LeetCode Minimum ASCII Delete Sum for Two Strings DP Solution

✏️ DSA Diary Day 16/100 📚🔤: Solving LeetCode’s “Minimum ASCII Delete Sum for Two Strings” 🚀✨ Today I worked on a Dynamic Programming + String Optimization problem on LeetCode 🔥👇 👉 Minimum ASCII Delete Sum for Two Strings This problem beautifully shows how DP helps minimize deletion cost while preserving the maximum common structure between two strings 🧠💡 🔹 My Approach 🛠️🧠 I used Bottom-Up Dynamic Programming (Tabulation) 🔽👇 🔸 Step 1: Define the DP Idea 📌 Instead of directly calculating what to delete, I focused on maximizing the ASCII value of the common subsequence between both strings. The more valuable the common part is, the less we need to delete 🔁✨ 🔸 Step 2: Decision Making 🔍 At every character comparison: 1️⃣ If both characters match → Add their ASCII value to the total 2️⃣ If they don’t match → Carry forward the maximum value from previous comparisons This is similar to the Longest Common Subsequence (LCS) concept, but with weights (ASCII values) instead of length 📈 🔸 Step 3: Getting the Final Answer 🧮 We calculate the total ASCII value of both strings, then subtract twice the value of the common subsequence. This ensures: ✔ Only unnecessary characters are deleted ✔ The deletion cost is minimum #LeetCode 🚀 #Java ☕ #DSA 🧠 #DynamicProgramming 📊 #ProblemSolving 💡 #CodingChallenge 💻 #100DaysOfCode 🔥 #DSADiaryByRethanya ✨ #Strings 🔤 #Tabulation 🧩 #LearnInPublic 📢 #TechJourney 🚀

📘 LeetCode Daily Challenge – Problem 712: Minimum ASCII Delete Sum for Two Strings Today I worked on a classic Dynamic Programming problem that focuses on making two strings equal by deleting characters with the minimum possible ASCII cost. This is a variation to the Classic Longest Common Subsequence problem. 🔍 How to approach this problem: The key idea is to think in terms of prefixes of both strings and ask: What is the minimum delete cost to make s1[i…] and s2[j…] equal? This naturally leads to a recursive + DP solution. 🧠 Core Observations: If both strings are exhausted, no cost is needed. If one string is exhausted, we must delete all remaining characters from the other string (sum of their ASCII values). If the current characters match, we move forward in both strings with no extra cost. If the characters don’t match, we have two choices: Delete the current character from the first string Delete the current character from the second string We take the option with the minimum total ASCII cost. ⚙️ Optimization: Since the same subproblems repeat, using a 2D DP table to store results avoids re - computation and reduces the time complexity from exponential to O(n × m). 🎯 Takeaway: This problem is a great example of how: Breaking a problem into smaller overlapping subproblems simplifies the logic Memoization transforms an inefficient recursive solution into an optimal one Clear base cases are crucial in DP problems Highly recommended for anyone strengthening their Dynamic Programming fundamentals. #LeetCode #DynamicProgramming #DP #ProblemSolving #DSA #Java #Learning #Algorithms

📌 Today’s DSA Series Topic: Linked List A Linked List is a linear data structure where elements (called nodes) are stored in non-contiguous memory locations. Each node contains: Data – the value stored Next – reference to the next node The list starts with a head node and ends with NULL. Unlike arrays, linked lists are dynamic in size and allow efficient insertions and deletions. 🔹 CRUD Operations on Linked List (Explained visually in the images below 👇) ✅ Create – Insert a new node (at beginning or end) ✅ Read – Traverse the list from head to NULL ✅ Update – Modify the data of an existing node ✅ Delete – Remove a node and adjust pointers 📊 Time Complexity Overview Insert at Beginning: O(1) Read / Traverse: O(n) Update: O(n) Delete: O(1) / O(n) (based on position) 💬 Want to learn Linked List CRUD operations in a specific programming language (C, C++, Java, Python, JavaScript)? Comment the language name, and I’ll explain it with code examples. 🚀 Keep learning. Keep growing. #DSASeries #LinkedList #DataStructures #CRUD #Programming #TimeComplexity #SoftwareEngineering

🎯 Day 24/100 of my #LeetCode Challenge — Problem Solved: Minimum ASCII Delete Sum for Two Strings Today's challenge was a classic dynamic programming problem with a twist: finding the minimum ASCII sum of deleted characters to make two strings equal. The Problem: Given two strings s1 and s2, we need to delete characters from both so they become identical — minimizing the sum of ASCII values of the deleted characters. Key Insight: This is essentially a variation of the Longest Common Subsequence (LCS) problem. Instead of maximizing length, we minimize ASCII deletion cost. My Approach: Used a 2D DP table where dp[i][j] represents the minimum deletion cost for s1[0..i-1] and s2[0..j-1]. Base cases: deleting all characters from one string when the other is empty. Transition: If characters match: carry over previous cost. If they differ: choose the cheaper deletion between removing from s1 or s2. Result: dp[m][n] gives the minimal ASCII deletion sum. Complexity: Time: O(m * n) Space: O(m * n) Another step forward in sharpening my DP skills and preparing for technical interviews! 💡 #LeetCode #100DaysOfCode #DynamicProgramming #Algorithm #CodingChallenge #InterviewPrep #Tech #SoftwareEngineering #ProblemSolving #Java

