🚀 Day 78 of- #100DaysOfCode Today’s problem: Check if the Sentence Is Pangram A pangram is a sentence that contains every letter of the English alphabet at least once. 🔹 Approach I Used: Loop from 'a' to 'z' Check if each character exists in the given string If any character is missing → return false If all characters are present → return true 💻 Java Solution: 🧠 Time Complexity: O(26 * n) → Since we check 26 letters and contains() takes O(n) Simplified: O(n) 📌 Space Complexity: O(1) (No extra data structure used) 💡 What I Learned: Sometimes simple brute-force solutions are clean and efficient enough. Always analyze time complexity even for small loops like 26 characters.
Checking if a sentence is a pangram in Java
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Refactoring for Clarity: Array Manipulation in Java 👨💻 I’ve just finished a practical exercise on array manipulation. While the goal was simple (summing two arrays), I used it as an opportunity to apply improvements. - Method decomposition: Separated concerns into specialized methods (Read, Calculate, Display). - Input validation: Built a robust loop to handle invalid inputs and prevent crashes. - Data Formatting: Used printf to create a clear, readable results table. Small improvements in logic and organization make a huge difference in software quality. #Java #Algorithms #CleanCode #Backend #LearningToCode
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Today I solved “Fibonacci Number” on LeetCode using Java. 💡 Problem Summary The Fibonacci sequence is defined as: F(0) = 0 F(1) = 1 F(n) = F(n-1) + F(n-2) We need to compute F(n). 🧠 Approach I Used (Iterative Optimization) Instead of recursion (which is slow and redundant), I used an iterative approach: Maintain two variables → previous two values Keep updating them in a loop Build the answer step by step ⚙️ Why Not Recursion? Recursion leads to repeated calculations Time complexity becomes O(2ⁿ) ❌ Iterative solution reduces it to O(n) ✅ 📊 Complexity Analysis ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) ⚡ Result ✅ Accepted ⚡ Runtime: 0 ms (Beats 100%) 📚 Key Learning Always try to optimize recursion → iteration Use variables smartly to reduce space Even simple problems teach important optimization patterns Slowly building strong fundamentals, one problem at a time 💯 Day 15 done. Let’s keep the streak alive 🔥 #DSA #LeetCode #Java #100DaysOfCode #Algorithms #ProblemSolving #CodingJourney #Consistency #LearningInPublic
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🚀 Day 98 - #100DaysOfCode Today’s problem was all about string comparison and operations simulation in Java. 💡 Problem Insight: Given a list of operations like "++X", "X++", "--X", "X--", we need to compute the final value of X after performing all operations. ⚠️ One key learning today: In Java, always use .equals() for string comparison instead of ==. Using == compares references, not actual content — a very common mistake! 🧠 Approach: Initialize x = 0 Traverse through each operation Increment or decrement based on the operation string 📌 What I Improved Today: Better understanding of string handling in Java Avoiding common pitfalls in comparisons Writing cleaner conditional logic #Java #CodingJourney #LeetCode #100DaysOfCode #ProblemSolving #Consistency
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Day 34 of #100DaysOfLeetCode 💻✅ Solved #110. Balanced Binary Tree on LeetCode using Java. Approach: • Used a bottom-up recursive approach to calculate height • Returned -1 immediately if any subtree is unbalanced • Compared left and right subtree heights at each node • Checked if the height difference is greater than 1 • Stopped early to optimize unnecessary computations Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) ✓ Memory: 45.33 MB (Beats 95.41% submissions) Key Learning: ✓ Understood how to combine height calculation with balance checking ✓ Learned early termination technique in recursion ✓ Improved problem-solving for tree-based recursive problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinaryTree #Recursion #TreeProblems #ProblemSolving #CodingJourney #100DaysOfCode
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📺 YouTube: https://lnkd.in/du8dCrS4 Watching this session gave me a powerful insight: Java code should not be seen only as text, but as a behavior model. For years we’ve focused on syntax. But this talk showed me that to truly understand code, we need to look at three layers: 🌳 AST (Abstract Syntax Tree) → the syntactic structure of the code 🔀 CFG (Control Flow Graph) → the execution flow 🔢 Symbolic Model → operations represented as mathematical objects Take this simple example: int sum = 0; for (int i = 0; i < n; i++) { sum += i; if (sum > 10) break; } return sum; As text, it’s just a loop and a condition. As a behavior model: • 🌳 AST → for, if, sum += i, return sum • 🔀 CFG → Entry → Loop check → Body → Condition → Break/Continue → Exit • 🔢 Symbolic Model → • var.load sum • var.load i • add sum i • var.store sum • gt sum 10 → if true → java.break What did this give me? • 🚀 Optimization – spotting and removing redundant operations • ✨ Refactoring – making code more readable and maintainable • 🛡️ Safety – stronger flow and type checks • 💡 Productivity – less time wasted on debugging and analysis My takeaway: Java code is not only written, it must also be modeled to be truly understood.
