Day 3 of my #30DayCodeChallenge: Handling "Infinite" Numbers! The Problem: Multiply Strings. The Logic: I went back to the basics-literally back to grade school. Since I couldn't rely on the * operator for these massive strings, I simulated the Long Multiplication algorithm: Digit-by-Digit: I broke the strings down, converted characters to integers, and multiplied them one by one. The Index Secret: The product of digits at positions i and j always lands at index (i+j+1). The Carry Phase: I handled the carries in a separate pass to keep the code clean and readable ensuring every digit remains between O and 9. Optimization Highlight: Using a StringBuilder for the final conversion instead of simple String concatenation. This avoids creating multiple immutable String objects in memory, keeping the time complexity at O(m x n). Small steps every day lead to big results. Onward to Day 4! #Java #CodingChallenge #ProblemSolving #Algorithms #StringManipulation #SoftwareDevelopment #150DaysOfCode
Java String Multiplication Challenge: Day 3
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🚀 Day 26/100 Days of #CodeChallenge Today’s problem: Isomorphic Strings (Leetcode 205) This problem helped me understand how to map characters between two strings while maintaining consistency and order. 💡 Key Concept: Two strings are isomorphic if characters in one string can be replaced to get the other string — with: ✔️ One-to-one mapping ✔️ No two characters mapping to the same character ✔️ Order preserved 🧠 What I Learned: How to use mapping (arrays/hashmaps) efficiently Importance of bidirectional checking Handling edge cases like unequal lengths ⚡ Approach: Compare lengths first Track character mappings using arrays Ensure consistency in both directions ⏱️ Complexity Analysis: Time Complexity: O(n) → We traverse the strings once Space Complexity: O(1) → Fixed-size arrays (256 characters) ✅ Successfully solved and understood the logic! Every day is a step closer to mastering problem-solving 💪 #Day26 #100DaysOfCode #Java #DSA #LeetCode #CodingJourney #ProblemSolving #TechLearning
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🚀100 Days of Code Day-26 LeetCode Practice – Remove Duplicates from Sorted Array Solved a classic problem using the Two Pointer Technique 💡 📌 Problem: Given a sorted array, remove duplicates in-place and return the number of unique elements. 🔍 Key Idea: Since the array is sorted, duplicates are adjacent. Using two pointers helps efficiently overwrite duplicates without extra space. ⚡ Complexity: Time → O(n) Space → O(1) 💻 Clean and optimized approach makes this problem a great example of in-place array manipulation! #LeetCode #Java #DataStructures #CodingPractice #ProblemSolving #100DaysOfCode
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Day 91 of #365DaysOfLeetCode Challenge Today’s problem: **Arithmetic Slices (LeetCode 413)** An interesting problem that focuses on identifying patterns in subarrays. The goal is to count all contiguous subarrays of length ≥ 3 where the difference between consecutive elements remains constant. 💡 **Key Insight:** Instead of checking every subarray (which would be inefficient), we track the current streak of arithmetic sequences. * If the current 3 elements form an arithmetic sequence → extend the streak * Keep adding the count of valid slices ending at current index * Reset when the pattern breaks 📌 **Approach:** * Use two variables: * `curr` → counts current valid extensions * `total` → accumulates final answer * Traverse from index 2 onward * Compare consecutive differences ⚡ **Time Complexity:** O(n) ⚡ **Space Complexity:** O(1) **What I learned today:** Sometimes, problems that look like they need nested loops can be optimized using pattern tracking and dynamic accumulation. Consistency is key — 91 days down, 274 to go! #LeetCode #DSA #CodingChallenge #Java #ProblemSolving #100DaysOfCode #TechJourney
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🚀 Day 69 — LeetCode Practice 🚀 Today’s focus was on string manipulation and careful indexing. ✅ Problem solved today: 🔹 Find Common Characters 💡 Key learnings from today: • Understood how to find common characters across multiple strings • Learned the importance of character frequency counting • Practiced handling nested loops efficiently • Realized how small mistakes in indexing can affect the entire logic • Improved attention to detail while working with strings Initially, I made a mistake by using the wrong variable inside charAt(). This caused incorrect indexing, which can either lead to wrong output or even StringIndexOutOfBoundsException. After fixing it, the logic worked perfectly by correctly comparing characters and tracking their minimum frequency across all words. This problem reinforced an important lesson: Sometimes errors are not in logic — but in small details like indexing and variable usage. Accuracy matters as much as logic. Step by step, getting better 💪 On to Day 70 🚀 #Day69 #DSA #LeetCode #ProblemSolving #Java #CodingJourney #Consistency #FutureEngineer
