭 What I Learned: Used binary search by treating the 2D matrix as a flattened sorted array. Key idea: 👉 Convert 1D index → 2D coordinates using: row = mid / number_of_columns col = mid % number_of_columns At each step: ✔️ Calculated row and column from mid ✔️ Compared matrix value with target ✔️ Adjusted search space accordingly This approach avoids nested loops and gives optimal performance. ⏱ Time Complexity: O(log (m × n)) 🧠 Space Complexity: O(1) ⚡ Key Takeaways: ✔️ 2D problems can often be reduced to 1D using indexing tricks ✔️ Binary search works whenever the data is globally sorted ✔️ Index mapping is a powerful technique in matrix problems 💻 Language Used: Java ☕ 📘 Concepts: Binary Search · Matrix · Index Mapping · Optimization #CodeEveryday #DSA #LeetCode #Java #BinarySearch #Matrix #ProblemSolving #Algorithms #CodingJourney #Consistency 🚀

𝗗𝗮𝘆 𝟲𝟳/𝟳𝟱 | 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝟳𝟱 𝗣𝗿𝗼𝗯𝗹𝗲𝗺: 338. Counting Bits 𝗗𝗶𝗳𝗳𝗶𝗰𝘂𝗹𝘁𝘆: Easy 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 𝗦𝘂𝗺𝗺𝗮𝗿𝘆: Given an integer n, return an array where each index i (0 ≤ i ≤ n) contains the number of 1’s in the binary representation of i. 𝗠𝘆 𝗔𝗽𝗽𝗿𝗼𝗮𝗰𝗵: This problem is solved efficiently using Dynamic Programming with bit manipulation. Instead of converting each number to binary separately, we reuse previously computed results. • Key Observation: – Right shifting a number (i >> 1) removes the last bit – (i & 1) tells whether the last bit is 1 or 0 • Transition: – ans[i] = ans[i >> 1] + (i & 1) This means: Take the count of 1’s from i/2 and add 1 if the current number is odd. • Base Case: – ans[0] = 0 • Final answer: – Array ans of size n + 1 This avoids repeated binary conversions and builds the solution in a bottom-up manner. 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆 𝗔𝗻𝗮𝗹𝘆𝘀𝗶𝘀: • Time Complexity: O(n) • Space Complexity: O(n) 𝗞𝗲𝘆 𝗧𝗮𝗸𝗲𝗮𝘄𝗮𝘆: Problems involving binary representations often have patterns that can be reused. Bit manipulation + DP is a powerful combination for optimizing such computations. 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻 𝗟𝗶𝗻𝗸: https://lnkd.in/gydkB_rv #Day67of75 #LeetCode75 #DSA #Java #Python #DynamicProgramming #BitManipulation #MachineLearning #DataScience #ML #DataAnalyst #LearningInPublic #TechJourney #LeetCode

## 🧩 Solved LeetCode 9 – Palindrome Number Today I revisited a fundamental integer manipulation problem: determining if a number reads the same forward and backward without converting it to a string. 🔍 **Problem Insight:** While string conversion makes this trivial, the real challenge is solving it mathematically. The goal is to reverse the integer (or half of it) and compare it to the original while being mindful of potential integer overflow. 💡 **Key Learnings:**  * **Negative Numbers:** Any negative number is immediately invalid (e.g., -121 becomes 121-).  * **Mathematical Reversal:** Using modulo % 10 and integer division / 10 allows us to "pop" and "push" digits.  * **Optimization:** We only need to reverse **half** of the number. Once the original number becomes less than or equal to the reversed half, we've reached the middle.  * **Edge Case Savvy:** Numbers ending in 0 (except 0 itself) cannot be palindromes. ⚙️ **Approach Used:**  1. Eliminate negative numbers and numbers ending in 0.  2. Build the reversed half of the number by iteratively taking the last digit.  3. Stop when the reversed part is \ge the remaining part.  4. Check for equality (handling odd-lengthed numbers by dividing the reversed part by 10). 📊 **Complexity:**  * **Time:** O(\log_{10}(n)) — we divide the input by 10 in every iteration.  * **Space:** O(1) — constant space used regardless of input size. 🔥 **Takeaway:** This problem is a perfect example of why we shouldn't always reach for the easiest data type conversion. Thinking in terms of pure logic and math often leads to a more memory-efficient and elegant solution. #LeetCode #Algorithms #Math #ProblemSolving #CodingJourney #Java #DataStructures #CompetitiveProgramming