Symbolic Modeling and Transformation of Java Code #JVMLS
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LeetCode Problem || Find Unique Binary String (1980)🚀 Today I solved the problem "Find Unique Binary String" using Java. 🔹 Problem: We are given an array of n binary strings, each of length n. The goal is to return a binary string of length n that does not exist in the array. 🔹 Approach (Diagonal Flip Technique): The idea is simple but powerful: Traverse the array using index i. Look at the i-th character of the i-th string (nums[i][i]). Flip the bit (0 → 1, 1 → 0). Append it to a result string. 💡 Time Complexity: O(n) Practicing problems like this strengthens logical thinking and problem-solving skills. #LeetCode #Java #CodingPractice #ProblemSolving #DSA
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✨ DAY-36: 🌳 Understanding Object Cloning in Java – Made Simple! This visual perfectly represents how object cloning works in Java using a tree analogy. The big tree represents the original object, while the smaller trees symbolize the cloned objects created using the .clone() method. Just like these mini trees look identical to the original, cloned objects also copy the properties of the original object. ✨ Key Idea: Cloning allows you to create duplicate objects without manually copying each value. 🌱 Think of it like: Instead of planting a new tree from scratch, you simply grow multiple identical trees from one! 💡 Bonus Insight: Shallow Copy → Copies only references (faster, but linked) Deep Copy → Creates fully independent objects (safer) 📌 Cloning helps improve performance and reduces repetitive code in Java development. #Java #Programming #Coding #JavaDeveloper #OOP #Learning #TechConcepts #SoftwareDevelopment
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Day 3/50 | #50DaysOfCode 📍 Platform: LeetCode 💻 Language: Java ✅ 977. Squares of a Sorted Array (Easy) Today’s problem focused on array manipulation and sorting logic. It helped me understand how negative numbers affect order after squaring and how to maintain sorted order efficiently. 🔎 Approach: Traverse the array and calculate the square of each element Store the squared values in a new array Sort the new array in non-decreasing order Return the sorted squared array 📌 Example: Input: nums = [-4, -1, 0, 3, 10] Output: [0, 1, 9, 16, 100] This problem strengthened my understanding of arrays, sorting, and handling negative values in Java. #DSA #LeetCode #Java #CodingJourney #ProblemSolving #Consistency #LearningJourney #50DaysOfCode #LinkedIn
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🚀 LeetCode Day 18 – Complement of Base 10 Integer Today I solved LeetCode Problem 1009: Complement of Base 10 Integer using Java. 🧠 Problem: Given a non-negative integer n, return the complement of its binary representation. The complement means flipping all bits in the binary form of the number. 📌 Example: Input: n = 5 Binary of 5 → 101 Complement → 010 Output → 2 💡 Approach: • Create a bitmask with all bits set to 1 up to the highest bit of n. • Use the XOR (^) operator with the mask to flip the bits. 💻 Java Code: class Solution { public int bitwiseComplement(int n) { if (n == 0) return 1; int mask = 0, temp = n; while (temp > 0) { mask = (mask << 1) | 1; temp >>= 1; } return n ^ mask; } } ⏱️ Time Complexity: O(log n) 📦 Space Complexity: O(1) 📚 Practicing Data Structures & Algorithms daily to improve problem-solving skills. #LeetCode #Java #DSA #ProblemSolving #CodingJourney #100DaysOfCode
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💡 If life had a Garbage Collector… While learning about the Garbage Collector in Java, I had a funny thought… Imagine if humans had a Garbage Collector too! 😄 ✔️ Unused worries → Automatically deleted ✔️ Old embarrassing memories → Deallocated ✔️ Negative thoughts → Marked as unreachable and cleaned In Java, the Garbage Collector automatically removes objects that are no longer referenced, freeing memory and keeping applications efficient. 🧠 Simple idea: If an object is not being used anymore and no reference points to it, the JVM eventually removes it from memory. 📌 Why it matters in real applications: • Prevents memory leaks • Optimizes memory usage • Developers don’t need to manually free memory (unlike some languages) So while we focus on writing clean code, the Garbage Collector quietly works in the background keeping our applications healthy. Sometimes the best systems are the ones we don’t even notice working. 😉 #Java #JVM #GarbageCollector #BackendDevelopment #ProgrammingHumor #LearningJourney
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