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Day 35 of Consistency 💻🔥 Solved Longest Substring Without Repeating Characters — a classic sliding window problem that really tests optimization skills. ✨ Key Takeaways: Mastered the sliding window technique Used HashMap for efficient lookup Improved understanding of two-pointer approach ⚡ Performance: Runtime: 5 ms Beat: 87%+ submissions Every day is about getting sharper, not just solving problems. #LeetCode #DataStructures #Algorithms #CodingJourney #Consistency #Java #ProblemSolving
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🚀 Day 3/100 – LeetCode Challenge 🔍 Problem: Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be inserted to maintain the sorted order. 💡 Key Concepts Used: • Binary Search • Time Complexity Optimization – O(log n) • Efficient searching in sorted arrays 📚 What I Learned: • How binary search reduces search time significantly compared to linear search • Handling edge cases when the target element is not present • Determining the correct insertion index while maintaining sorted order hashtag #100DaysOfCode #LeetCode #Java #DataStructures #Algorithms #BinarySearch #CodingJourney #ProblemSolving
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🚀 Today's Coding Challenge: Rotate Array Took on the Rotate Array problem and focused on getting both correctness and efficiency right. ✅ Implemented an in-place solution with O(n) time complexity and O(1) space ✅ Leveraged the array reversal technique to avoid unnecessary extra memory ✅Performance beats 42.38% of submissions This problem looks simple at first glance, but the real test is optimizing space while keeping the logic clean and bug-free. It’s a solid reminder that understanding underlying patterns always beats brute force. Consistency > intensity. Showing up daily. 💪 #DataStructures #Algorithms #CodingChallenge #Java #ProblemSolving #LeetCode
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🚀 #60DaysOfLeetCode – Day 32 Today’s problem was about removing all adjacent duplicates in a string. 🔹 Approach Used: Stack To efficiently handle consecutive duplicates, I used a stack-based approach: Traverse the string character by character. If the stack is not empty and the top element matches the current character, pop it (this removes the duplicate pair). Otherwise, push the current character onto the stack. Finally, build the result string from the stack and reverse it to maintain the correct order. 🔹 Key Learning: Using a stack helps simulate the process of removing adjacent duplicates in a single pass, achieving O(n) time complexity and avoiding unnecessary reprocessing. This problem highlights how stacks are powerful for handling sequential patterns and cancellations, especially in string manipulation problems. On to Day 33! 💻🔥 #LeetCode #DSA #Java #CodingChallenge #ProblemSolving #LearningInPublic #60DaysOfCode
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🚀 LeetCode – Subsets | Backtracking Approach Given an integer array nums of unique elements, return all possible subsets (the power set). The solution set must not contain duplicate subsets. Return the solution in any order. 💡 Approach Used – Backtracking I used a recursive backtracking strategy where: 1. Start with an empty subset 2. Add the current subset to the result 3. Choose an element and explore further combinations 4. Backtrack by removing the element to try other possibilities This ensures that we explore all possible combinations systematically. Since each element has two choices (include or exclude), the total subsets become 2ⁿ. #LeetCode #DSA #Backtracking #Java #CodingPractice #Algorithms #ProblemSolving #SoftwareEngineering #TechLearning
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Day 33 of #100DaysOfCode 🚀 Today, I learned about Counting Sort in Data Structures and Algorithms and solved a coding problem based on it. Here’s what I explored in more detail: Finding the frequency of each unique element Understanding and analyzing the time complexity of arrays Performing dry runs to better understand code execution #100DaysOfCode #DSA #Java #CountingSort #Algorithms #CodingProblem
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