Matrix Chain Multiplication with Dynamic Programming

✅ Day 57/75 – Dynamic Programming | Edit Distance 📌 Problem: Given two strings word1 and word2, find the minimum number of operations required to convert word1 into word2. 🔧 Allowed Operations: 🔺 Insert a character 🔺 Delete a character 🔺 Replace a character 💡 Approach: 🔺 Defined dp[i][j] as the minimum operations needed to convert the first i characters of word1 into the first j characters of word2 🔺 Used Dynamic Programming (Tabulation) to build the solution 🔺 Applied optimal substructure:     🔹 If characters match → carry forward dp[i-1][j-1]     🔹 If characters differ → 1 + min(insert, delete, replace) 📊 Complexity Analysis: 🔺 Time Complexity: O(n × m) 🔺 Space Complexity: O(n × m) #Day57 #75DaysOfCode #DynamicProgramming #EditDistance #LeetCode #Java #DSA #ProblemSolving #SoftwareEngineering

📘 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

🚀 Day 467 of #500DaysOfCode 📌 Problem Solved: 1458. Max Dot Product of Two Subsequences (Hard) Today’s problem was a reminder that not all subsequence problems are greedy-friendly. We’re given two arrays and asked to find non-empty subsequences whose dot product is maximized. Sounds simple at first — until negative numbers enter the picture. 💡 Key Realization: This problem is similar to LCS, but instead of matching characters, we’re: pairing numbers, multiplying them, and deciding whether to take or skip each element. The tricky part? 👉 We cannot choose empty subsequences, and sometimes the best answer is negative. 🧠 Approach Used: Used 2D Dynamic Programming dp[i][j] represents the maximum dot product using prefixes of both arrays At each step, we decide to: take the current pair, skip an element from the first array, or skip one from the second A key trick was using max(previous, 0) to safely start a new subsequence when needed ⚙️ Concepts Reinforced: Dynamic Programming on two sequences Handling negative values correctly Non-empty subsequence constraints Why initialization matters in DP 🔥 Hard problems like this really sharpen DP intuition and edge-case thinking. On to Day 468 🚀 #LeetCode #DynamicProgramming #HardProblems #ProblemSolving #Java #DSA #Consistency #CodingJourney #500DaysOfCode

🚀 DSA Journey – Day 16 📌 Problem: Word Break (String Segmentation) Today, I worked on the Word Break problem, which helped me understand how Dynamic Programming can be used to solve string-based problems efficiently. 🔍 Problem Summary: Given a string and a dictionary of valid words, determine whether the string can be segmented into a space-separated sequence of dictionary words. The same word can be reused multiple times. 💡 Approach I Learned: Used a Dynamic Programming (DP) array dp[i] represents whether the substring s[0…i-1] can be segmented For each position, checked all possible previous splits If the previous part is valid and the current substring exists in the dictionary, mark it as valid 🧠 Key Learnings: ✔ Large problems can be broken into smaller subproblems ✔ DP helps avoid repeated work ✔ Using a HashSet improves lookup efficiency ✔ Understanding the meaning of dp[i] is the core of the solution ⏱ Complexity: Time: O(n²) Space: O(n) Slowly getting comfortable with Dynamic Programming concepts 💪 On to Day 17 🚀 #DSA #Day16 #DynamicProgramming #Strings #Java #ProblemSolving #BeginnerInDSA #CodingJourney #LearningInPublic

✏️ 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 🚀

Day 5/365 – Word Break (Dynamic Programming) 📌 Problem: Given a string and a dictionary of words, determine if the string can be segmented into a sequence of dictionary words. 🧠 Key Challenge: Brute-force recursion leads to exponential time. This problem is about identifying the right DP state. 🔍 Approach: Use a boolean DP array where dp[i] = can the substring s[0…i) be segmented? Initialize dp[0] = true (empty string) For each position i, check valid split points j Optimize by limiting checks using the maximum word length 💡 Key Learning: Dynamic Programming converts repeated substring checks into a linear scan over valid states. ⏱ Time Complexity: O(n²) (optimized by max word length) 📦 Space Complexity: O(n) Day 5 done — strengthening DP fundamentals step by step #LeetCode365 #DSA #DynamicProgramming #WordBreak #Java #ProblemSolving #LearningInPublic

🚀 Day 144 | Dynamic Programming & String Optimization Today’s problem was a great example of how DP transforms a deletion problem into a common-subsequence optimization task. 🧩 Problem Solved: 712. Minimum ASCII Delete Sum for Two Strings 🔍 Approach: • Calculated the total ASCII sum of both strings. • Used Dynamic Programming to compute the maximum ASCII sum of a common subsequence between the two strings. • Built a DP table similar to LCS, but instead of length, stored ASCII value sums. • Final result = totalSum − 2 × (max common ASCII sum) ✨ Key Insight: Instead of directly minimizing deletions, it’s easier to maximize what you keep. Once the highest-value common subsequence is known, the minimum delete cost follows naturally. 📚 Topics: Dynamic Programming · Strings · LCS Variant 💻 Platform: LeetCode #LeetCode #DSA #ProblemSolving #DailyCoding #DynamicProgramming #Strings #Java #Consistency