Consistency is starting to compound 💻🌱🔥 Solved another problem today—this time from Binary Trees 🌳🚀 Completed Root to Leaf Paths with 100% accuracy (1115/1115 test cases passed) 💯✅ 🔍 Problem Insight: The task was to return all possible paths from the root node to every leaf node in a binary tree 🌳➡️🍃 🧠 My Approach: 👉 Used DFS (Depth-First Search) with recursion 🔁 👉 Maintained a currentPath list to track nodes while traversing 🛤️ 👉 At each node, added the current value to the path ➕ 👉 If a leaf node was reached, stored a copy of the current path in the final answer 📌 👉 Used backtracking by removing the last node after returning from recursion 🔙 ✨ Why this approach works well: ✔ Natural fit for tree traversal 🌳 ✔ Explores every root-to-leaf path efficiently 🚀 ✔ Backtracking keeps memory controlled 🧠 ✔ Clean recursive structure and easy to follow 🧼 📊 Complexity: ⏱️ Time Complexity: O(n) 💾 Space Complexity: O(h) recursion stack (plus output space for storing all paths) 📦 💡 Key Takeaway: Tree problems become much simpler when you think in terms of: 👉 traversal pattern 👉 path tracking 👉 backtracking DFS + recursion continues to be one of the cleanest ways to solve path-based tree problems 🎯🌳 One more problem solved, one more concept strengthened 💪📈🔥 #BinaryTree #Trees #Recursion #DFS #Java #DataStructures #DSA #ProblemSolving #CodingJourney #100DaysOfCode #SoftwareEngineering #Programming #TechGrowth 💻🌳🚀

#Day74 of my second #100DaysOfCode Binary search variations getting more interesting now. DSA • Solved Find Minimum in Rotated Sorted Array (LeetCode 153) – Brute: linear scan → O(n) – Optimal: binary search with an early check for already sorted part → O(log n) • Key idea: the minimum always lies in the unsorted portion, and if a part is already sorted, the answer can be taken directly • Difference from previous problems: instead of searching for a target, we’re tracking the minimum while narrowing the search space • Edge cases: – already sorted array – single element case – careful updates while narrowing the range This one was more about understanding the pattern than just applying binary search. #DSA #BinarySearch #LeetCode #Algorithms #Java #100DaysOfCode #WomenWhoCode #BuildInPublic #LearningInPublic

🚀 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

 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐃𝐚𝐢𝐥𝐲 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞 – 𝐃𝐞𝐜𝐨𝐝𝐞 𝐭𝐡𝐞 𝐒𝐥𝐚𝐧𝐭𝐞𝐝 𝐂𝐢𝐩𝐡𝐞𝐫𝐭𝐞𝐱𝐭 (𝐌𝐞𝐝𝐢𝐮𝐦) Today’s problem was a 𝐟𝐮𝐧 𝐦𝐚𝐭𝐫𝐢𝐱 𝐭𝐫𝐚𝐯𝐞𝐫𝐬𝐚𝐥 + 𝐬𝐢𝐦𝐮𝐥𝐚𝐭𝐢𝐨𝐧 𝐜𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐋𝐢𝐧𝐤 : https://lnkd.in/gsSCrJbr 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐋𝐢𝐧𝐤 : https://lnkd.in/gMhqhD6M The idea behind the encoding process: The original text is written 𝐝𝐢𝐚𝐠𝐨𝐧𝐚𝐥𝐥𝐲 𝐢𝐧 𝐚 𝐦𝐚𝐭𝐫𝐢𝐱 with a fixed number of rows. The matrix is then read row by row to produce the encoded string. To decode it, we reverse the process: 🔹 𝐊𝐞𝐲 𝐎𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧𝐬 If n = encodedText.length() and we know rows, then cols = n / rows Characters of the original text lie on diagonals starting from each column in the first row. Traverse diagonally (row++, col++) from each starting column. 🔹 𝐒𝐭𝐞𝐩𝐬 Compute the number of columns. For every column c, traverse diagonally down-right. Map the matrix index using r * cols + c. Build the result string. Remove trailing spaces at the end. 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲 Time: O(n) Space: O(n) 𝐓𝐚𝐤𝐞𝐚𝐰𝐚𝐲: Whenever you see a matrix encoding problem, think about how the data was written and reverse the traversal pattern. #LeetCode #CodingInterview #DataStructures #Algorithms #Java #ProblemSolving #DailyCodingChallenge