Finished the core IDE practice for Multidimensional Arrays today. The focus here was on searching inside sorted matrices and understanding how the approach changes performance, not just correctness. I worked on: - basic search through full traversal - optimized search starting from a matrix corner - handling matrix updates like Set Matrix Zeroes using better space strategies These problems made it clear that matrix questions are often about choosing the right starting point, not just writing more loops. int[][] matrix = {   {1, 4, 7, 11},   {2, 5, 8, 12},   {3, 6, 9, 16},   {10, 13, 14, 17} }; int target = 9; int i = 0; int j = matrix[0].length - 1; boolean found = false; // Efficient search from top-right corner while (i < matrix.length && j >= 0) {   if (matrix[i][j] == target) {     found = true;     break;   } else if (matrix[i][j] > target) {     j--;   } else {     i++;   } } System.out.println(found); // Output: true What became clearer from this stage: - better starting position can reduce time complexity - optimization is often about direction, not new logic - matrix problems reward thinking before coding - many advanced questions reuse the same core ideas This felt like a good checkpoint before moving to LeetCode matrix problems. #Java #DSA #Matrices #Arrays #LearningInPublic #ProblemSolving #CodingJourney #Programming #JavaDeveloper

I spent some time benchmarking C++ vs Rust across RAII, lock-free data structures, async I/O, and zero-copy string processing. Rigorous methodology, idiomatic implementations in both languages. The results? For 90% of systems programming work, the performance difference is under 5%. C++ wins: SSO for small strings, raw coroutine overhead, custom memory layouts. Rust wins: mature async runtime (tokio destroys hand-rolled C++ executors), simpler Arc semantics, guaranteed move semantics. Tie: lock-free structures, string views, allocation performance. The real difference isn't performance—it's the borrow checker catching bugs that would otherwise require careful review and Valgrind. I can confirm: performance isn't the reason to resist Rust. Learning ownership semantics is the cost, but it pays for itself in preventing use-after-free and data races. Full writeup and benchmarks: Source: https://lnkd.in/gzqKnbdA Article: https://lnkd.in/g3M6ukA3

I built a programming language with algebraic effects  KryhtaJS is a small language with JavaScript-like syntax that supports algebraic effects. Effects let you write code that "asks" for  things without knowing how they'll be provided. The handler decides.  let result = handle {    let x = perform Ask!("value");    x * 2  } with {    Ask!(label, resume) -> resume(21)  };  // result = 42  This makes generators, async/await, exceptions, and state all expressible as patterns instead of language primitives.  The interpreter is written in Rust, compiles to WebAssembly, and runs in your browser. Try it out, link in comments!

Stop memorizing recursion solutions. Master the "Take / No-Take" pattern instead. Recursion and backtracking problems (subsets, knapsack, dynamic programming) used to be challenging. I often felt confused about where to place the return statement or how to backtrack. Then I discovered that 90% of these problems follow the same pattern: the Take / No-Take Method. At each step of the array, you have two choices: 1. TAKE: Include the element. 2. NOT TAKE: Skip the element. Recursion explores both paths. Here is the universal template that solves 14+ LeetCode hard problems: function(index) { // Base Case if (index == n) return result; // 1. TAKE add(arr[index]); take = function(index + 1); // 2. BACKTRACK (Undo the change) remove(arr[index]); // 3. NOT TAKE notTake = function(index + 1); return combine(take, notTake); } I’ve compiled a comprehensive guide of 14 classic patterns solved using this logic in Java, including: - Subsets (Power Set) - Permutations - Combination Sum I & II - N-Queens - Sudoku Solver - Word Search - Palindrome Partitioning If you want the full clean PDF with code, input/output examples, and explanations, like this post and comment "Recursion" below. Follow for more high-impact engineering guides. I’ll DM you the link directly! #Java #Programming #DataStructures #LeetCode #SoftwareEngineering #CodingInterview #FAANG #Developer

🚀 Day 46/50 – LeetCode POTD 🔍 3. Longest Substring Without Repeating Characters 📊 Medium 🧠 Key Idea (Sliding Window + Two Pointers) Whenever a problem asks for a substring without repeating characters, the best approach is 👉 Sliding Window. ✔️ Use two pointers: left (l) and right (r) ✔️ Use a boolean array (arr[128]) to track characters (ASCII set) ✔️ If the current character is not present in the window ➡️ expand the window (r++) ➡️ update the maximum length ✔️ If a duplicate character is found ➡️ shrink the window from the left (l++) ➡️ remove characters until the duplicate is gone ✔️ The window always contains unique characters 📌 Example s = "abcabcbb" ➡️ Longest substring = "abc" ➡️ Output = 3 ⚙️ How the Code Works (Java) arr[ch] = true → character exists in the current window On duplicate → move the left pointer and clear characters Track the answer using maxLen = Math.max(maxLen, r - l + 1) ⚙️ Complexity Analysis ⏱ Time Complexity: O(n) 💾 Space Complexity: O(1) (fixed-size array of 128 characters) 💡 Key Takeaway “Substring + uniqueness constraint” ➡️ Think Sliding Window ➡️ Two pointers give an optimal linear-time solution ➡️ Avoid brute force and nested loops 🔁 Consistency beats motivation 🚀 #LeetCode #DSA #SlidingWindow #TwoPointers #Strings #Java #ProblemSolving #CodingJourney #50DaysChallenge

🖥️ Automate HTML Data Extraction with Python & PyQt I recently built a desktop application that makes parsing HTML files simple and efficient. With a PyQt interface, users can select HTML and Excel files, while Selenium and BeautifulSoup extract structured data automatically. The results are saved directly into Excel (XLSX), ready for analysis or reporting. This tool saves hours of manual work, reduces errors, and makes web-based data accessible to everyone — no coding required. If you’re looking to automate HTML data extraction or streamline your workflow, feel free to reach out! 🔗 More projects: https://lnkd.in/dpiy69BF #Python #Automation #PyQt #WebScraping #ExcelAutomation #DataProcessing #TechTools #Portfolio